A Hostetler Handbook...vergence theorem. Using the divergence theorem on a sphere of radius Rabout the origin gives us V r ~V d˝ = r^ r2 d~a = 2ˇ 0 d˚ ˇ 0 r^ r2 R2 sin r^d = 4ˇ:

A Hostetler Handbook

v 0.8

Contents

Preface v

1 Electrostatics 11.1 Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Electric Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Electric Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.4 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.5 Work and Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.6 Conductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251.7 Capacitors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.8 Summary: Electrostatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2 Special Techniques 342.1 Laplace’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342.2 Method of Images . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.3 Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382.4 Multipole Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472.5 Summary: Special Techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

3 Polarization and Dielectrics 553.1 Polarization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553.2 Electric Displacement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593.3 Linear Dielectrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593.4 Summary: Polarization and Dielectrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

4 Magnetostatics 674.1 Lorentz Force Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 674.2 Biot-Savart Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 704.3 Ampere’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 744.4 The Vector Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 774.5 Magnetic Fields in Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 804.6 Summary: Magnetostatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

5 Electrodynamics 895.1 Electromotive Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 895.2 Electromagnetic Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 925.3 Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 975.4 Summary: Electrodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

6 Conservation Laws 1016.1 Charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1016.2 Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1016.3 Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1036.4 Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1086.5 Summary: Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

7 Electromagnetic Waves 1127.1 One-dimensional Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

iii

iv CONTENTS

7.2 EM Waves in a Vacuum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1167.3 EM Waves in Linear Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1187.4 Absorption and Dispersion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1237.5 Guided Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1267.6 Summary: Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

8 Potentials and Fields 1358.1 The Potential Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1358.2 Continuous Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1388.3 Point Charges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1428.4 Summary: Potentials and Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

9 Radiation 1499.1 Electric Dipole Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1499.2 Magnetic Dipole Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1519.3 Arbitrary Radiation Sources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1539.4 Point Charges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1539.5 Radiation Reaction and Self-Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1549.6 Summary: Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156

10 Relativistic Electrodynamics 15810.1 Special Theory of Relativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15810.2 Relativistic Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16010.3 Relativistic Electrodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16310.4 Summary: Relativistic Electrodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168

11 Check Your Understanding 171

Preface

About Me

My name is Leon Hostetler. I am currently a student at Florida State University majoring in physics as well asapplied and computational mathematics. Feel free to download, print, and use these class notes. If you find themuseful, consider buying me a coffee.

All of my class notes can be found at www.leonhostetler.com/classnotes

Please bear in mind that these notes will contain errors. If you find one, please email me at [email protected] the name of the class notes, the page on which the error is found, and the nature of the error. If you includeyour name, I will probably list your name on the thank you page if I decide to compile and sell my notes.

This work is currently licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 In-ternational License. That means you are free to copy and distribute this document in whole for noncommercial use,but you are not allowed to distribute derivatives of this document or to copy and distribute it for commercial reasons.

Last Updated: June 6, 2018

About These Notes

Cover image: NOAA Photo Library

These are my class notes from a course on electricity and magnetism (PHY 4323) that I took at Florida StateUniversity. Our textbook was the third edition of “Introduction to Electrodynamics” by David J. Griffiths.

Conventions

I will represent a vector by a bold letter topped with an arrow. For example: ~F .

I will represent a unit vector (i.e. a direction vector with magnitude 1) by a bold leter topped with a hat. Forexample: x.

In physics, a time derivative is often represented with a dot. For example, instead of writing dxdt or d2x

dt2 , onemight write x or x. This convention will generally be used in these notes.

v

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mailto:[email protected]

Chapter 1

Electrostatics

1.1 Vector Fields

Divergence Theorem

In 1D, the integral of a derivative over a length is equalto the value of the function at the endpoints. More gen-erally, the integral of a derivative over a region is equalto the value of the function at the boundary. In our case,for the integral of a divergence (a derivative) over a vol-ume (a region), is equal to the value of the function atthe surface (a boundary). This is the divergence theorem

ˆV

(~∇ · ~v

)dτ =

˛S

~v · d~a.

The divergence theorem is also sometimes called Green’stheorem.

Remember the divergence theorem because it will beused again and again. It can be simply stated as theintegral of a vector field (e.g. ~E) over a surface is thesame as the integral of the divergence of the vector field(e.g. ~∇ · ~E) over the volume enclosed by the surface.

Tip:

Suppose we want to calculate the divergence of

~V (~r) =r

r2.

We can calculate the divergence using the divergenceoperator in spherical coordinates. Since ~v has no θ or φcomponents, we can just use the r part.

~∇ · ~V =1

r2

∂

∂r

(r2 vr

)=

1

r2

∂

∂r

(r2 1

r2

)=

1

r2

∂

∂r(1)

= 0.

However, if we graph the vector field ~V (~r), we have vec-tors pointing radially outward from the origin. This sug-gests a positive divergence at the origin.

We can also calculate the divergence using the Di-vergence theorem. Using the divergence theorem on asphere of radius R about the origin gives usˆ

V

(~∇ · ~V

)dτ =

˛r

r2· d~a

=

ˆ 2π

0

dφ

ˆ π

0

r

r2·R2 sin θ r dθ

= 4π.

We get that the volume integral of the divergence is 4π,but earlier, we found that ~∇ · ~V = 0, so the volume inte-gral should be zero. The volume integral is correct. Theproblem occurs at r = 0 when we calculate the divergenceusing the formula for divergence. That formula is wrongfor the case r = 0 since ~V blows up there. In other words,~∇ · ~V = 0 only for r 6= 0.

Notice that our result using the divergence theoremˆV

(~∇ · ~V

)dτ = 4π,

does not depend on the radius R. In other words, thedivergence of ~V is zero everywhere except at r = 0, butthe integral of the divergence over any volume centeredat r = 0 is 4π.

We can write this result as

~∇ · rr2

= 4π δ(~r),

where δ(~r) is the three-dimensional Dirac delta func-tion

δ(~r) =

∞ : ~r = ~0 = 〈0, 0, 0〉0 : otherwise.

.

Note: We only have to worry about

~∇ · rrn,

1

2 CHAPTER 1. ELECTROSTATICS

when calculating divergence in spherical coordinates inthe case n = 2. For all other n, this problem does not oc-cur, and we can calculate the divergence using the normalformula.

Stokes’ Theorem

For the divergence theorem, we treated the divergenceoperator as a kind of derivative. We can do the samewith the curl operator. In our case, the integral of a curl(a derivative) over a surface (a region), is equal to thevalue of the function at the perimeter of the surface (aboundary). This is Stokes’ theorem

ˆS

(~∇× ~v

)d~a =

˛L

~v · d~l.

Potentials

In physics, we have four fundamental forces. The strongforce holds nuclei together, the weak force is involvedin beta decay, the electromagnetic force holds atoms to-gether, and the gravitational force holds larger systemstogether. Other forces, such as friction, are usually man-ifestations of the electromagnetic force.

In this course, we focus on fields instead of particles.A scalar field is a field that associates a number (i.e.scalar) with each point in the field. A vector field asso-ciates a vector with each point in space. One example ofa vector field is the wind velocity field of a hurricane ortornado.

Maxwell’s equations, the equations of electromag-netism, are in terms of the divergence and curl of theelectric and magnetic fields.

If we know the divergence of a vector field ~∇ · ~F ,and the curl of a vector field ~∇ × ~F , is this sufficient touniquely determine the vector field ~F ? It turns out thatit is not. We must also know the boundary conditions.Given the divergence and curl of a vector field along withits boundary conditions, the Helmholtz theorem tellsus that we can uniquely determine the field.

If the curl of a vector field is everywhere zero, thenthe field can be written as the gradient of a scalar poten-tial. That is,

~∇× ~F = 0 ⇔ ~F = −~∇V.

In fact, for such a curl-less field or irrotational field,there is a theorem that states that all of the followingstatements are equivalent.

1. ~∇× ~F = 0 everywhere2. The line integral

ˆ b

a

~F · d~,

is independent of the path for given endpoints.3. The line integral of the field about a closed loop is

zero ˛~F ·d~l = 0.

4. The vector field ~F is the gradient of some scalarfunction

~F = −~∇V.

There is a similar theorem for the divergence. If thedivergence of a vector field is everywhere zero, then thefield can be expressed as the curl of a vector potential.That is,

~∇ · ~F = 0 ⇔ ~F = ~∇× ~A.

In fact, for such a divergence-less field or solenoidalfield, there is a theorem that states that all of the follow-ing statements are equivalent.

1. ~∇ · ~F = 0 everywhere2. The surface integral ˆ

~F · d~a,

is independent of the surface for any given bound-ary.

3. The surface integral of the field over a closed surfaceis zero ˛

~F ·d~a = 0.

4. The vector field ~F is the curl of some vector

~F = ~∇× ~A.

Note that any vector can be written as two parts

~F = −~∇V + ~∇× ~A,

for any ~F . This is always valid.

Line Integrals

Suppose we have the electric (vector) field

~E = y2 x+ (2xy + z2)y + 2yz z,

and we want to calculate the electric potential at an ar-bitrary point (x0, y0, z0) using the line integral

V (x, y, z) = −ˆC

~E · d~l,

along a straight line P starting at the origin (0, 0, 0) andending at (x0, y0, z0).

The common method of breaking the path intostraight lines along the coordinate axis cannot be usedhere because we’ve specified that the line should bestraight. There are several different methods we coulduse. With any method, since this is a line integral, ourintegral will end up being a 1-dimensional integral. Thatis, it will not be a double or triple integral.

1.1. VECTOR FIELDS 3

Method 1

The first method we’ll use is the formulaˆC

~F · d~l =

ˆ 1

0

~F (~r(t)) · ~r ′(t) dt,

where the curve C is parametrized as ~r(t) with 0 ≤ t ≤ 1.

In our case, we have ~F = ~E.

Since the curve is a straight line, we can write it asthe vector equation

~r(t) = (1− t)~vi + t~vf ,

where ~vi is the starting point, and ~vf is the ending point.In our case, we have

~r(t) = (1− t)〈0, 0, 0〉+ t〈x0, y0, z0〉= 〈x0t, y0t, z0t〉.

Plugging this into our electric field gives us

~E (~r(t)) = 〈y20t

2, 2x0y0t2 + z2

0t2, 2y0z0t

3〉.

The time derivative of our parametrized line is

~r ′(t) = 〈x0, y0, z0〉.

Then our dot product simplifies to

~E (~r(t)) · ~r ′(t) =(3x0y

20 + 3y0z

20

)t2.

Plugging this into the integral gives us

ˆ 1

0

~E (~r(t)) · ~r ′(t) dt =(3x0y

20 + 3y0z

20

) ˆ 1

0

t2 dt

= x0y20 + y0z

20 .

So the electric potential at (x0, y0, z0) is

V (x0, y0, z0) = −x0y20 − y0z

20 .

This is valid for any point (x0, y0, z0), so we might as wellwrite it as

V (x, y, z) = −xy2 − yz2.

Method 2

We want to compute the line integral

V (x, y, z) = −ˆC

~E · d~l,

In Cartesian coordinates, the line element is

dl = dx x+ dy y + dz z.

The dot product is

~E · d~l = y2 dx+ (2xy + z2)dy + 2yz dz.

However, y and z are not independent of x.

Along a straight line from (0, 0, 0) to (x0, y0, z0), weknow that

y =y0

x0x, dy =

y0

x0dx

z =z0

x0x, dz =

z0

x0dx

Plugging these into the dot product we found above, weget

Substituting these in for y, dy, z, and dz in our ex-pression for the dot product, and simplifying gives us

~E · d~l =

[3y2

0 + 3z20

y0

x0

]x2

x20

dy.

Integrating, to get the potential, we get

V (x0, y0, z0) = −ˆC

~E · d~l

= −[3y2

0 + 3z20

y0

x0

]1

x20

ˆ x0

0

x2 dx

= −x0y20 − y0z

20 .

This is valid for any point (x0, y0, z0), so we might as wellwrite it as

V (x, y, z) = −xy2 − yz2.

Method 3

We will do the same problem a third way. We use againthe Cartesian line element

dl = dx x+ dy y + dz z.

The dot product is

~E · d~l = y2 dx+ (2xy + z2)dy + 2yz dz.

Now we treat the path as a pair of functions of x

y = f(x), f(0) = 0

z = g(x), g(0) = 0.

We can now write the differentials as

dy = f ′(x) dx

dz = g′(x) dx.

Plugging these into the dot product gives us

~E · d~l = f2 dx+ (2xf + g2)f ′ dx+ 2fg g′ dx.

Note that the product rule of differentiation gives us

d

dx

(xf2

)= f2 + 2xff ′

d

dx

(g2f)

= 2gg′f + g2f ′.


Returning to the dot product, we can regroup theterms as

~E · d~l =

[d

dx

(xf2

)+

d

dx

(g2f)]dx

=d

dx

[xf2 + g2f

]dx.

Now we integrate to cancel out the derivative and getˆC

~E · d~l =

ˆd

dx

[xf2 + g2f

]dx

= xf2 + g2f.

Backsubstituting gives us

xf2 + g2f = xy2 + yz2.

So our potential is again

V (x, y, z) = −xy2 − yz2.

1.2. ELECTRIC FIELD 5

1.2 Electric Field

Discrete Charges

Given discrete source charges q1, q2, . . ., what is the forceexerted on a test charge Q? Note that electrostatics isthe special case when the source charges are stationary,and that’s all we’ll be dealing with in this first chapter.

We denote the position of the ith source charge bythe vector ~ri

′ and the position of the test charge (or ob-server) with the vector ~r. We can then denote the vectorfrom the ith source charge to the test charge as

~u = ~r − ~ri ′,

as shown in the diagram below. Notice that ~r = 0 and~u = −~ri ′ if the test charge Q (or observer) is at the originof our coordinate system.

y

z

x

Q

q3

q2

q1

~r

~r1′

~u

The force that each source charge q exerts on the testcharge Q is given by Coulomb’s law which states thatthe force is proportional to the two charges and inverselyproportional to the square of the distance between them

~F =1

4πε0qQ

~r − ~r ′

|~r − ~r ′|3.

We can write this simpler in terms of ~u as

~F =1

4πε0

qQ

u2u.

This is in SI units where a charge q has units of Coulomb(C). Coulomb’s law in Gaussian units (also called the

CGS system) is

~F =qQ

u2u.

In this system, charge has units of statcoulomb (statC)and force has units of dyne. In this course, we will usethe more standard SI units.

So the force on Q due to particle qi

~F i =1

4πε0

qiQ

u2i

ui.

To compute the force on Q due to n particles, we can usethe principle of superposition and directly add thecontributions of the individual particles. That is,

~F = ~F 1 + ~F 2 + · · ·+ ~F n =

n∑i=1

1

4πε0

qiQ

u2i

ui. (1.1)

The principle of superposition and Coulomb’s law are thekey principles of electrostatics.

Note that

ε0 = 8.85× 10−12 C2

Nm2,

is the permittivity of free space.

Returning to Eq. (1.1), notice that the test chargeQ is constant with respect to the sum. Therefore, we canfactor the Q out as

~F = Q

n∑i=1

1

4πε0

qiu2i

ui.

The sum is now defined to be the electric field at P dueto the discrete charges qi. That is,

~F = Q~E,

where the electric field is

~E =1

4πε0

n∑i=1

qiu2i

ui.

The electric field ~E is defined for all positions ~r. It isnot a contact force. It is rather, an ‘action at a distance’force. The electric field itself has energy, momentum, andangular momentum.


What is the electric field at a point P at a heightz above the middle of a pair of source charges q shownin the diagram below?

x

P

q qd2

d2

z

We begin by noting the vectors ~u from each particleto the observer at P . Since the particles both have pos-itive charge, the electric field vectors point away fromthe particles.

x

~E

P

q1 q2

~E1~E2

~u2~u1

θ

We know that

~E = ~E1 + ~E2 =1

4πε0

(q1

u21

u1 +q2

u22

u2

).

We also know that the magnitude of both charges arethe same. By symmetry, we note that the horizontalcomponents of the field cancel, so the only componentwe have to calculate is the one along z. The z directionis given by cos θ. We know that ~E2 for example, is avector of some magnitude β in the direction u. Thatis, ~E2 = βu. To obtain the vertical component of thisvector, we simply multiply its magnitude by cos θ. Thatis,

~E =2

4πε0

q

u2cos θz.

Since cos θ = zu and u =

√z2 +

(d2

)2by the

Pythagorean theorem, we have that

~E =q

2πε0

z

u3z

=1

4πε0

2qz(z2 +

(d2

)2) 32

z.

What happens as z →∞? As z gets large,

z(z2 +

(d2

)2) 32

∼ 1

z2,

so as z →∞~E =

1

4πε0

2q

z2z.

This is the electric field of a single particle of charge2q at a distance z. This makes sense because as yougo further and further away from the pair of particles,they will look more and more like a single particle.

What happens as d→ 0? Again, we get the limit

~E =1

4πε0

2q

z2z,

which is what we should expect.

Example:

When computing ~E for a given charge distribution,always try to simplify the problem using symmetry.

Tip:

After you get a solution check the limit of your so-lution as various values approach ∞ and 0. Are thelimits what you should expect to see?

Tip:


Consider the same problem as in the previous ex-ample except with one of the charges being negative.

x

~E

P

q −q

~E1

~E2

This time the vertical components of the field can-cel, and we are left only with the horizontal component

given by

~E =2

4πε0

q

u2sin θ x

=2

4πε0

q

u2

d2

ux

=1

4πε0

qd(z2 +

(d2

)2) 32

x.

What if z >> d?

Notice that if z >> d, then ~E → 0, since there isa plus charge near a minus charge.

The form~E =

1

4πε0

qd

z3x,

is characteristic of a dipole.

Also notice that if d → 0, then ~E → 0. That is,if the two opposite charges are brought together, theycompletely cancel out.

Example:

Continuous Charge Distribution

For a continuous charge distribution the calculation of theelectric field is essentially the same, except now instead ofsumming over the discrete charges, we have to integrateover the continuous charge distribution.

~E(~r) =1

4πε0

ˆ1

u2u dq.

For a linear charge distribution, that is, for chargedistributed along a curve with linear charge density λ,the charge differential is dq = λ dl′. That is, if you take asmall length element dl′ of the line and multiply it by thecharge per unit length λ, you get the elemental charge dq.The integral becomes then

~E(~r) =1

4πε0

ˆL

λ(~r ′)

u2u dl′.

Notice that λ could be a function of position.

For a surface charge distribution with a surfacecharge density of σ, the charge differential is dq = σ da′,and the integral becomes

~E(~r) =1

4πε0

ˆS

σ(~r ′)

u2u da′.

For a volume charge distribution with a volumecharge density of ρ, the charge differential is dq = ρ dτ ′,and the integral becomes

~E(~r) =1

4πε0

ˆV

ρ(~r ′)

u2u dτ ′.

These are the three situations that can occur and thelast one, being the most general, is typically calledCoulomb’s law. In each case, the integral is taken overthe entire charge distribution.

Note in each case that the source is always indicatedwith a “prime” symbol.


Calculate the electric field at a height z above themiddle of a straight wire of length 2L with a constantcharge density λ.

x

z

dq

d~E

~u

−L L

The first step is to identify the vectors to the sourcepoint and the field point. The source point is the in-finitesimal charge dq, and the field point is the locationat which we want to calculate the electric field. We seethat

~r ′ = (x, 0, 0) (Source point)

~r = (0, 0, z) (Field point)

Always including this step will make it a lot easier tocalculate the electric field for a continuous charge dis-tribution. Now we have that

~u = ~r − ~r ′ = (−x, 0, z).

Our integral is then

~E =1

4πε0

ˆ L

−L

λ

u2u dx,

where

u =~u

|~u|=−xx+ zz√x2 + z2

.

By the symmetry (equal charges at equal distanceson both sides of x = 0), we know that the x-componentof the integral is zero. The only part that remains isthe z-component, so

~E =1

4πε0

ˆ L

−L

λ

u2

−xx+ zz√x2 + z2

dx

=1

4πε0

ˆ L

−L

λ

u2

zz√x2 + z2

dx

=λz

4πε0z

ˆ L

−L

1

(x2 + z2)3/2dx

We can integrate this after making a trigonometric sub-stitution to get

~E =1

4πε0

2λL

z√L2 + z2

z.

What happens if z >> L? If z >> L then L2

z2 ∼ 0.Factoring z2 from the radical gives us

~E =1

4πε0

2λL

z2

√1 + L2

z2

z,

so for z >> L,

~E ∼ 1

4πε0

2λL

z2z.

This is the electric field for a point charge λ2L at a dis-tance z, which is to be expected. The wire has a totalcharge of λ2L and from a great distance, it just lookslike a point charge.

What if L → ∞? Taking the limit, we see that ifL→∞, then

~E =1

4πε0

2λ

zz.

This is the electrical field a distance z from an infiniteline of charge. Make sure to remember this.

Example:


Calculate the electric field at a point P a distancez above the center of a disk with a uniform charge dis-

tribution σ.

z

R

P

By symmetry, we know that the horizontal compo-nents of the field cancel, and we are left with only thecomponent in the z direction. Choosing an infinitesimalring of width dr at the distance r from the center of thedisk, by Coulomb’s law, we have that the field due tothe ring is

Ering =1

4πε0

σ2πr dr

(r2 + z2)

z√r2 + z2

,

where z√r2+z2

is the cosine of the angle. That is, it is

the projection that removes the horizontal components.

We add up all the rings by integrating with respectto r from 0 to R to get the whole field

~Edisk =σ2πz

4πε0

ˆ R

0

r

(r2 + z2)32

drz

=1

4πε02πσz

[1

z− 1√

z2 +R2

]z.

If z >> R, we should expect to see a point charge.However, if z →∞, we get that E ∼ 0. To get the limitthat is useful, we need to do a Taylor expansion of thesecond term.

1√z2 +R2

=1

z− 1

2

R2

z3+ · · · .

Plugging the first two terms of this Taylor expansioninto Edisk to approximate it, we get

~Edisk ≈1

4πε0

πσR2

z2z =

1

4πε0

Q

z2z,

where Q is the total charge of the disk. This limit isthat of a point charge Q.

If R→∞, we get the electric field due to an infiniteplane

~Eplane =σ

2ε0z.

Notice that the magnitude of the electric field near aninfinite plane of charge does not depend on the distancez from the plane, and its direction is always perpendic-ular to the plane.

Example:

Gauss’s Law

Electric field lines cannot start or stop in midair. Theycan only start at positive charges and end at negativecharges or at infinity. Field lines cannot cross each other.If they did, they would imply that the electric field hastwo different directions at the intersection.

There are two common ways of representing fieldlines. One is with arrows whose lengths denote the mag-nitude of the field there. The other way of representingfield lines is with unadorned curves. In this case, the mag-nitude of the field in a region can be determined by thedensity of the field lines there.

Recall that for a point charge q, the electric field is

~E =1

4πε0

q

r2r.

If you enclose the charge in a spherical shell of radius rthen the flux (i.e. the field lines through the surface of theshell) is given by the surface area times E, the magnitude

of ~E,

4πr2 1

4πε0

q

r2=

q

ε0.

The flux of ~E is defined as

ΦE =

ˆS

~E · d~a,

where´S

denotes an integral over a surface. Note that ona closed surface, d~a is a vector that points outward and isperpendicular to the surface, so ~E · d~a is the componentof ~E that is perpendicular to the surface.

For a positive charge enclosed in a spherical shellwith no other charges, all the field lines from the chargemust pass through the surface of the shell and never re-turn. If the charge is outside the closed surface, the flux iszero since every field line that goes into the surface mustalso go out of the surface. There is no net flux if everyfield line entering a closed surface also leaves it.

For a charge a at the origin in a spherical coordinatesystem, we can compute the flux passing through an en-


closing spherical shell of radius r by integrating over thesurface

ΦE =

˛S

~E · d~a.

Recall that in spherical coordinates, da, the area differ-ential is da = r2 sin θ dφ dθ. Since d~a is the vector withmagnitude da and perpendicular to the area differentialda, we have that d~a = r2 sin θ dφ dθ r. So our integralbecomes

ΦE =

˛S

1

4πε0

q

r2r · r2 sin θ dφ dθ r.

Since r · r = 1, this simplifies to

ΦE =q

4πε0

˛S

sin θ dφ dθ.

Integrating over the surface, we get

ΦE =q

4πε0

ˆ 2π

0

dφ

ˆ π

0

sin θ dθ

=q

ε0.

Notice that the flux is independent of r. That is, we getthe same value for the flux no matter how big our spher-ical shell is. In fact, we can deform the surface any waywe like and the result is the same. This means we can usean arbitrary surface. Just remember that this only workswhen q is inside the closed surface.

Thus, for a point charge q enclosed by an arbitrarysurface S, ˛

S

~E · d~a =q

ε0.

For multiple charges inside the closed surface, by the prin-ciple of superposition, we have that ~E = ~E1 + ~E2 + · · ·+~En. This tells us that

˛S

~E · d~a =

∞∑i=1

˛S

~Ei · d~a =

∞∑i=1

qiε0

=Qencε0

,

where Qenc is the total charge enclosed by the surface.Thus, for an arbitrary surface S, we get the integral formof Gauss’s law ˛

S

~E · d~a =Qencε0

.

So in general, the flux through a closed surface is propor-tional to the charge enclosed and is not dependent on thesize of the closed surface. With this derivation, we cansee why. Coulomb’s law tells us that ~E ∼ 1

r2 . But surfacearea goes as A ∼ r2. These two cancel each other whencalculating flux, so flux is independent of r.

To obtain the differential form of Gauss’s law, weapply the divergence theorem˛

S

~E · d~a =

ˆV

~∇ · ~E dτ =Qencε0

,

where the integral on the right is over the volume V en-closed by the surface S.

We can calculate the enclosed charge by integratingthe charge density ρ over the volume as

Qenc =

ˆV

ρ(~r) dτ,

where ρ is the charge density.

So our integral becomes

ˆV

~∇ · ~E dτ =1

ε0

ˆV

ρ(~r) dτ.

Since this is true for any volume V, the integrands have tobe equal. This gives us the differential form of Gauss’slaw

~∇ · ~E =ρ(~r)

ε0.

This is also the first of Maxwell’s equations for a vacuum.

A second way of getting this result is by directly tak-ing the divergence of Coulomb’s law. Recall Coulomb’slaw for a continuous charge distribution in 3D

~E(~r) =1

4πε0

ˆρ(~r ′)

u2u dτ ′.

Taking the divergence of both sides, we get

~∇ · ~E =1

4πε0

ˆ~∇ · ρ(~r ′)

u2u dτ ′.

Note that ~∇ acts only on ~r, not ~r ′ since ~E is a functionof ~r. The only part of the integral that depends on ~r is~u = ~r − ~r ′. So ~∇ acts only on the ~u part of the integral

~∇ · ~E =1

4πε0

ˆρ(~r ′)~∇ · u

u2dτ ′.

We know that

~∇ · uu2

= 4πδ3(~u),

so our integral becomes

~∇ · ~E =1

ε0

ˆρ(~r ′)δ3(~r − ~r ′) dτ ′,

which simplifies to

~∇ · ~E =ρ(~r)

ε0.


Given~E = kr3 r,

find the charge density ρ(~r).

From the differential form of Gauss’s law, we knowthat

ρ(~r) = ε0~∇ · ~E.

Using spherical coordinates since it’s a radial field,we get

ρ(~r) = ε0~∇ ·(kr3 r

)= ε0

1

r2

∂

∂r

(r2 kr3

)= ε0

1

r25kr4

= 5ε0kr2.

Now that we have the charge density, we can alsocalculate the total charge that is producing this field.

Q =

ˆV

ρ(~r) dτ

= 5ε0k

ˆ 2π

0

ˆ π

0

ˆ R

0

r2 r2 sin θ dθd φ dr

= 4πε0kR5.

This is the charge enclosed in a sphere of radius R.

Example:

Calculate the charge density ρ(~r) if the electric fieldis

~E(~r) = e−λrr

r2.

We can obtain the charge density from the electric fieldby using the differential form of Gauss’s law

~∇ · ~E =ρ(~r)

ε0.

We start by applying an identity from vector cal-culus. If f is a scalar function and ~F is a vector field,then

~∇ ·(f ~F)

= f ~∇ · ~F + ~F ·(~∇f).

In this case, it is convenient to let e−λr be the scalarfunction and let r

r2 be the vector field so

~∇ · ~E = e−λr ~∇ · rr2

+r

r2·(~∇e−λr

).

But we know that

~∇ · rr2

= 4π δ(~r).

Furthermore, since the delta function is zero except at~r = ~0, we know that

e−λr4π δ(~r) = e−λ(0)4π δ(~r) = 4π δ(~r).

Therefore, we have that

~∇ · ~E = 4π δ(~r) +r

r2·(~∇e−λr

)= 4π δ(~r) +

r

r2·(∂

∂re−λr

)r

= 4π δ(~r)− λe−λr rr2· r

= 4π δ(~r)− λe−λr

r2.

So our charge density is

ρ(~r) = ε0

[4π δ(~r)− λe−λr

r2

].

Example:

If the charge is uniformly distributed in a volume, thenyou’re not dealing with a conductor since the chargesare not free to move.

Tip:

When working in spherical or cylindrical coordinates,beware, the divergence of a vector field denoted ∇ · ~Fis not just the dot product of the gradient in that coor-dinate system with the vector field as it is in cartesiancoordinates. Since the unit vectors are not constantin spherical and cylindrical coordinates, special caremust be taken when applying the gradient operatorsince it contains derivatives. When in doubt, look upthe divergence for the coordinate system you are using.

Tip:


Symmetries play an important role in the applicabil-ity of Gauss’s law. Gauss’s law is always valid, but it’snot always useful. Gauss’s law is useful with three kindsof symmetry:

1. Spherical symmetry: Use a concentric sphere asyour Gaussian surface.

2. Cylindrical symmetry: Use a coaxial cylinder asyour Gaussian surface.

3. Plane symmetry: Use a Gaussian box straddling theplane as your Gaussian surface.

Note, a plane must be infinite to apply Gauss’s law. Oth-erwise, you would get fringe effects from the edges of theplane. The same is true for cylindrical symmetry. Theline of charge must be very long and the coaxial cylinderrelatively small and near the center in order to get a goodapproximation of the field using Gauss’s law.

When applying Gauss’s law remember that the sur-face integral is over the Gaussian surface, not over thesurface of the charged object.

Tip:

If the charge enclosed is zero, the field inside theGaussian surface is not necessarily zero. That is,¸~E · d ~A = Qenc/ε = 0, does not imply that ~E = 0.

It only implies that the flux through the closed sur-face is zero. There could be a nonzero field that is notradially symmetric in the enclosed area.

Tip:

Calculate the electric field outside of a sphere ofradius R with charge q distributed uniformly.

We start with Gauss’s law

˛S

~E · d~a =Qencε0

,

and our Gaussian surface is a concentric sphere withradius greater than R. We know that the field is radial,therefore the field lines are perpendicular to our Gaus-sian surface. In other words, ~E ‖ d~a, which tells us that~E · d~a = E da. We also know that Qenc = q. So, usingspherical coordinates and plugging in the area element

da = r2 sin θ dθ dφ law gives us

ˆ 2π

0

ˆ π

0

Er2 sin θ dθ dφ =q

ε0

Er2

ˆ 2π

0

dφ

ˆ π

0

sin θ dθ =q

ε0

Er24π =q

ε0

E =q

4πε0r2.

And since we know that the direction of the field is ra-dially outward, we have that

~E =1

4πε0

q

r2r.

This is the same as the field for a point charge q.For any radially, uniformly distributed charge, we cancalculate the electric field by pretending that the entirecharge is concentrated at the center point.

Example:


Calculate the electric field inside of a long cylinderwith a charge density of ρ = kr where r is the distancefrom the cylinder’s axis.

By Gauss’s law, we know that˛S

~E · d~a =Qencε0

,

where S is the surface of a Gaussian cylinder of radius Rthat is coaxial with and inside of the charged cylinder.We can calculate the total charge within the Gaussiancylinder by integrating the charge density over the vol-ume of the cylinder

Qenc =

ˆV

ρdτ

=

ˆ 2π

0

ˆ L

0

ˆ R

0

(kr)r2 dr dφ dz

=2

3πkLR3.

There is no flux through the ends of our Gaussiancylinder, so we can ignore that part of the surface. On

the curved part of the Gaussian cylinder, the electricfield is perpendicular to the surface so ~E · d~a = E da.So Gauss’s law becomes

˛S

~E · d~a =23πkLR

3

ε0

E

ˆS′da =

2πkLR3

3ε0

E2πRL =2πkLR3

3ε0

E =kR2

3ε0.

We used the fact that´S′da is just the surface area of

S′, which in this case is the curved part of the Gaussiancylinder.

We now have the magnitude of the electric fieldinside the cylinder, and we know that its direction isradially outward, so

~E =kR2

3ε0r.

Example:

Calculate the electric field due to an infinite planewith a uniform surface charge density σ.

We start by putting a Gaussian box of infinitesimalheight and area A on the top and area A on the bot-tom. The Gaussian surface should straddle the chargedsurface so that the top of the Gaussian box is above thecharged surface and the bottom of the Gaussian box isbelow the charged surface.

The area of the charged surface that is enclosed bythe box is A, so the charge enclosed is Qenc. The onlysurfaces of the Gaussian box that matter are the topand the bottom. Since the electric field of the chargedsurface is perpendicular to the surface, there is no fluxthrough the sides of the box—only through the top andthe bottom, and there, the electric field is perpendicular

to the surface, so ~E · d~a = E da. So by Gauss’s law

˛S

~E · d~a =Qencε0

E

ˆS′da =

σA

ε0

E 2A =σA

ε0

E =σ

2ε0.

Here, S′ is the surface that matters (i.e. the top andbottom), and we used the fact that

´S′da = 2A is just

the surface area. So the electric field due to an infiniteplane is

~E =σ

2ε0n,

where n is the normal vector to the plane. Notice thatthe electric field due to an infinite plane does not de-pend on the distance from the plane. This makes sensebecause if we saw an infinite plane in front of us, therewould be no way to tell the distance to the plane.

Example:


Calculate the electric field everywhere due tocharged spherical shell with inner radius a and outerradius b. Inside the shell, that is, between a and b, thecharge density is ρ = k

r2 .

On the inside, we use a Gaussian surface with ra-dius 0 < r < a. There is no charge enclosed in thisGaussian surface, so the electric field is E = 0 here.

~E = 0, 0 < r < a.

Between the inner and outer surfaces of the shell,we calculate the electric field by using a Gaussian spherewith radius a < r < b. The enclosed charge is computedas

Qenc =

ˆ r

a

ρ dτ

=

ˆ 2π

0

ˆ π

0

ˆ r

a

k

r2r2 sin θ dr dθ dφ

= 4πk(r − a).

The field is perpendicular to the Gaussian surface so

~E · d~a = E da, and by Gauss’s law, the electric field is

E

˛S

da =4πk(r − a)

ε0

E4πr2 =4πk(r − a)

ε0

E =k(r − a)

ε0r2.

The electric field within the charged shell is

~E =k(r − a)

ε0r2r, a < r < b.

To find the field outside the spherical shell, we usea Gaussian surface with r > b. The total charge en-closed can be found by replacing r by b in the formulawe found earlier for the enclosed charge. Then the elec-tric field outside the shell is just the field of a pointparticle with the total charge of the sphere.

~E =k(b− a)

ε0r2r, r > b.

Example:

The curl of the electric field is calculated as

~∇× ~E.

The curl may also be called the rotator.

Consider the electric field of a point charge at theorigin

~E =1

4πε

q

r2r.

Visually, we know the field lines radiate outward instraight lines. There’s no curving of the field lines, sothere is zero curl. More formally, we take the curl of thisfield in spherical coordinates as

~∇× ~E =1

r sin θ

[∂

∂θ(sin θ Eφ)− ∂

∂φEθ

]r

+1

r

[1

sin θ

∂

∂φEr −

∂

∂r(rEφ)

]θ

+1

r

[∂

∂r(rEθ)−

∂

∂θEr

]φ

=1

r sin θ

∂Er∂φ

θ − 1

r

∂Er∂θ

φ

= 0.

We found that the curl of a charge at the origin is zero,but this is actually true for a charge anywhere.

The field due to a collection of n charges is

~E = ~E1 + ~E2 + · · ·+ ~En.

Taking the curl gives us

~∇× ~E = ~∇×(~E1 + ~E2 + · · ·+ ~En

)= ~∇× ~E1 + ~∇× ~E2 + · · ·+ ~∇× ~En

= 0 + 0 + · · · 0.

This tells us that ~∇× ~E = 0 for any charge distribution.

More generally, we can take the curl of an arbitraryelectric field via Coulomb’s law

~∇× ~E = ~∇× 1

4πε0

ˆρ(~r ′)

u2u dτ

=1

4πε0

ˆρ(~r ′)

(~∇× u

u2

)dτ

= 0.

We can move the curl operator inside the integral pastthe density since the derivatives in the curl operator arewith respect to ~r and not ~r ′. In the end, we used thefact that

~∇× u

u2= 0.

So we have that

~∇× ~E = 0 .

We say therefore, that the electric field is curl-less orirrotational.

Stokes’ theorem tells us then that˛L

~E · d~l = 0,

1.3. ELECTRIC POTENTIAL 15

where the closed curve L is the boundary of the surface.The line integral of the electric field over any closed pathis zero, which implies that the line integral over a non-closed path

ˆ b

a

~E· d~l,

is independent of the path taken between b and a. Thisline integral is not zero in general, it is just independentof the path taken. This occurs because ~E is curl-less.

Let’s come at this from the other direction by cal-culating

´~E· d~l from point a to point b. In spherical

coordinates, the line element is

d~l = dr r + r dθ θ + r sin θ dφ φ,

then

~E · d~l =1

4πε0

q

r2dr.

Now our line integral becomes

ˆ b

a

~E· d~l =q

4πε0

ˆ b

a

1

r2dr

= − q

4πε0

1

r

∣∣∣rbra

= − q

4πε0

(1

rb− 1

ra

).

If we do the integral about a closed path, then ra = rb,and

˛~E · d~l = 0.

Then by Stokes’ theorem, we have ~∇× ~E = 0.

Determine if

~E = k(xyx+ 2yzy + 3xyz),

is a valid electric field.

Remember that the curl of the electric field mustbe zero. That is,

~∇× ~E = 0.

To check that these are valid fields, we take the curlof it and see if it’s zero.

~∇× ~E1 = k

∣∣∣∣∣∣∣∣x y z∂∂x

∂∂y

∂∂z

xy 2yz 3xy

∣∣∣∣∣∣∣∣= (3x− 2y)x− 3y y − x z6= 0.

Since ~∇× ~E 6= 0, this is clearly not a possible electricfield.

Example:

1.3 Electric Potential

We now define a function called the electric potentialcalculated as a line integral

V (~r) = −ˆ ~r

O

~E· d~l.

Here, O is some standard reference point. The negativesign is by convention to relate the electric potential V topotential energy.

The units of potential are

volt =Joule

Coulomb=

Newton ·meter

Coulomb.

The potential difference between two points~b and~a is given by

V (~b)− V (~a) = −ˆ ~b

O

~E· d~l+

ˆ ~a

O

~E· d~l

= −ˆ ~b

O

~E· d~l−ˆ O

~a

~E· d~l.

This simplifies to

V (~b)− V (~a) = −ˆ ~b

~a

~E· d~l.


The fundamental theorem for gradients statesthat

V (~b)− V (~a) =

ˆ ~b

~a

~∇V · d~l.

Equating this to what we found above gives us

ˆ ~b

~a

~∇V · d~l = −ˆ ~b

~a

~E· d~l.

This implies that the electric field is the negative of thegradient of the electric potential

~E = −~∇V.

We now have the means to calculate the electric potentialgiven the electric field and also the means to calculate theelectric field given the electric potential.

When calculating the electric potential given theelectric field, the reference point O is chosen for conve-nience for a given problem. Changing the reference pointchanges V , but not the underlying field ~E.

If you calculate the electric potential given the field~E using the line integral above, check your work byverifying that ~∇V = −E(~r).

Tip:

Since ~E is curl-less, the three differential equationsof ~E reduce to the single scalar function V (~r).

We will generally fix the potential at infinity equalto zero since this is usually the most convenient choice.

V (~r) = −ˆ ~r

∞~E· d~l.

This is a line integral from infinity to the point ~r. It isimportant to realize that it is a line integral starting atinfinity and coming in to ~r. For the radially symmetriccase, ~r = r, this simplifies to

V (r) = −ˆ r

∞~E· d~l.

Note that if r is inside a sphere for example, then thereis a disconitnuity in the electric field between ∞ and r.Specifically, there is a discontinuity at the surface of thesphere. In that case, you have to break the integral intotwo parts. One goes from infinity to the surface of thesphere, and the second goes from the surface of the sphereto r.

For an infinite plane, choosing the reference pointas infinity does not work since the charge also goes toinfinity.


Calculate the potential at a point (x, y, z) due tothe electric field

~E = k(y2 x+ [2xy + z2] y + 2yz z).

Recall that the potential at a point ~r is calculatedby the line integral

V (~r) = −ˆ ~r

O

~E· d~l.

If we let the reference point O be the origin of a carte-sian coordinate system, then we can calculate the lineintegral by breaking it into three pieces—one along aline in each component direction.

V (x, y, z) = −ˆL1

Ex dx−ˆL2

Ey dy −ˆL3

Ez dz.

y

z

x

(x, y, z)

(0, 0, 0)

(x, 0, 0)(x, y, 0)

Plugging in the components gives us

V = −kˆL1

y2 dx− kˆL2

[2xy + z2] dy − kˆL3

2yz dz.

Now we look at each of the different curves. Along L1,y = z = 0 and x goes from 0 to x. Along L2, z = 0, xis constant, and y goes from 0 to y. Along L3, x and yare constant and z goes from 0 to z. Now our integralsbecome

V = −kˆ x

0

(0) dx− kˆ y

0

2xy dy − kˆ z

0

2yz dz.

Evaluating the integrals, we get

V (x, y, z) = −k(xy2 + yz2).

We can always check our result by verifying that

~∇V = −~E.

Example:


Calculate the potential inside and outside of a uni-formly charged spherical shell of radius R with surfacecharge density σ.

We know that

V (~r) = −ˆ ~r

O

~E· d~l.

For r > R, by Gauss’ law, we can treat all the chargeas being at the origin. We know that the field outsideis

~E =1

4πε0

q

r2r,

and the total charge of the shell is q = σ4πR3. So tak-ing infinity to be the reference point, outside the shellthe potential is

V (r) = −ˆ ~r

∞~E· d~l

= −ˆ r

∞

1

4πε0

q

r2dr

=1

4πε0

q

r.

Notice that we get the same potential as that due to apoint charge.

Inside the sphere where r < R, we know that~E = 0. The potential is calculated as the line inte-gral from ∞ to r where r is inside the sphere, and sowe have to consider two regions where the electric fieldis different.

V (r) = −ˆ ~r

∞~E· d~l

= −ˆ r

∞E dr

= −ˆ R

∞E dr −

ˆ r

R

E dr

= −ˆ R

∞

1

4πε0

q

r2dr −

ˆ r

R

(0) dr

=q

4πε0R.

Notice that the potential is constant inside the shellwhere the electric field is zero.

Also, notice that although the field is discontinu-ous at r = R, the potential is continuous. In fact, V isalways continuous.

Example:

Calculate the potential at any point in space dueto an infinite wire with linear charge density λ.

We know that the electric field due to an infinitewire is

~E =1

4πε0

2λ

ss,

where s is the distance from the wire.

We know that

V (~s) = −ˆ ~s

O

~E· d~s.

We cannot use infinity as our reference point since thecharge extends to infinity, so we use an arbitrary dis-tance a as our reference point. Since our differentialdistance d~s is parallel to the electric field, we know that

~E · d~s = E ds, so

V (s) = −ˆ s

a

1

4πε0

2λ

sds′

= − λ

2πε0ln( sa

)=

λ

2πε0ln(as

).

To verify, we take the negative of the gradient

~E = −~∇V

=λ

2πε0ss.

Notice that the arbitrary reference point dropped out.

Example:

So far we have learned that

~∇ · ~E =ρ

ε0~∇× ~E = 0

~E = −~∇V.

If we plug the third equation into the first, we get Pois-

son’s equation

∇2V = − ρ

ε0.

This is a second order differential equation satisfied bythe potential V . If there are regions where the chargedensity ρ is 0, then in those regions Poisson’s equationreduces to Laplace’s equation

∇2V = 0.


We’ve now gone from a pair of first order differential equa-tions (i.e. ~∇ · ~E = ρ

ε0and ~E = −~∇V ) to a single second

order differential equation ∇2V = − ρε0

.

It takes four equations to calculate ~E using ~∇ · ~E =ρε0

and ~∇× ~E = 0. Calculating V only requires one dif-

ferential equation, namely, ∇2V = − ρε0

. A good strategy

is to first obtain V using either ∇2V = − ρε0

or ∇2V = 0,

and then calculate ~E using ~E = −~∇V

Superposition Principle

The superposition principle also holds for the poten-tial V . That is, we can add potentials.

For a single isolated charge q at the origin, we knowthat

~E =1

4πε0

q

r2r,

then the potential is

V (~r) = −ˆ r

∞~E· d~l =

1

4πε0

q

r.

Remember that the potential is a scalar, so it has noassociated direction.

Tip:

If the charge is not at the origin, then we have to usethe vector ~u = ~r − ~r ′, and we get

V (~r) =1

4πε0

q

u,

where u is just the distance between the charge and ~r.This is the potential at ~r due to the charge.

If we have n charges qi, we can use the principle ofsuperposition to add their contributions to the potential

V (~r) =1

4πε0

n∑i=1

qiui,

where ui = |~r−~ri ′|. For a continuous charge distribution,the sum becomes an integral

V (~r) =1

4πε0

ˆ1

udq.

In the continuous linear case, we get

V (~r) =1

4πε0

ˆλ(~r ′)

|~r − ~r ′|dl′.

For the continuous surface case, we get

V (~r) =1

4πε0

ˆσ(~r ′)

|~r − ~r ′|da′.

For the continuous volume case, we get

V (~r) =1

4πε0

ˆρ(~r ′)

|~r − ~r ′|dτ ′.

After finding V using the charge distribution, the elec-tric field can easily be calculated as ~E = −~∇V .

Tip:

We can now solve six different electrostatics prob-lems. Given any one of ~E, V , or ρ, we should be able tocalculate the other two.


Calculate again the potential inside and outside ofa uniformly charged spherical shell of radius R with sur-face charge density σ. This time, use

V (~r) =1

4πε0

ˆσ

uda′,

instead of Gauss’s law.

We take the sphere to be centered at the origin ofa spherical coordinate system. To compute the electricfield at an arbitrary point, we can take that point to beon the z-axis and at a distance z from the origin. Thefield point is then at

~r = 〈0, 0, z〉.

Taking a small arbitrary piece of the surface of thespherical shell to be our source point, we have

~r ′ = 〈R sin θ′ cosφ′, R sin θ′ sinφ′, R cos θ′〉.

Then

~u = 〈−R sin θ′ cosφ′,−R sin θ′ sinφ′, z −R cos θ′〉,

and its magnitude is

u =(R2 sin2 θ′ sin2 φ′ +R2 sin2 θ′ cos2 φ′

+z2 +R2 cos2 θ′ − 2zR cos θ′) 1

2

=(R2 sin2 θ′ + z2 +R2 cos2 θ′ − 2zR cos θ′

) 12

=√R2 + z2 − 2zR cos θ′.

The area element on a spherical surface is

da′ = R2 sin θ′ dθ′dφ′.

Putting it all together, we have

V (z) =σ

4πε0

ˆR2 sin θ′ dθ′dφ′√

R2 + z2 − 2zR cos θ′

=σR2

4πε0

ˆ 2π

0

dφ

ˆ π

0

sin θ′√R2 + z2 − 2zR cos θ′

dθ′

=σR2

2ε0

√R2 + z2 − 2zR cos θ′

Rz

∣∣∣π0

=σR

2ε0z

[√(R+ z)2 −

√(R− z)2

].

Outside the shell we have z > R, which impliesthat

√(R− z)2 = z −R. This gives us

V (r) =σR

2ε0z[R+ z − (z −R)] =

σR2

ε0z.

Inside the shell we have z < R, which implies that√(R− z)2 = R− z. This gives us

V (r) =σR

2ε0z[R+ z − (R− z)] =

σR

ε0.

The total charge is q = 4πR2σ. Using this, we canwrite the potential inside and outside in the same formthat we calculated earlier using Gauss’ law

V (r > R) =1

4πε0

q

r

V (r < R) =1

4πε0

q

R.

Example:

1.4 Boundary Conditions

If we zoom in on any charged surface, we can approximateit as a flat sheet of charge. Consider a Gaussian box thatstraddles this surface. The thickness of the box is in-finitesimal, and the top and bottom have area A. We cantake the thickness of the box to zero, then there will bezero flux through the sides of the box. By Gauss’ law,then the perpendicular component of the electric fieldabove and below the surface of charge is related to thecharge distribution on the surface by

E⊥above − E⊥below =σ

ε0.

For the component of the electric field that is parallelto the surface, we do a line integral that straddles thesurface. We find that

~E‖above = ~E

‖below.

Notice that here we are using the vector form of the fieldbecause there’s not just one direction that is parallel tothe surface.

We can combine these two boundary conditions as

~Eabove − ~Ebelow =σ

ε0n. (1.2)

Here, n is the unit vector perpendicular to the surfaceand points from below to above.

For the potential, we do a line integral from a pointa below the surface to a point b above the surface, then

Vabove − Vbelow = −ˆ b

a

~E· d~l.

If we let the path length shrink to zero, then the integralgoes to zero, which implies that

Vabove = Vbelow.

1.5. WORK AND ENERGY 21

But since ~E = −~∇V , from Eq. (1.2) we know that

~∇Vabove − ~∇Vbelow = − σε0n.

You may also see this relation written as

∂Vabove∂n

− ∂Vbelow∂n

= − σε0,

where∂V

∂n≡ ~∇V · n.

In fact, electric potential is always continuous.If there were a discontinuity in V , there would be an in-finity in ~E since ~E = ~∇V .

1.5 Work and Energy

Given a distribution of stationary source charges qi, whatwork does it take to move a test charge Q from a point ain the distribution to a point b along some path?

q1

q2

q3

q4

a

b

The force you have to exert on Q at any point in theelectric field in order to move Q is

~F = −Q~E.

The force exerted on Q by the field is ~F = Q~E, butthe force you exert on Q to move it is the negative ofthat. Notice that ~F = Q~E is just the electric analogueof ~F = m~a.

The work required to move Q is defined in the sameway as it is defined in mechanics as a line integral

W =

ˆ b

a

~F · d~l.

Making the substitution ~F = −Q~E gives us

W = −Qˆ b

a

~E· d~l.

Recalling that V (~b)− V (~a) = −´ ba~E· d~l, we have that

W = Q[V (~b)− V (~a)

].

This tells us that the potential difference between pointa and b is equal to the work per unit charge required tomove a particle from a to b against the electric forces.

If a is at infinity, that is, you’re bringing your testcharge in from infinity to the point b = r, then V (~a) = 0,and we have

W = QV (~r).

What is the work required to assemble a group ofcharges in some configuration, bringing each one at a timein from infinity? The work to bring the first charge infrom infinity and place it at ~r1 is W1 = 0 since thereare no other charges around. The work to bring in eachsubsequent charge qi and place it at ~ri depends on thecharges already existing in the region.

W2 = q21

4πε0

(q1

u12

)W3 = q3

1

4πε0

(q1

u13+

q1

u23

)W4 = q4

1

4πε0

(q1

u14+

q1

u24+

q1

u34

).

where uij is of course the distance between charges i andj. The energy/work required to assemble the whole groupof four particles is

W = W1 +W2 +W3 +W4.

The work required to assemble n discrete charges qiby bringing them in from infinity is

W =1

4πε0

n∑i=1

n∑j=i+1

qiqjuij

.

The second sum starts at j = i + 1 so that the termsare not counted twice. By counting every term twice andthen dividing by two, we can also write this as

W =1

8πε0

n∑i=1

n∑j=1

qiqjuij

.

Rearranging, we get

W =1

2

n∑i=1

qi

1

4πε0

n∑j=1j 6=i

qjuij

.

Notice that the quantity in parentheses is the potentialat ~ri due to all of the other charges. We can thereforewrite the work as

W =1

2

n∑i=1

qiV (~ri).

Note that V (~ri) is the potential at the position of the ithparticle due to all the other particles—not just the ones


that were there when this particle was brought in frominfinity.

This is the work required to assemble the configu-ration of n charges. It is also the same as the potentialenergy stored in the configuration, and since q ~E is a con-servative force, it is also the same as the work we get fromthe system when we allow it to fly apart.

What is the work required to assemble four pointcharges in a square as in the image?

q4 = q

q1 = −q q2 = q

q3 = −qa

The work to bring the first charge, the top leftone, is W1 = 0. The work to bring in the top rightright charge given that only the top left charge is al-ready in place is

W2 = q21

4πε0

(q1

u12

)= − q2

4πε0a.

The work to bring in the third charge is

W3 = q31

4πε0

(q1

u13+

q2

u23

)= − q2

4πε0a

(−1√

2+ 1

).

The work to bring in the fourth charge is

W4 = q41

4πε0

(q1

u14+

q2

u24+

q3

u34

)=

q2

4πε0a

(−1 +

1√2− 1

).

The total work required to assemble the chargesis then

W = W1 +W2 +W3 +W4.

Example:

To compute the work required to construct a contin-uous charge distribution, we could extend the summationinto an integral as

W =1

2

ˆρV dτ. (1.3)

Here, the integral is over the charge distribution. Thisis true, but it’s not very useful. It’s better if we expressthe density in terms of the electric field via Gauss’s lawρ = ε0~∇ · ~E.

W =ε02

ˆV(~∇ · ~E

)dτ.

From the product rule, we know that

~∇ ·(V ~E

)= V ~∇ · ~E + ~E ·

(~∇V

).

Applying the divergence theorem to the first termgives us ˆ

V

~∇ ·(V ~E

)dτ =

˛S

V ~E · d~a.

So combining the product rule and the divergence theo-rem gives us

˛S

V ~E · d~a =

ˆV

V ~∇ · ~E dτ +

ˆV

~E ·(~∇V

)dτ.

Rearranging this, we get that the work to construct acontinuous charge distribution is

W =ε02

ˆV(~∇ · ~E

)dτ

=ε02

˛S

V ~E · d~a− ε02

ˆV

~E ·(~∇V

)dτ.

Recalling that ~E = −~∇V , we get

W =ε02

˛S

V ~E · d~a+ε02

ˆV

E2 dτ.

This is true for any volume V, but surface integrals aren’talways easy. If we let the volume approach infinity, thenthe surface integral takes place at infinity where V = 0,so the first term disappears, and we get

W =ε02

âllspace

E2 dτ. (1.4)

1.5. WORK AND ENERGY 23

Find the energy of a spherical shell of radius R witha uniform surface charge density

σ =q

4πR2.

We have to adapt our work formula for a surfaceinstead of a volume, but that is easily done.

W =1

2

ˆσV da.

This V is the potential V at the surface.

We know that from the outside, the spherical shelllooks like a point charge, so

V (r) =1

4πε0

q

r, r > R.

Inside the shell, there is no electric field, so we knowthat the potential is constant. What is its constantvalue there? We know that the potential has to be con-tinuous at the surface, so

V (r) =1

4πε0

q

R, r < R.

In our integral, we can use either of these values. Sincewe’re only integrating over the surface, we can approachit from either the inside or outside.

Approaching from the inside, the energy is

W =1

2

ˆσV da

=1

2

q

4πR2

ˆ 2π

0

ˆ π

0

1

4πε0

q

RR2 sin θ dφ

=q2

8πε0R.

We could also use the second equation to calculatethe work for a continuous charge distribution. To usethe other equation, we have to integrate over all space.We know that the field outside the spherical shell is

~E =1

4πε0

q

r2r, r > R.

Inside the shell, the field is ~E = 0. So its energy is

W =ε02

âllspace

E2 dτ

=ε02

ˆr<R

(0) dτ +ε02

(q

4πε0

)2 ˆr>R

1

r2r · r dτ

=q2

32π2ε0

ˆ 2π

0

dφ

ˆ π

0

sin θ dθ

ˆ ∞R

1

r4r2 dr

=q2

8πε0R.

This is the energy stored in this charge distribution.Alternatively, we can think of it as the work required toassemble it by bringing each charge in from infinity.

Example:

What is the energy stored in a solid sphere of radiusR with a uniform charge distribution

ρ =3q

4πR3.

We know that outside, the sphere looks like a pointcharge so

~E =1

4πε0

q

r2r.

For the inside, we can use a Gaussian surface to get

~E =1

4πε0

qr

R3r.

So the work to assemble the sphere is

W =ε02

âllspace

E2 dτ

=ε02

ˆr<R

(1

4πε0

qr

R3r

)2

dτ

+ε02

ˆr>R

(1

4πε0

q

r2r

)2

dτ

=ε02

4π

(q

4πε0R3

)2 ˆ R

0

r4 dr

+ε02

4π

(q

4πε0

)2 ˆ ∞R

1

r2dr

=1

4πε0

3q2

5R.

Example:


So far, we’ve used equations (1.3) and (1.4) inter-changeably, but they are not interchangeable. There is asubtle distinction between the two. If we try to use thesecond one for a point charge, we get infinity as the result.The reason is that it does not make sense to assemble apoint charge. So Eq. (1.4) includes the energy to “make”the charge. We do not have this problem with Eq. (1.3).So for a discrete set of charges, we have to use

W =1

2

ˆρV dτ.

For continuous charge distributions, we can use eitherequation because the integrals only differ at a single point.

The potential V has to be continuous as it crosses sur-faces, unlike the electric field.

Tip:

Remember V and ~E for all the geometric cases welooked at.

Tip:

Beware that equations (1.3) and (1.4) are not linear.That is, we cannot just add the work/energy of differentthings. Note for example that(

~E1 + ~E2

)2

= ~E2

1 + ~E2

2 + 2~E1~E2.

Because of the “interference” term 2~E1~E2 we clearly can-

not just add the work/energy of different things. If our

field is the sum of two fields ~E = ~E1 + ~E2, then

Wtotal =ε02

âllspace

E2 dτ

=ε02

âllspace

(E2

1 + E22 + 2~E1

~E2

)dτ

= W1 +W2 + ε0

âllspace

~E1 · ~E2 dτ.

1.6. CONDUCTORS 25

Calculate the work required to assemble a pair ofnested spherical shells where the inner shell has radiusa and a uniformly distributed surface charge −q, andthe outer sphere has radius b and uniformly distributedsurface charge q.

a

b

−q

+q

The field inside the inner sphere is ~E = 0 as it isinside any spherical shell. The field outside the outersphere is also ~E = 0 since the two charges look like apoint charge of q−q = 0. The field between the spheresdepends only on the inner sphere, by Gauss’s law, andit is

~E = − 1

4πε0

q

r2r.

So the energy store in the system is

W =ε02

âllspace

E2 dτ

=ε02

(q

4πε0

)2

4π

ˆ b

a

1

r4r2 dr

=q

8πε0

(1

a− 1

b

).

This is the energy stored in the system.

We can also calculate it in a different way. Thefield due to the inner shell is

~E1 = − 1

4πε0

q

r2r, r > a.

The field due to the outer shell is

~E2 =1

4πε0

q

r2r, r > b.

From a previous example, we know that the work toassemble these shells individually is

W1 =q2

8πε0

1

a

W2 =q2

8πε0

1

b.

We can’t just add these two to get the work required toassemble the entire system. We also have to calculatethe work for the interference term ~E1

~E2

~E1~E2 = −

(1

4πε0

)2q2

r4, r > b.

The work due to the interference term is

W3 = ε0

ˆ~E1~E2 dτ = − q2

4πε0

1

b.

Now the energy of the entire system is

W = W1 +W2 +W3

=q

8πε0

(1

a− 1

b

).

Example:

1.6 Conductors

In a conductor, the electrons are free to move around.We’ll assume perfect conductors—no resistivity.

• The field within a conductor is ~E = 0. If there was afield within the conductor, the charges would move,and it would no longer be an electrostatics problem.If we put a conductor in an electric field, the fieldcauses the charges to move in the conductor, theymove until the electric field they set up cancels theexternal electric field. In the real world, this hap-pens very quickly. In an idealized conductor, wetake it to be instantaneous.

• Since ~E = 0 in a conductor, the charge density is

ρ = ε0~∇ · ~E = 0.

• If the conductor has any net charge, it must there-fore reside on the surface.

• The conductor is an equipotential. That is, for twopoints ~a and ~b within a conductor, the potentialdifference is

V (~b)− V (~a) = −ˆ ~b

~a

~E· d~l = 0.

So any two points within a conductor have the samepotential, V (~b) = V (~a).

• The field immediately outside a conductor is per-pendicular to the surface. If there were a tangentialcomponent, the charges at the surface would rear-range until the tangential component was cancelledout.

• In general, the charge on a conductor is not uni-formly distributed. A spherical conductor is a spe-cial case in which the charges on the surface areuniformly distributed.


Energetics can explain why charges in a conductorreside on the surface. For charges on the surface of asphere, the energy is

E =1

8πε0

q2

R.

If the charge is distributed uniformly throughout thesphere, the energy is

E =3

20πε0

q2

R.

Since 320 >

18 , it is energetically favorable for the charge

to be distributed over the surface.

Bringing another charge near a conductor, as il-lustrated in the conducting sphere below, will inducea charge rearrangement in the conductor. In fact, thecharges in the conductor will rearrange until the elec-tric field in the conductor is zero. If a positive chargeis brought near a conductor, the charges in the conduc-tor will rearrange so that the side closest to the outsidecharge is more negative. Since the negative charges in theconductor are closer to the outside charge than the posi-tive charge, this results in a net attractive force betweenthe conductor and the outside charge.

−−

−+

+

+

+q

Now let us consider a conductor like a large chunkof metal. Suppose there is a cavity within within thisconductor and this cavity does not contain any electriccharges. We know that the electric field within the con-ducting material is zero, but what about the field withinthe cavity? For any arbitrary closed loop, we know that¸~E · d~l = 0. Suppose we did a line integral over a closed

loop such that part of the loop goes through the conduct-ing material and part of it goes through the cavity. Since~E = 0 in the conducting material, that part contributesnothing. However, if there is an electric field within thecavity, we would get a positive value. Since we must getzero, we can conclude that there is no electric field withinthe cavity.

There is no electric field in a cavity surrounded by aconducting material, regardless of the electric fields thatmay or may not be on the outside. This means we canshield stuff like sensitive electroncs against electric fieldssimply by wrapping them in a conducting material. Thisis the principle behind Faraday cages.

If we have a conductor with a cavity and that cav-ity contains an electric charge then the field within the

cavity will no longer be zero. Gauss’s law can give us theelectric field within the cavity. However, the conductingmaterial surrounding the cavity cannot have an electricfield, so what happens to the field lines within the cavity?If there is a +q charge inside the cavity, this will inducea charge of qinduced = −q on the surface of the cavity. Sothe positive field lines emanating from the positive chargein the cavity end at negative charges on the surface of thecavity. But since this negative charge moved to the cavitysurface from within the conductor, there is now a positivecharge imbalance in the conductor. This positive charge+q moves to the surface of the conductor.

More specifically, consider an uncharged sphericalconductor like the one shown below containing a cavitywith a point charge +q. What is the electric field outsideof the conductor? The shape of the cavity, the locationof the point charge within the cavity, and the locationof the cavity within the conductor are all irrelevant. Inorder for the field in the conductor to be zero, the chargein the cavity induces a charge of −q on the surface of thecavity. The charge imbalance of +q must reside on thesurface of the conductor. Since the surface of the con-ductor is spherical, and we know that the electric field atthe surface of a conductor is perpendicular to the surface,this implies that the charge +q on the surface is uniformlydistributed. Thus, for an outside observer, there is only asphere with a uniformly distributed surface charge. Thisproduces a field of

~E =1

4πε0

q

r2r,

which is just the field of a point charge +q.

In this case, we had three different charge distribu-tions and three different fields due to those charge distri-butions: ~Eq of the charge in the cavity, ~Einduced of the

induced charge on the surface of the cavity, and ~Erest ofthe charge on the surface of the conductor. By superposi-tion, all of these must have added up to produce the fieldof a point charge given above.

+q +

++

++

+++++

++

+

++

++

+ + + ++

+++

+–––––

–– – – –

Recall the boundary condition


ε0n.

1.7. CAPACITORS 27

Since ~E = 0 inside a conductor, this implies that theelectric field outside of a conducting surface is

~E =σ

ε0n,

where σ is the surface charge density. We can write thisin terms of the potential as

σ = −ε0∂V

∂n.

With this pair of equations, we can calculate the surfacecharge density σ given either the field ~E or the potentialV .

A surface charge density in the presence of an electricfield experiences a force per unit area of

~F = σ ~E.

But the electric field is discontinuous as it crosses acharged surface, so which field should we use? Shouldwe use the field below the sheet or the field above thesheet. We need to use the average of the two, so

~F =1

2σ(~Eabove + ~Ebelow

).

Inside of a conductor, we have ~E = 0. Outside, we have~E = σ

ε0n. Plugging these in, gives the force acting on the

conductor per unit area as

~F =σ2

2ε0n.

This force results in an electrostatic pressure on thesurface of the conductor that pushes it toward the fieldoutside. In terms of the field E just outside the surface,this pressure is

P =ε0E

2

2.

Suppose you had a charged spherical conducting shell. Allof the charge will be on the surface and it will be exertingan outward pressure regardless of the sign of the charge.You could think of it as all the charges on the surface are

trying to get away from each other, and this results in anoutward pressure like there would be if there was a gasinside the shell. If your spherical shell was flexible andstretchable (think of a balloon painted with a conduct-ing surface), then this electric pressure would cause theballoon to expand.

1.7 Capacitors

To calculate the potential difference between two con-ductors, we can choose any points in the conductors sinceconductors are equipotentials.

Consider two arbitrary conducting solids, one withcharge +Q and the other with charge −Q. Then thevoltage difference between the two is

V = V+ − V− = −ˆ +

−~E· d~l.

The electric field ~E can in principle be obtained byCoulomb’s law, but it is virtually impossible for arbitraryshapes. However, we know by Coulomb’s law that thiswill be proportional to Q, so

V = −ˆ +

−~E· d~l =

Q

C,

where 1C is the proportionality constant, and C is the ca-

pacitance. The unit of capacitance is the farad (F), andit is defined as

1 farad = 1 Coulomb/Volt.

Capacitance is always a positive quantity.

Capacitance is the ability for an object to storecharge. As a measure, it gives the charge per volt thatthe object holds. An object with a higher capacitancecan hold more charge at a given voltage.

Keep in mind that capacitance is a geometric quan-tity. It is determined entirely by the sizes, shapes, andseparations of the conductors involved.


What is the capacitance of a parallel plates capac-itor where the conducting plates each have area A, andthey are held apart a distance d?

If one plate is given a charge +Q and the otherplate is given the charge −Q then both charges willspread evenly over the surface, and each plate will havea surface charge density of

σ =A

Q.

Outside the plates, the field is ~E = 0. Between theplates, we know that the field is a constant

~E =σ

ε0=

Q

Aε0,

pointing from the positively charged plate to the neg-atively charged one. The values come from the factthat the electric field due to an infinite plane of uni-form charge is E = σ

2ε0and the assumption that the

area of the plates used here is much larger than the

distance d between them. In that case, we can approxi-mate the field due to the parallel plates by treating theplates as infinite planes. Then the field on the outsideis 0 because the plates have opposing fields there, andso they cancel out. Between the plates, the fields due toboth plates point in the direction of the negative plate.So the strength of the field between the plates is twicethe strength of the field due to a single infinite plane ofcharge.

Then the potential difference between the plates is

V =Qd

Aε0.

From this, we can conclude that the capacitance of theparallel plates capacitor is

C =Aε0d.

If we added a dielectric, we could enhance its capaci-tance.

Example:

Consider a spherical capacitor with an outer shellof radius b and charge −Q and an inner shell with radiusa and charge +Q.

Inside and outside, we know that the electric fieldis zero. Between the shells, the electric field can readilybe calculated by Gauss’s law as the field caused by apoint charge +Q at the center of the inner shell.

~E =1

4πε0

Q

r2r.

The potential difference between the inner andouter shells is

V = −ˆ +

−~E· d~l.

To get a positive potential difference, we always inte-grate from the negatively charged surface to the posi-tively charged surface.

V = − Q

4πε0

ˆ a

b

1

r2r· d~l

= − Q

4πε0

ˆ a

b

1

r2dr

=Q

4πε0

b− aab

.

So the capacitance is

C =Q

V= 4πε0

ab

b− a> 0.

Example:

What is the work required to charge a capacitor? Atsome intermediate stage, let the current charge be q, thenV = q

C . To add additional charge dq, recall that the workneeded to bring a charge in from infinity is equal to thecharge already there times the potential difference, so

dW =q

Cdq.

Integrating this from q = 0 to q = Q, gives us the totalwork to charge a capacitor

W =1

2

Q2

C=

1

2CV 2.

This is also the potential energy stored by a capacitor.

In a capacitor, the +Q is canceled out by the −Q, sofar away, the capacitor appears neutral and there is noelectric field.

Tip:

1.7. CAPACITORS 29

Calculate the capacitance of a pair of coaxialcylinders of length L. The inner cylinder has radius aand charge +Q, and the outer shell has radius b andcharge −Q.

We only care about the field between the cylin-ders, which is

~E =Q

2πε0

1

sLs.

Then the potential difference between the cylinders is

V = −ˆ a

b

~E· d~l

=Q

2πε0

1

L

ˆ a

b

1

sds

=Q

2πε0

1

Lln

(b

a

).

So the capacitance is

C =Q

V=

2πε0L

ln(ba

) .The length L of the cylinders is arbitrary, so we

should look at CL , the capacitance per unit length.

C

L=

2πε0

ln(ba

) .

Example:


1.8 Summary: Electrostatics

Electric Force

The force exerted by a charge q on a test charge Q is

~F =1

4πε0

qQ

u2u.

Remember that ~u = ~r − ~r ′, where ~u is the vector fromq to the observer at Q, ~r is the vector from the origin toQ, and ~r ′ is the vector from the origin to q. If we have ncharges qi acting on our test charge Q, then the net forceis given by the sum

~F =1

4πε0

n∑i=1

qiQ

u2i

ui.

Electric Field

Since ~E is an irrotational field, all of the following areequivalent:

1. ~∇× ~E = 0 everywhere2. The line integral

ˆ b

a

~E· d~,

is independent of the path for given endpoints.3. The line integral of the field about a closed loop is

zero ˛~E·d~l = 0.

4. The vector field ~E is the gradient of a scalar poten-tial function

~E = −~∇V.

To check if a given field is a valid electric field, it is suffi-cient to check that ~∇× ~E = 0.

The electric field at a point P due to n discretecharges qi is

~E =1

4πε0

n∑i=1

qiu2i

ui.

For a continuous charge distribution we get

~E(~r) =1

4πε0

ˆ1

u2u dq.

For a volume charge distribution with a volume chargedensity of ρ, the charge differential is dq = ρ dτ , andyou integrate over the volume. When computing ~E for agiven charge distribution, always try to simplify the prob-lem using symmetry. After you get a solution, check thelimit of your solution as various values approach ∞ and0. Are the limits what you should expect to see?

The electric force experienced by a charge q in thepresence of an electric field ~E is

~F = q ~E.

The flux of ~E through a surface S is the surfaceintegral

ΦE =

ˆS

~E · d~a,

For an arbitrary surface S, we get the integral formof Gauss’s law

˛S

~E · d~a =Qencε0

.

Gauss’s law is useful when our problem has planar, spher-ical, or cylindrical symmetry. We can calculate the en-closed charge by integrating the charge density over thevolume as

Qenc =

ˆV

ρ(~r) dτ,

although it is usually easier to use the symmetry of theproblem to calculate Qenc.

The electric field can easily be calculated as the gra-dient of the potential function

~E = −~∇V.

Given the electric field ~E, you can calculate the po-tential and the charge density using

V = −ˆ~E · d~l

~∇ · ~E =ρ

ε0.

Charge Density

The differential form of Gauss’s law allows us to cal-culate the charge density by taking the divergence of theelectric field

~∇ · ~E =ρ(~r)

ε0.

Given the charge density ρ, you can calculate thepotential and the field using

V =1

4πε0

ˆρ

udτ

~E =1

4πε0

ˆρ

u2u dτ.

1.8. SUMMARY: ELECTROSTATICS 31

Electric Potential

The potential difference between two points ~a and ~b isthe negative of the line integral of ~E from ~a to ~b

V (~b)− V (~a) = −ˆ b

a

~E· d~l.

In general, the potential at a point ~r is calculated by theline integral

V (~r) = −ˆ ~r

O

~E· d~l,

where O is a reference point. Usually, O is taken to be∞. For radially symmetric cases, ~r = r, which makes theintegral a lot easier to evaluate. Note that if r is insidea sphere for example, then there is a discontinuity in theelectric field between ∞ and r. Specifically, there is adiscontinuity at the surface of the sphere. In that case,you have to break the integral into two parts. One goesfrom infinity to the surface of the sphere, and the secondgoes from the surface of the sphere to r, with a different~E for each region.

Given ~E = E(x, y, z), calculate V (x, y, z) by lettingthe origin be the reference point and then breaking theline integral into one along each coordinate direction. If~E = E(r), calculate V (r) by letting ∞ be the referencepoint, and do a line integral coming in from infinity.

If you’re asked to calculate the potential everywhere,you may have to calculate V separately for different re-gions. For example, to calculate the potential everywheredue to a uniformly charged shell, you have to compute Vfor two different regions since V is constant inside theshell, but not outside. In both cases, though, you haveto compute the line integral coming in from infinity, so ifthat line passes through a charged surface, you have tobreak the line integral into multiple integrals, since ~E isdifferent in the two regions.

After finding V given ~E, check your results in theend to confirm that ~E = −~∇V .

If we have n discrete charges qi, we can use the prin-ciple of superposition to add their contributions to thepotential

V (~r) =1

4πε0

n∑i=1

qiui.

For the continuous volume case, we get

V (~r) =1

4πε0

ˆ1

udq.

For a volume charge distribution with a volume chargedensity of ρ, the charge differential is dq = ρ dτ , and youintegrate over the volume.

The potential V has to be continuous as it crosses asurface, unlike the electric field.

Given the potential V , you can calculate the chargedensity and the field using

∇2V = − ρ

ε0~E = −~∇V.

Work and Energy

The work to move a charge Q from a to b against the field~E is

W = −Qˆ b

a

~E· d~l = Q[V (~b)− V (~a)

].

If a is at infinity, that is, you’re bringing your charge Qin from infinity, then W = QV (~b).

The work to bring n charges in from infinity andplace them at ~ri is

W =1

2

n∑i=1

qiV (~ri).

Note that V (~ri) is the potential at the position of the ithparticle due to all the other particles.

The work required to construct a continuous chargedistribution is

W =1

2

ˆρV dτ,

where the integral is over the charge distribution. Pro-vided that the charges are not discrete, we can also usethe more useful

W =ε02

âllspace

E2 dτ.

Conductors

In a conductor, the electrons are free to move around.

• The field within a conductor is ~E = 0.• The charge density within a conductor is ρ = ε0~∇ ·~E = 0.

• Any net charge resides at the surface.• The conductor is an equipotential. That is, for two

points ~a and ~b within a conductor, the potentialdifference is

V (~b)− V (~a) = −ˆ ~b

~a

~E· d~l = 0.

• The field immediately outside a conductor is per-pendicular to the surface and is

~E =σ

ε0n.


The electric field below a surface and the electric fieldabove a surface are related to the charge distribution onthe surface as


ε0n.

Here, n is defined to point from below to above. Theforce acting on the conductor per unit area due to thesurrounding electric field is

~F =σ2

2ε0n.

This force results in an electrostatic pressure on the sur-face of the conductor of

P =ε0E

2

2,

that pushes it toward the field outside. Here E is themagnitude of the field outside.

Look for the “metal” keyword, which implies thatyou’re dealing with a conductor, and all charge resides onthe surface.

Capacitors

A capacitor consists of two conducting solids, one withcharge +Q and the other with charge −Q. In general,the capacitance is calculated as

C =Q

V,

where Q is the magnitude of the charge on one of theconductors, and V is the potential difference between thetwo conductors.

The potential difference between the two conductorsis calculated as

V = −ˆ +

−~E· d~l.

To get a positive potential difference, we always inte-grate from the negatively charged surface to the positivelycharged surface.

The work required to charge a capacitor with a finalpotential difference V from q = 0 to q = Q is

W =1

2CV 2 =

1

2

Q2

C.

This is also the energy stored by a capacitor.

For a parallel plate capacitor with plates of area Aseparated by a distance d, we have

C =ε0A

d, V = Ed, E =

Q

ε0A.

Symmetric Objects

In addition to the electric field and electric potential fordiscrete charges and continuous distributions, there are afew symmetric cases that should be remembered. Theyare often useful to compare with the limiting cases of morecomplicated distributions. In each of the following cases,the charge is uniformly distributed. The work W whenit is given is the work required to assemble the object bybringing point charges in from infinity.

1. Infinite line:

~E =1

4πε0

2λ

zz

2. Infinite plane:

~E =σ

2ε0n

3. Spherical Shell with radius R and charge q

~E = 0, r < R

=1

4πε0

q

r2r, r > R

V (r) =1

4πε0

q

R, r < R

=1

4πε0

q

r, r > R

W =1

4πε0

q2

2R.

4. Solid Sphere with radius R and total charge q:

~E =1

4πε0

qr

R3r, r < R

=1

4πε0

q

r2r, r > R

W =1

4πε0

3q2

5Rr, r > R.

Miscellaneous

The divergence theorem states that

ˆV

~∇ · ~E dτ =

˛S

~E · d~a,

where S is the surface of the volume V.

If you’re dealing with a radial charge distribution,then the following may be helpful. They are the radialterms of the gradient, divergence, and Laplacian in spher-ical coordinates. If there is any θ or φ dependence, thenthese are not sufficient.

~∇V =∂V

∂rr

~∇ · ~E =1

r2

∂

∂r

(r2Er

)∇2V =

1

r2

∂

∂r

(r2 ∂V

∂r

).

1.8. SUMMARY: ELECTROSTATICS 33

If f is a scalar function and ~F is a vector field, then

~∇ ·(f ~F)

= f ~∇ · ~F + ~F ·(~∇f).

The divergence of rr2 is

~∇ · rr2

= 4π δ(~r),

The Taylor expansion of f(x) about x = x0 is

f(x) =

∞∑n=0

f (n)(x0)

n!(x− x0)n.

This might be useful when finding the limit of a function.

Remember the following integration shortcuts:

ˆda = area

ˆ 2π

0

dφ

ˆ π

0

sin θ dθ = 4π.

The last one is the solid angle of a sphere.

If you have two adjacent sides a and b and the in-cluded angle C of any triangle, then the law of cosinesgives

c2 = a2 + b2 − 2ab cosC.

Chapter 2

Special Techniques

2.1 Laplace’s Equation

To find the electric field ~E, we can use

~E(~r) =1

4πε0

ˆρ(~r)

u

u2dτ.

However, this integral is too difficult to solve except in afew special cases since u keeps changing unless there issymmetry to exploit.

In general, it is much easier to calculate the potentialV using the integral

V (~r) =1

4πε0

ˆρ(~r)

udτ.

This integral is generally much easier to evaluate since itis a scalar integral rather than a vector integral.

In general, to compute V , we can also use Poisson’sequation

∇2V = − 1

ε0ρ,

but it can be difficult since this is a partial differentialequation. If there is no charge in the volume, Poisson’sequation reduces to the much simpler Laplace’s equa-tion

∇2V = 0.

Solutions of Laplace’s equation are called harmonicfunctions. Most of this section focuses on solvingLaplace’s equation.

In 1D, Laplace’s equation is

d2V

dx2= 0,

and its solution is the line

V (x) = mx+ b.

Notice that if your boundaries are x = −a and x = a,then the solution can be written as the average valueV (x) = 1

2 (V (x+ a) + V (x− a)). Second, a maximum or

minimum of V can only occur at one of the two bound-aries (i.e. endpoints). A local extreme cannot occur inthe middle of the region. To determine the exact solu-tion, we need two boundary conditions to determine theunknowns m and b.

In 2D, Laplace’s equation is

∂2V

∂x2+∂2V

∂y2= 0.

There’s no general form of the solution for this PDE. Thesolutions are called harmonic functions. Consider thesolution at a point (x, y)

V (x, y) =1

2πR

˛circle

V dl.

We have to do the line integral over a circle. That is,the value at the center of the circle can be found by av-eraging the values along the circle. This is a property ofharmonic functions, and it tells us that we can tell what’sgoing on in the inside by knowing only what’s happeningon the boundary. Second, V (x, y) has no local maximumor minimum inside the circle. If it has extrema, they mustbe on the boundary of the circle. We can also use othershapes, but the solution simplifies only for a circle.

In 3D, Laplace’s equation becomes

∂2V

∂x2+∂2V

∂y2+∂2V

∂z2= 0.

The value of V at any point ~r is now the average of Vover a spherical surface of radius R centered at ~r:

V (~r) =1

4πR2

˛sphere

V da.

Again, by the properties of harmonic functions, V hasno local extrema inside the sphere. They can only occuron the boundary, that is, on the surface of the enclosingsphere.

For example, let’s calculate the average potentialover a spherical surface of radius R due to a single pointcharge q located outside the spherical surface.

34

2.1. LAPLACE’S EQUATION 35

y

z

x

q

θ R

u

In general,

V =1

4πε0

q

u.

From the law of cosines, we have

u2 = z2 +R2 − 2Rz cos θ.

Our source is at

r′ = 〈0, 0, z〉.

Our field point is at

r = 〈R sin θ cosφ,R sin θ sinφ,R cos θ〉.

Then we can get u from ~u = ~r − ~r ′. Putting everythingtogether gives us

Vavg =1

4πR2

q

4πε0

ˆR2 sin θ√

z2 +R2 − 2zR cos θdθ dφ

=q

8πε0zR

[√(z +R)2 −

√(z −R)2

]=

q

4πε0z.

This is the same as the potential at (0, 0, 0) due to thepoint charge. If it works for a point charge, then by theprinciple of superposition, it will work for any charge dis-tribution.

First Uniqueness Theorem

Recall that in 1D, the solution to Laplace’s equation hasthe form V = mx+ b. Specifying V at two points is suf-ficient to find the specific solution. We could also specifyV and dV

dx at one point or specify V at one point and dVdx

at another point.

Once we have a solution, how do we know it’s unique?

The first uniqueness theorem tells us that the so-lution to Laplace’s equation in some volume V is uniquelydetermined if the potential V is specified on the boundarysurface S of the volume V.

We can prove uniqueness theorems by contradiction.In this case, assume there are two different solutions V1

and V2 to Laplace’s equation in some volume contain-ing zero charge. Then ∇2V1 = 0 and ∇2V2 = 0, whereV1 = V2 on the surface of the volume. Then we defineVdiff ≡ V1 − V2, which must also satisfy Laplace’s equa-tion ∇2Vdiff = 0. Since V1 = V2 on the boundary of thesurface, Vdiff = 0 on the surface. If Vdiff = 0 on thesurface, then Vdiff = 0 everywhere. This implies thatV1 = V2 everywhere (not just on the surface), and so thesolution is unique. In other words, if you find any solutionthat works, then it must be the unique solution.

A useful corollary is: The potential in a volume V isuniquely determined if the charge density ρ throughoutthe volume and the potential V on the boundaries arespecified.

We again use proof by contradiction. If there are twosolutions to Poisson’s equation in this region ∇2V1 = −ρεand ∇2V2 = −ρε , we can define Vdiff = V1 − V2, whichmust also satisfy Poisson’s equation, ∇2Vdiff = −ρε . Butsince Vdiff = 0 on the boundaries, it must be zero every-where, and so V1 = V2.

This first uniqueness theorem is for the general case.Since conductors act differently, we have to construct asecond uniqueness theorem specifically for conductors.

Second Uniqueness Theorem

In a volume V containing conductors and a specifiedcharge density ρ, the electric field is uniquely determinedif the total charge on each conductor is specified. To pic-ture this, think of some volume containing a variety ofconducting objects as well as empty space. This unique-ness theorem requires that we know the charge densityof the overall volume as well as the total charge on eachindividual conductor.

This means if we find a solution for the electric fieldin such a setup, then it must be the solution.

To prove this, we assume that there are two fieldssatisfying the conditions of the problem. That is, we as-sume that the solution is not unique. Then ~∇ · ~E1 = ρ

ε ,

and ~∇ · ~E2 = ρε . For each conductor in this volume, we

have

˛~E1 · d ~Ai =

Qiε0˛

~E2 · d ~Ai =Qiε0.

Here, both integrals are over the surface of the ith con-ductor, so for each conductor in our volume, we have apair of these integrals.

36 CHAPTER 2. SPECIAL TECHNIQUES

On the outer boundary of the volume, we have˛~E1 · d ~A =

Qtotalε0˛

~E2 · d ~A =Qtotalε0

,

where the surface integrals are over the outer boundaryof the volume.

Imagine a third field ~E3 = ~E1 − ~E2. This impliesthat ~∇ · ~E3 = 0 and

¸~E3 · d ~A = 0 over each boundary

surface.

Each of the conductors is an equipotential, so V3 is aconstant over each boundary surface. Next, we considerthe vector product rule

~∇ ·(V3~E3

)= V3

~∇ · ~E3 + E3 · ~∇V3.

We know that ~∇· ~E3 = 0, and we know that ~E3 = −~∇V3,so this reduces to

~∇ ·(V3~E3

)= −

(~E3

)2

.

ThenˆV

~∇ ·(V3~E3

)dτ =

˛S

~E3V3 · d ~A = −ˆV

E23dτ.

We know that¸~E3 · d ~A = 0, so the middle term is zero

implying ˆV

E23dτ = 0.

We know that E23 ≥ 0, so ~E3 = 0, and this implies that

~E1 = ~E2 and our solution is unique.

2.2 Method of Images

The method of images exploits the first uniqueness the-orem detailed in the previous section. Suppose we havesome complicated setup in which calculating V directlyis difficult. However, if we construct a simple artificialsetup that satisfies Laplace’s equation and has the sameboundary conditions as the complicated setup, then bythe first uniqueness theorem, the solution to the artifi-cial setup is the same as the solution to the complicatedsetup.

Example: Plane Conductor

Consider a point charge q a distance d above a groundedconducting plate. What is the potential in the regionabove the plane? Note that the point charge induces acharge on the plane, which also contributes to the poten-tial V in the region of interest.

Let the conducting plane form the xy-plane of a 3Dcartesian coordinate system, and charge q be at z = d.We need to solve Poisson’s equation in the region z > 0subject to the boundary conditions

1. V = 0 when z = 0. That is, the conducting plate isgrounded, so it is at V = 0.

2. V → 0 when x2 + y2 + z2 >> d2. That is, the po-tential approaches zero when you go far away fromthe charge.

If we can compute V for an artificial setup that satis-fies Laplace’s equation and has the same boundary con-ditions, then by the uniqueness theorems, this must bethe same solution V as the one for the problem we aretrying to solve.

Consider now an artificial setup with two charges.We have a charge q at (0, 0, d) and a charge −q at(0, 0,−d) and no conducting plane. The potential ev-erywhere due to this pair of charges is

V (x, y, z) =1

4πε0

q√x2 + y2 + (z − d)2

+1

4πε0

−q√x2 + y2 + (z + d)2

,

where the first term is the potential due to q and the sec-ond term is the potential due to −q. Notice that at z = 0,the two cancel and V = 0, satisfying our first boundarycondition. Second, if x2 + y2 + z2 >> d2, then V → 0, soour second boundary condition is satisfied. Therefore, bythe uniqueness theorems, the solution V (x, y, z) for thisartificial setup is also the solution to the original problem.

Keep in mind that this solution is only valid for theregion z ≥ 0. We don’t know what the potential is in theregion z < 0.

So the effect of this conducting plate with q at z = dis the same as if there was a charge −q at z = −d insteadof a conducting plate. That is, q induces a charge in theconducting plate at z = 0 that looks like a charge −q atz = −d.

Since we know the potential above the conductingplate, we can compute the surface charge induced on theplate by the charge q above the plate by using

σ = −ε0∂V

∂n.

In this case, our surface is at z = 0, and the normal di-rection is z, so n = z, and we have

σ = −ε0∂V

∂z

∣∣∣z=0

.

That is, the induced surface charge is calculated by firsttaking the partial derivative of V with respect to the di-rection that is normal to the conducting surface and then

2.2. METHOD OF IMAGES 37

evaluating the result at the conducting surface. We findthat

σ(x, y) = −ε0∂V

∂z

∣∣∣z=0

= −ε01

4πε0

−q(z − d)

[x2 + y2 + (z − d)2]32

∣∣∣z=0

−ε01

4πε0

q(z + d)

[x2 + y2 + (z + d)2]32

∣∣∣z=0

.

Evaluating this at z = 0, we find that the surface chargedistribution on the conducting plate that we induced withour charge q is

σ(x, y) = − q

2π

d

(x2 + y2 + d2)32

.

Notice that the induced charge has opposite sign and peakvalue at d = 0 when the point charge is directly above it.This is the expected result.

The total charge that is induced in the conduct-ing plate by our charge q can be found by integratingσ over the entire plane using polar coordinates. We letr2 = x2 + y2 then da = r dr dφ, and

Q =

ˆσ da

=

ˆ−qd

2π(r2 + d2)32

da

= − qd2π

ˆ 2π

0

dφ

ˆ ∞0

r

(r2 + d2)32

dr

=qd√r2 + d2

∣∣∣∞0

= −q.

As expected, the induced charge is equal in magnitudeand opposite in sign to the point charge. So we havea charge q that induces a charge −q in the conductingplate. The charge −q corresponds to the image charge−q at z = −d.

In general, the total charge induced in the conductingsurface should be the negative of your charge doing theinducing.

Tip:

The force between the charge q at z = d and theimage charge −q at z = −d is

~F = − 1

4πε0

q2

(2d)2z.

This is the same as the force between the charge q andthe conducting plane. It’s easier to use the image charge,hence why we use them. We could also compute the forceusing

dF =1

4πε0

ˆσ(x, y)q da,

where da = dx dy.

The work required to set up the charge q and theimage charge −q is

W = − 1

4πε0

q2

2d.

This is not correct for the charge q and the conductingplane. The image method doesn’t work to compute thework because of the Faraday cage effect. The true fieldon the other side of the conducting plane is zero. Theactual value (i.e. for q and the conducting plane) is halfof the above or

W = − 1

4πε0

q2

4d.

We can verify this by directly calculating the work re-quired to bring the charge q in from infinity to a distanced from the infinite conducting plane. Recall that the forcerequired to move a particle against a field is the oppositeof the force that a field exerts on a particle.

W =

ˆ d

∞~F · d~

=1

4πε0

q2

4

ˆ d

∞

1

z2dz

= − 1

4πε0

q2

4d.

Example: Spherical Conductor

Suppose you have a point charge q located a distance afrom the center of a grounded conducting sphere of radiusR. What is the potential everywhere outside the sphere?

qR

q′

P

We need to solve Poisson’s equation with the bound-ary conditions V (R) = 0, and V → 0 as r →∞. We wantto know V at an arbitrary point P outside the sphere.

q′

P

q

u′u

~r

bθ


We place an image charge q′ inside the sphere a dis-tance b from the center. We then ignore the sphere, andour potential at the point P outside the sphere is then

V =1

4πε0

(q

u+q′

u′

).

The law of cosines gives us u2 = a2 + r2 − 2ar cos θ and(u′)2 = b2 + r2 − 2br cos θ. Then

V (r, θ) =1

4πε0

q√a2 + r2 − 2ar cos θ

+1

4πε0

q′√b2 + r2 − 2br cos θ

.

We know the potential on the surface of the sphere mustvanish, so

V (R, θ) =1

4πε0

q√a2 +R2 − 2aR cos θ

+1

4πε0

q′√b2 +R2 − 2bR cos θ

= 0.

This is true for all θ, so to get a pair of equations to solvefor the two unknowns, we choose two convenient angles,θ = 0 and θ = π. This gives us the pair of equations

V (R, 0) =1

4πε0

q√a2 +R2 − 2aR

+1

4πε0

q′√b2 +R2 − 2bR

= 0

V (R, π) =1

4πε0

q√a2 +R2 + 2aR

+1

4πε0

q′√b2 +R2 + 2bR

= 0.

This gives us the unknown values

q′ = −Raq

b =R2

a.

We can calculate the induced surface charge densityusing

σ(θ, φ) = −ε0∂V

∂n.

When we do so, we get

σ(θ, φ) =− qR

4πR2a

(1− R2

a2

)[1 + R2

a2 −2Ra cos θ

] 32

.

Integrating this over the surface of the sphere should giveus the total induced charge q′ = −Ra q.

Note that as we bring the test charge closer to thesurface of the sphere, the image charge also gets closerand in the limit that a = R, we get q = q′.

We can calculate the force between the charge andthe sphere as

F =1

4πε0

q(−Ra q

)(a− b)2

= − q2Ra

4πε0(a2 −R2)2.

Image charges are never placed in the region of spacewhere we seek the solution V . If you do that, youchange the problem.

Tip:

For some configurations, there is no way to choose asingle image charge that will solve the problem. Some-times, you need to use multiple image charges.

2.3 Separation of Variables

To perform separation of variables on Laplace’s equation

∇2V = 0,

follow this general procedure:

1. Study the problem statement. Draw a picture.Should you use cartesian or spherical coordinates?In what region(s) are you asked to find the poten-tial? Is the configuration independent of one ormore coordinates?

2. Identify and list the boundary conditions.3. Is it a 1D, 2D, or 3D problem? Identify the variables

of which the solution will be independent.4. Write down Laplace’s equation. Assume a separable

solution and separate variables, writing Laplace’sequation as a set of ODEs.

5. Solve the set of ODEs using the boundary condi-tions identified in step 2.

6. Write the general solution as the linear combinationof the products of the solutions to the ODEs.

7. Use the final boundary condition (i.e. the “initialcondition”) and orthogonality to write a formula toobtain the Fourier coefficients.

You need to know the orthogonality conditions of dif-ferent eigenfunctions by heart.

Tip:

2D Cartesian Coordinates

This is Example 3.3 in Griffiths.

Two semi-infinite grounded metal plates are lyingparallel to the xz-plane, one at y = 0, and the otherat y = a. The left end at x = 0 is closed off with an in-finite strip insulated from the two plates and maintained

2.3. SEPARATION OF VARIABLES 39

at a specific potential V (y). Find the potential inside theslot.

Since this problem is independent of z, we solveLaplace’s equation in two dimensions

∂2V

∂x2+∂2V

∂y2= 0,

subject to the following boundary conditions

1. V = 0 when y = 02. V = 0 when y = a3. V = V0(y) when x = 04. V → 0 as x→∞

We begin by assuming the solution is a separableproduct

V (x, y) = X(x)Y (y).

Taking the derivatives and plugging into the 2D Laplaceequation gives us

Yd2X

dx2+X

d2Y

dy2= 0.

Dividing through by XY gives us

1

X

d2X

dx2= − 1

Y

d2Y

dy2.

The left hand side is independent of y, and the right handside is independent of x. This implies that both sides areequal to a constant. If we let this constant by C1, thenwe have the pair of ODEs

1

X

d2X

dx2= C1

− 1

Y

d2Y

dy2= C1.

Rearranging a little gives us

d2X

dx2= C1X

d2Y

dy2= −C1Y.

For the X equation, we try the solution X = eαx.Then,

d2X

dx2= α2eαx.

This implies that α2eαx = C1eαx, which gives us α =

±√C1. Let’s define

√C1 ≡ k. Then the general solution

to this ODE is

X(x) = Aekx +Be−kx.

For the Y equation, we try the solution Y = eβy.Then,

d2Y

dy2= β2eβy.

This implies that β2eβy = −C1eβy, which gives us β2 =

−√C1, or β = ±i

√C1 = ±ik. Then the general solution

to this ODE is

Y (y) = C ′eiky +D′e−iky = C sin(ky) +D cos(ky).

Note: We could also have chosen +C1 for the Y equa-tion, then we would end up with the solutions X(x) =A sin(kx) + B cos(kx) and Y (y) = Ceky + De−ky. How-ever, the fourth boundary condition implies that we can-not use sin(kx) and cos(kx). Hence, we go with the ex-ponentials for X(x).

Now we look at the boundary conditions. The fourthboundary conditions says we need V → 0 as x→∞. Thisimplies that A = 0, so our X ODE reduces to

X(x) = Be−kx.

Our solution is now

V (x, y) = Be−kx [C sin(ky) +D cos(ky)] .

The first boundary condition tells us that V = 0 wheny = 0. Plugging this in, shows that this can only be trueif D = 0. We now have the solution

V (x, y) = Ce−kx sin(ky),

where C is a new unknown constant. The second bound-ary condition tells us that V = 0 when y = a. Pluggingthis in gives us

V (x, a) = Ce−ka sin(ka) = 0.

This implies that sin(ka) = 0, which implies

k =nπ

a, n = 1, 2, 3, . . .

We can’t go further until V0(y) is specified. The mostgeneral form of the solution is

V (x, y) =

∞∑n=1

Cne−nπxa sin

(nπya

),

and that

V (0, y) =

∞∑n=1

Cn sin(nπy

a

)= V0(y).

This last one is just a Fourier series.

We know that the orthogonality condition for thefunctions sin

(nπya

)is

ˆ a

0

sin(nπy

a

)sin(mπy

a

)dy =

a

2δnm.


Multiplying both sides of the Fourier series bysin(mπya

)and integrating from y = 0 to y = a gives

us∞∑n=1

Cn

ˆ a

0

sin(nπy

a

)sin(mπy

a

)dy

=

ˆ a

0

V0(y) sin(mπy

a

)dy.

But, by the orthogonality condition,

∞∑n=1

Cna

2δnm =

ˆ a

0

V0(y) sin(mπy

a

)dy.

From the series, we now only get a contribution whenn = m. All other terms are zero. We are left with

Cma

2=

ˆ a

0

V0(y) sin(mπy

a

)dy.

Solving for Cm and then switching the m’s back to n’s(since they’re just an index), gives us

Cn =2

a

ˆ a

0

V0(y) sin(nπy

a

)dy.

Our general solution is, therefore,

V (x, y) =

∞∑n=1

Cne−nπxa sin

(nπya

),

where

Cn =2

a

ˆ a

0

V0(y) sin(nπy

a

)dy.

Suppose V0(y) = V0 is a constant. Then the coeffi-cients become

Cn =2V0

a

ˆ a

0

sin(nπy

a

)dy

=2V0

nπ[1− cos(nπ)]

=2V0

nπ[1− (−1)n]

=

4V0

nπ if n is odd

0 if n is even.

Only the odd terms contribute, so we can write our finalsolution as

V (x, y) =4V0

π

∞∑nodd

1

ne−

nπxa sin

(nπya

).

This series can be explicitly summed by writing thesummand as the imaginary part of an exponential func-tion and then using the formula for the sum of an infinitegeometric series. The final result is

V (x, y) =2V0

πtan−1

(sin(πya

)sinh

(πxa

)) .

It is easy to check that this solution satisfies Laplace’sequation and all of our boundary conditions.

Our method used two important properties of thesine function

• completeness• orthogonality

What we’re using is the theorem that if we have a com-plete set of orthogonal functions fn(x), then we can writeany function f(x) as

f(x) =

∞∑n=1

Cnfn(x).

Then, if the functions are orthogonal on the interval [0, a],

ˆ a

0

fn(x)fm(x) dx = 0 if n 6= m.

The functions sin(nπya

), for example, are complete

and orthogonal on the interval [0, a].


Find the potential inside a metal (i.e. conducting) cube ofside a if the sides and bottom are grounded, and the topis insulated from the other sides and held at a potentialV0.1

We start by setting up our Cartesian coordinate sys-tem such that one corner of the box is at the origin, andour boundary conditions are

1. V = 0 at x = 02. V = 0 at x = a3. V = 0 at y = 04. V = 0 at y = a5. V = 0 at z = 06. V = V0 at z = a

Since there are no charges in the box, the potentialinside the box satisfies the 3D Laplace equation

∂2V

∂x2+∂2V

∂y2+∂2V

∂z2= 0.

Assuming a separable solution of the form

V (x, y, z) = X(x) · Y (y) · Z(z),

the Laplace equation becomes

X ′′

X+Y ′′

Y+Z ′′

Z= 0.

1This is problem 3.15 in Griffiths


Each term of this equation must separately be equal to aconstant, so

C1 + C2 + C3 = 0,

where C1 is the constant associated with the x term, C2

is the constant associated with the y term, and C3 is theconstant associated with the z term.

Notice that in the x and y directions, the potentialmust be zero at the edges of the box. This implies thatC1 and C2 must be negative. If they were positive, the so-lution would involve exponentials, which cannot be madeto go to zero at both boundaries simultaneously. If theyare negative, the solution will involve sines and cosines.So we have that

C1 = −k2

C2 = −`2

C3 = k2 + `2,

where k and ` are positive.

So the Laplace equation becomes the set of ODEs

X ′′ = −k2X

Y ′′ = −`2YZ ′′ =

(k2 + `2

)Z.

The x equation has the solution

X(x) = A cos(kx) +B sin(kx).

From the first boundary condition, we get that A = 0.The second boundary condition implies ka = nπ for

n = 1, 2, 3, . . ., so

X(x) = B sin(nπx

a

), n = 1, 2, 3, . . .

The y equation has the solution

Y (y) = C ′ cos(`y) +D sin(`y).

From the third boundary condition, we get that C ′ = 0.The fourth boundary condition implies à = mπ form = 1, 2, 3, . . ., so

Y (y) = D sin(mπy

a

), m = 1, 2, 3, . . .

The z equation has the solution

Z(z) = Eerz + Fe−rz,

where

r =√k2 + `2 =

√(nπa

)2

+(mπa

)2

=π

a

√n2 +m2.

The fifth boundary condition gives us F = −E, so oursolution for the z equation is

Z(z) = Eerz − Ee−rz

= 2E sinh(πa

√n2 +m2 z

).

Combining the solutions, we have that

V (x, y, z) = Cnm sinh(πa

√n2 +m2 z

)sin(mπy

a

)sin(nπx

a

), n,m = 1, 2, 3, . . . .

The constants B, D, and 2E have been absorbed into the constant C which could depend on n and m. The generalsolution is then

V (x, y, z) =

∞∑n,m=1

Cnm sinh(πa

√n2 +m2 z

)sin(mπy

a

)sin(nπx

a

).

From the final boundary condition, we get

V (x, y, a) = V0 =

∞∑n,m=1

Cnm sinh(π√n2 +m2

)sin(mπy

a

)sin(nπx

a

).

Multiplying both sides by sin(m′πya

)sin(n′πxa

)and integrating from 0 to a with respect to x and y, we have

V0

ˆ a

0

ˆ a

0

sin

(m′πy

a

)sin

(n′πx

a

)dy dx =

∞∑n,m=1

Knm

ˆ a

0

ˆ a

0

sin(mπy

a

)sin(nπx

a

)sin

(m′πy

a

)sin

(n′πx

a

)dy dx.


where Knm = Cnm sinh(π√n2 +m2

)for brevity. By the orthogonality of sine functions, we know that RHS = 0

unless m′ = m and n′ = n, so we get

V0

ˆ a

0

ˆ a

0

sin(mπy

a

)sin(nπx

a

)dy dx = Knm

ˆ a

0

sin2(mπy

a

)dy

ˆ a

0

sin2(nπx

a

)dx

V0

ˆ a

0

sin(mπy

a

)dy

ˆ a

0

sin(nπx

a

)dx = Knm

a2

4

V0

[− a

mπcos(mπy

a

)] ∣∣∣a0

[− a

nπcos(nπx

a

)] ∣∣∣a0

= Cnm sinh(π√n2 +m2

) a2

4

V0a

mπ[1− (−1)m]

a

nπ[1− (−1)n] = Cnm sinh

(π√n2 +m2

) a2

44V0

nmπ2[1− (−1)m] [1− (−1)n] = Cnm sinh

(π√n2 +m2

)

Cnm =

0 if either or both n or m are even

16V0

nmπ2 sinh(π√n2+m2)

if n and m are both odd.

So the complete solution is

V (x, y, z) =16V0

π2

∞∑m=1odd

∞∑n=1odd

1

nm sinh(π√n2 +m2

) sinh(πa

√n2 +m2 z

)sin(mπy

a

)sin(nπx

a

).

At the center of the cube, the potential is

V(a

2,a

2,a

2

)=

16V0

π2

∞∑m=1odd

∞∑n=1odd

1

nm sinh(π√n2 +m2

) sinh(π

2

√n2 +m2

)sin(mπ

2

)sin(nπ

2

).

Using a trig identity, we can rewrite the sines as

sin(mπ

2

)sin(nπ

2

)= cos

(π2

[m− n])− cos

(mπ2

)cos(nπ

2

).

Keeping in mind that n and m are odd, we note that the second term on the right is always zero, and that m− n iseven. We can simplify our result a little

V(a

2,a

2,a

2

)=

16V0

π2

∞∑m=1odd

∞∑n=1odd

1

nm sinh(π√n2 +m2

) sinh(π

2

√n2 +m2

)cos(π

2[m− n]

).

3D Spherical Coordinates

In spherical coordinates, the Laplacian of the potential is

∇2V =1

r2

∂

∂r

(r2 ∂V

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂V

∂θ

)+

1

r2 sin2 θ

∂2V

∂φ2.

In most cases, the potential has no dependence on φ,

so the third term is generally zero, and we have

∇2V =1

r2

∂

∂r

(r2 ∂V

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂V

∂θ

).

We want to solve ∇2V = 0 in spherical coordinates,so setting the above equation equal to zero and multiply-ing both sides by r2 gives us

∂

∂r

(r2 ∂V

∂r

)+

1

sin θ

∂

∂θ

(sin θ

∂V

∂θ

)= 0.


Assuming a separable solution, we write

V (r, θ) = R(r) ·Θ(θ).

Plugging this in and simplifying gives us

1

R

d

dr

(r2 dR

dr

)+

1

Θ sin θ

d

dθ

(sin θ

dΘ

dθ

)= 0.

Notice that the first term is now a function only of r, andthe second term is a function only of θ. This implies thatboth are constant, and the fact that the pair adds to zeroimplies that one is the negative of the other.

Looking at the r equation, a convenient constant is`(`+ 1), so we have the ODE

d

dr

(r2 dR

dr

)= `(`+ 1)R.

To solve this ODE, we start by letting R = rn, then weget

d

dr

(r2nrn−1

)= `(`+ 1)rn

nd

dr

(rn+1

)= `(`+ 1)rn

n(n+ 1)rn = `(`+ 1)rn

n(n+ 1) = `(`+ 1).

There are two solutions to this, n = ` and n = −` − 1.These are the two powers of n that we can use. So oursolution for the r equation is

R`(r) = Ar` +B1

r`+1.

For the θ equation, we get the ODE

1

sin θ

d

dθ

(sin θ

dΘ

dθ

)= −`(`+ 1)Θ.

The solutions to this ODE are called Legendre poly-nomials when ` = 0, 1, 2, 3, . . .. There are also solutionswhen ` < 0 and for ` that are not integers, but they sufferfrom divergences, so there are not useful for us. The firstseveral Legendre polynomials are

P0(x) = 1

P1(x) = x

P2(x) =1

2(3x2 − 1)

P3(x) =1

2(5x3 − 3x)

P4(x) =1

8(35x2 − 30x2 + 3)

P5(x) =1

8(63x5 − 70x3 + 15x).

Note that when ` is even, then P`(x) is even, and if ` isodd, then P`(x) is odd. That is, the Legendre polynomi-als satisfy

P`(−x) = (−1)`P`(x).

We need to memorize at least P0, P1, and P2. Theycorrespond to a monopole, a dipole, and a quadrapole,

respectively. The Legendre polynomials can be generatedby the Rodrigues formula

P`(x) =1

2``!

(d

dr

)`(x2 − 1)`.

In our case, the natural variable is cos θ, so our Legendrepolynomials are given by P`(cos θ).

Our solutions are then the product of the r solutionsand the θ solutions

V`(r, θ) =

(Ar` +B

1

r`+1

)P`(cos θ).

Our general solution for an azimuthally symmetricsystem is then

V (r, θ) =

∞∑`=0

(A`r

` +B`1

r`+1

)P`(cos θ).

Remember this general form of the solution.

Example

Suppose we want to find the potential inside and out-side of a hollow sphere of radius R if the potential at thesurface is some function of theta, V0(θ).2

Inside the sphere, r → 0, which means 1r`+1 → ∞.

This means we have to require that B = 0 to preventour solution from blowing up at r = 0. So our generalsolution for the inside becomes

Vin(r, θ) =

∞∑`=0

A`r`P`(cos θ).

To compute the coefficients A`, we have to use the initialcondition V0(θ) = V (R, θ). This occurs when r = R, so

V (R, θ) =

∞∑`=0

A`R`P`(cos θ) = V0(θ).

Next, we use the orthogonality of the Lengendre poly-nomials since they are orthogonal and complete. Theorthogonality condition isˆ 1

−1

P`(x)P`′(x) dx =

ˆ π

0

P`(cos θ)P`′(cos θ) sin θ dθ

=2δ``′

2`+ 1.

So if we multiply both sides of V0 above by P`′(cos θ) sin θand integrate from 0 to π with respect to θ then all termsdrop out on the right except those for which ` = `′. Sowe get ˆ π

0

V0(θ)P`(cos θ) sin θ dθ = A`R` 2

2`+ 1.

This implies that the coefficients A are given by

A` =2`+ 1

2R`

ˆ π

0

V0(θ)P`(cos θ) sin θ dθ.

2This is Example 3.6 and 3.7 in Griffiths.


Find the potential inside of a sphere of radius R ifthe potential at the surface is

V (R, θ) = k cos(3θ).

We know that the general solution inside of asphere is

V (r, θ) =

∞∑`=0

A`r`P`(cos θ),

and at the surface, the solution has to satisfy

k cos(3θ) =

∞∑`=0

A`R`P`(cos θ),

where the coefficients are given by

A` =2`+ 1

2R`

ˆ π

0

V (R, θ)P`(cos θ) sin θ dθ.

We want to write V (R, θ) in terms of Legendre polyno-mials. Using trig identities, we can write

cos(3θ) = cos(2θ + θ)

= cos(2θ) cos θ − sin(2θ) sin θ

= (2 cos2 θ − 1) cos θ − (2 sin θ cos θ) sin θ

= 2 cos3 θ − cos θ − 2 cos θ(1− cos2 θ)

= 4 cos3 θ − 3 cos θ.

This isn’t exactly what we want since we want to writethis as Legendre polynomials, so we have to do a little

algebraic manipulation.

5

8cos(3θ) =

5

2cos3 θ − 15

8cos θ

=5

2cos3 θ − 3

2cos θ − 3

8cos θ

= P3(cos θ)− 3

8P1(cos θ)

k cos(3θ) =8

5kP3(cos θ)− 3

5kP1(cos θ).

So we can write the boundary condition as

8

5kP3(cos θ)− 3

5kP1(cos θ) =

∞∑`=0

A`R`P`(cos θ),

This implies that we have only two terms and

A1 = − 3k

5R

A3 =8k

5R3.

All other A` must be zero.

So our solution for the potential inside the sphereis

V (r, θ) = A1rP1(cos θ) +A3r3P3(cos θ)

= − 3k

5RrP1(cos θ) +

8k

5R3r3P3(cos θ)

= − 3kr

5R3

(R2 + 4r2

)cos θ +

4k

R3r3 cos3 θ.

Example:

Now we deal with the outside of the sphere. We knowthat as r →∞, the A`r

` term will blow up, so it must bethat A` = 0 for all `. Now our solution becomes

Vout(r, θ) =

∞∑`=0

B`1

r`+1P`(cos θ).

Given the boundary condition V (R, θ), we know that

V (R, θ) =

∞∑`=0

B`1

R`+1P`(cos θ) = V0(θ).

We can apply the orthogonality condition again to get

B` =2`+ 1

2R`+1

ˆ π

0

V0(θ)P`(cos θ) sin θ dθ.

Find the potential outside of a sphere of radiusR if the potential at the surface is

V (R, θ) = k cos(3θ).

We can use the same procedure as in the previousexample, but now our general solution is

V (r, θ) =

∞∑`=0

B`1

r`+1P`(cos θ),

and our boundary condition gives us

8

5kP3(cos θ)− 3

5kP1(cos θ) =

∞∑`=0

B`1

R`+1P`(cos θ).

The only difference between this result and the onefrom the previous example is that the exponents on rand r are now −`− 1 instead of `. So our solution is

V (r, θ) = −3kR2

5r4

(r2 + 4R2

)cos θ +

4kR4

r4cos3 θ.

Example:


Find the surface charge density σ(θ) for the pre-vious example.

We can assume there is no charge inside or out-side of the sphere. All the charges producing the po-tentials of the previous two examples occur at the sur-face of the sphere.

To calculate the surface charge density, we use

∂Vab∂n− ∂Vbe

∂n= − 1

ε0σ,

where Vab is the potential above (outside the sphere),and Vbe is the potential below (inside the sphere). Inour case, the normal direction is r. Calculating thederivative of the potentials found in the previous twoexamples and evaluating them at r = R gives us

∂Vab∂r

=54k

5Rcos θ − 16k

Rcos3 θ

∂Vbe∂r

= −39k

5Rcos θ +

12k

Rcos3 θ.

Plugging these in and solving for σ gives us

σ(θ) =ε0k

Rcos θ

(28 cos2 θ − 93

5

).

Example:

The general procedure for solving Laplace’s equationin spherical coordinates is as follows:

1. Use this when you’re trying to find the potentialV (r, θ) when there is spherical symmetry. Typi-cally, you will be given V0, which may be a functioninvolving cos θ, on some surface. Start by writingthe general solution to Laplace’s equation in spher-ical coordinates

V (r, θ) =

∞∑l=0

(Alr

l +Bl1

rl+1

)Pl(cos θ).

2. Next, consider your region(s). The potential can-not blow up to infinity, so if your region includesr = 0, then Bl = 0 for all l. If your region includesr → ∞, then Al = 0 for all l. You will often haveone or more regions, and you will end up with adifferent solution for each region.

3. If you have two regions, you can eliminate Al orBl by setting your solutions equal to each other atthe boundary between the regions. We know thepotential must be continuous everywhere.

4. If you’re given V0 on a boundary, set your solutionat the boundary equal to V0. Make sure that V0 iswritten in terms of the Legendre polynomials.

5. Multiply both sides by Pm(cos θ) and integrate from0 to π, exploiting the orthogonality of the Legendre

polynomials with orthogonality conditionˆ π

0

Pl(cos θ)Pm(cos θ) sin θ dθ =2

2l + 1δlm.

6. Once you’ve found the potential inside and outsideof a surface, you can find the charge distributionσ(θ) using(

∂Vout∂r

− ∂Vin∂r

) ∣∣∣r=R

= −σ(θ)

ε0.

Notice that the differentiation is with respect to r,and in the end, you evaluate it at the surface of in-terest r = R. Your charge distribution may dependon θ. If the potential on the inside is constant, thenthis reduces to

∂Vout∂r

∣∣∣r=R

= −σ(θ)

ε0.

3D Cylindrical Coordinates

The Laplacian in cylindrical coordinates is

∇2V =1

s

∂

∂s

(s∂V

∂s

)+

1

s2

∂2V

∂φ2+∂2V

∂z2,

Assuming the potential has no z-dependence, this reducesto

∇2V =1

s

∂

∂s

(s∂V

∂s

)+

1

s2

∂2V

∂φ2.

We want to solve Laplace’s equation ∇2V = 0. Aftermultiplying both sides by s2, we get

s∂

∂s

(s∂V

∂s

)+∂2V

∂φ2= 0.


V (s, φ) = S(s)Φ(φ),

we plug this in and simplify to get

s

S

d

ds

(sdS

ds

)+

1

Φ

d2Φ

dφ2= 0.

Notice that the first term is now a function only of s, andthe second term is a function only of φ. This implies thatboth are constant, and the fact that the two add to zeroimplies that one is the negative of the other.

If we let the constant be n2, then from the s part ofthe equation, after expanding it using the product rule,we get the ODE

sd

ds

(sdS

ds

)= n2S,

and from the φ part, we get the ODE

d2Φ

dφ2= −n2Φ.


The φ equation has the well-known solution

Φn(φ) = A cos(nφ) +B sin(nφ).

For the s equation, if we assume a solution of the formS = sk, then

sd

ds

(sksk−1

)= n2sk

ksd

ds

(sk)

= n2sk

k2sk = n2sk

k = ±n.

So our solution for the s equation is

Sn(s) = Csn +Ds−n.

When n = 0, this reduces to a constant, so we treat thatcase separately. Plugging n = 0 back into the s ODE,

gives us

d

ds

(sdS

ds

)= 0

sdS

ds= E

S0(s) = E ln s+ F.

Combining the solutions, we get

V0(s, φ) = A0(E ln s+ F ),

and for n 6= 0, we have

Vn(s, φ) = [An cos(nφ) +Bn sin(nφ)][Cns

n +Dns−n] .

So our general solution is

V (s, φ) =

∞∑n=1

[An cos(nφ) +Bn sin(nφ)][Cns

n +Dns−n]

+A0(E ln s+ F ).

2.4. MULTIPOLE EXPANSION 47

2.4 Multipole Expansion

We know that far from a charge distribution, the po-tential looks like

V ∼ 1

4πε0

Q

r,

where Q is the total charge in the distribution. But whatif the total charge in the distribution is zero. Triviallythen, V ∼ 0. We want more information than that.

Consider a dipole, which is a pair of point charges+q and −q separated by some distance d.

−q

+q P

d

u+

u−

r

θ

The potential at a point P due to this dipole is

V (r) =1

4πε0

(q

u+− q

u−

).

By the law of cosines,

u2+ = r2 +

1

4d2 − 2r

d

2cos θ

= r2

(1− d cos θ

r+

d2

4r2

)u2− = r2 +

1

4d2 + 2r

d

2cos θ

= r2

(1 +

d cos θ

r+

d2

4r2

).

We can plug these into V (r) to see what the potential

looks like for r >> d. Dropping the d2

4r2 terms from bothgives us

1

u±=

1

r

(1∓ d cos θ

r

) 12

' 1

r

(1± d cos θ

2r

).

Then the potential becomes

V (r) ' q

4πε0

[1

r

(1 +

d cos θ

2r

)− 1

r

(1− d cos θ

2r

)]=

q

4πε0

d cos θ

r2.

So the potential of an electric dipole at r >> d is

V (r) =1

4πε0

qd cos θ

r2.

Now consider a general charge distribution like theone shown below.

O

dτ ′P~u

~r~r ′θ

The potential at a point P due to this charge distri-bution is

V (~r) =1

4πε0

ˆ1

uρ(~r ′)dτ ′.

By the law of cosines,

u2 = r2 + (r′)2 − 2rr′ cos θ′

= r2

[1 +

(r′)2

r2− 2r′

rcos θ′

]u = r

√1 + ε,

where

ε ≡ r′

r

(r′

r− 2 cos θ′

).

Then

1

u=

1

r(1 + ε)−

12 =

1

r

(1− ε

2+

3ε2

8− · · ·

).

If we expand, plug in the definition of ε, and then groupthe terms in powers of r′

r , we find that

1

u=

1

r

∞∑n=0

(r′

r

)nPn(cos θ′).

where Pn(cos θ′) are the Legendre polynomials.

Plugging this into V (~r) gives us the multipole ex-pansion of V for a continuous charge distribution

V (~r) =1

4πε0

∞∑n=0

1

rn+1

ˆ(r′)nρ(~r ′)Pn(cos θ′) dτ ′.

The first three terms of this are

V (~r) =1

4πε0

1

r

ˆρ(~r ′)dτ ′

+1

4πε0

1

r2

ˆρ(~r ′)r′ cos θ′ dτ ′

+1

4πε0

1

r3

ˆρ(~r ′)(r′)2

(3

2cos2 θ′ − 1

2

)dτ ′.


These are the monopole, dipole, and quadrupole terms.Notice that monopoles fall off as 1

r , dipoles fall off as 1r2 ,

quadrupoles fall off as 1r3 and so on.

Point Charges

The monopole term for a collection of n point charges isjust

Vmon(r) =1

4πε0

Q

r,

where

Q =

ˆρ(r′) dτ ′,

is the total charge of the collection, and r is your distancefrom the origin of your coordinate system.

Note: The monopole term is very simple to calculate.Just find the total charge and plug into the equationabove. Don’t make the mistake of thinking the r inthe monopole equation means you need to computeyour distance from each charge. No matter where thecharges are, this r is simply your distance from theorigin of your coordinate system.

Tip:

For a point charge at the origin, this monopole termis the exact potential. If the net charge is zero, the firstnonzero term in the multipole expansion is the dipoleterm, unless it too vanishes.

Vdip(r) =1

4πε0

1

r2

ˆρ(r ′)r′ cos θ′ dτ ′.

If the total charge of a group of particles is zero, thatis Q = 0, then the multipole expansion for that groupof charges has no monopole term since it is zero.

Tip:

We can write r′ cos θ′ = r · ~r ′. Then

Vdip(r) =1

4πε0

r

r2·ˆρ(r ′)~r ′ cos θ′ dτ ′.

The integral is called the dipole moment of the volumecharge distribution

~p =

ˆρ(r ′)~r ′ cos θ′ dτ ′.

Then we can write

Vdip =1

4πε0

r · ~pr2

.

The dipole moment of a collection of n point chargesis

~p =

n∑i=1

qi~ri′,

where ~ri′ is the vector from the origin of your coordinate

system to the ith particle.

A physical dipole is a real dipole formed by a posi-tive charge and a negative charge separated by a distanced. A pure or ideal dipole is an idealized version of sucha dipole in which d → 0, q → ∞ and p = dq remainsunchanged. The difference is that for the physical dipole,the multipole terms will contain the dipole term but alsothe higher order terms. The multipole expansion of theideal dipole contains only the dipole term because all thehigher order terms are zero. The ideal dipole does notexist in nature unlike the physical dipole. At large dis-tances, the physical dipole looks like the ideal dipole, andthat’s why we use multipole expansion to approximate aphysical (i.e. real) dipole at a distance.

For a physical dipole formed by a pair of charges +qand −q held a distance d from each other, the dipolemoment is

~p = q~d,

where ~d is the vector from the−q charge to the +q charge.Note the odd convention for this vector’s direction.

The dipole term is a vector, and we can have twodipoles ~p1 and ~p2, then the total dipole is just the vectorsum

~p = ~p1 + ~p2.

If you arrange four charges in the corners of a squareso that you have negative charges on opposite cornersand positive charges on the other opposite corners, thenthe total dipole is zero. This setup of charges is aquadrupole.

Note that the monopole term in a multipole expan-sion is a number, the dipole term involves a vector, andthe quadrupole term involves a tensor.

For a point charge at the origin, the potential ispurely the monopole term. However, a point charge thatis not at the origin also has a dipole moment (as well ashigher order terms). Recall that multipole expansion isin terms of distance from the origin, so if you move thecharge away from the origin, you change the multipoleexpansion.

2.4. MULTIPOLE EXPANSION 49

Remember, if you have a charge distribution near theorigin of your coordinate system, then to find the ap-proximate potential far from the origin, just use mul-tipole expansion.

Tip:

Calculate the first two terms of the multipole ex-pansion for the configuration of point charges shownbelow if the distance of each charge from the origin isd.

y

z

x

−q−qq

The total charge of the combined three chargesis −q. Since this is nonzero, we know that we have amonopole term

Vmon = − 1

4πε0

q

r.

Our dipole moment is

~p =

n∑i=1

qi~ri′ = −qd(−y)− qd y + qd z = qd z.

So our dipole term is

Vdip =1

4πε0

r · qd zr2

=1

4πε0

qd

r2r · z

=1

4πε0

qd

r2cos θ.

Here θ is the angle between ~r and the z axis as itnormally is in spherical coordinates.

Example:

Continuous Charge Distribution

For continuous charge distribution, the monopole term isagain

Vmon(~r) =1

4πε0

Q

r,

where Q is the total charge of the distribution. The dipoleterm is again

Vdip(~r) =1

4πε0

r · ~pr2

,

but now we have to integrate to find the dipole moment

~p =

ˆ~r ′ρ(~r ′) dτ ′.

Note: For the dipole term, these formulas are for thecase in which the dipole moment ~p is at the origin of ourcoordinate system, and pointing in the z direction.

Electric Field

After calculating the potential, we can calculate the fieldusing ~E = −~∇V . For an ideal dipole, for example, wewould compute ~E = −~∇Vdip(~r).This gives us the follow-ing formula for calculating the electric field in sphericalcoordinates given the dipole moment ~p.

~Edip = −~∇Vdip(r, θ)

= −~∇ 1

4πε0

r · ~pr2

= − 1

4πε0~∇(p cos θ

r2

)=

p

4πε0r3

(2 cos θ r + sin θ θ

).

This is valid in spherical coordinates with the dipole mo-ment in the z direction.

By noting that in spherical coordinates, ~p =p cos θ r − p sin θ θ, we can write this in the coordinatefree form

~Edip =1

4πε0r3[3(~p · r)r − ~p] .

If you have a charge distribution near the origin ofyour coordinate system, then to find the approximateelectric field far from the origin, use multipole expan-sion to find the approximate potential, and then use~E = −~∇V to find the approximate electric field farfrom the origin.

Tip:


Find the approximate electric field ~E(r, θ) for asolid sphere of radius R centered at the origin. Theupper half has a uniform charge density of ρ0, and thelower half has a uniform charge density of −ρ0.

Here we approximate the potential with a multi-pole expansion and then calculate the field from that.

Since the total charge is zero, we know that thereis no monopole term in the multipole expansion. Wehave to go directly to the dipole term. We start bycalculating the dipole moment for the sphere

~p =

ˆ~r ′ρ(~r ′) dτ ′

=

ˆ R

0

ˆ 2π

0

ˆ π2

0

(rρ0)r2 sin θ dθ dφ dr

+

ˆ R

0

ˆ 2π

0

ˆ π

π2

(−rρ0)r2 sin θ dθ dφ dr

= ρ0

ˆ R

0

r3 dr

ˆ 2π

0

dφ

ˆ π2

0

sin θ dθ

−ρ0

ˆ R

0

r3 dr

ˆ 2π

0

dφ

ˆ π

π2

sin θ dθ

=2πρ0R

4

4+

2πρ0R4

4= πρ0R

4.

So the dipole moment has magnitude

p = πρ0R4,

and it points in the z direction.

We can now calculate the electric field due to thedipole moment as

~Edip =p

4πε0r3


)=

ρ0R4

4ε0r3


).

Example:

2.5. SUMMARY: SPECIAL TECHNIQUES 51

2.5 Summary: Special Techniques

Method of Images

The method of images exploits the uniqueness theoremsassociated with Laplace’s equation

∇2V = 0.

If we can compute V for an artificial setup that satis-fies Laplace’s equation and has the same boundary con-ditions, then by the uniqueness theorems, this must bethe same solution V as the one for the problem we aretrying to solve. Typical image problems are problems inwhich a conductor (V = 0) forms the boundary. Thenthe potential is specified on the boundary surface, so bythe first uniqueness theorem, any potential that satisfiesLaplace’s equation and the boundary condition(s) is thesolution.

The general procedure for a method of images prob-lem is as follows:

1. Determine the boundary conditions on the poten-tial. If you’re dealing with conductors, you knowthey’re equipotentials.

2. Guess the configuration of images charges such thatthe potential of the setup satisfies the boundaryconditions. Image charges are never in the regionof interest.

3. When calculating the potential of your setup, treatthe image charges like actual charges and calculatethe potential at a point (in the region of interest)due to both the real charges and the image charges.Recall that the formula for calculating the potentialdue to point charges is

V (r) =1

4πε0

n∑i=1

qiui,

where ui is the distance from your test point (x, y, z)to the ith point charge.

4. The force acting on the real charge(s) is the sameas the force acting on the charge(s) due to the othercharges including the image charges. That is, to cal-culate the force on a real charge, treat all the imagecharges as real charges and calculate the force dueto all the charges

~F =Q

4πε0

n∑i=1

qiu2i

ui.

5. The surface charge induced on a conducting surfaceby a charge q above the surface is

σ = −ε0∂V

∂n

∣∣∣surface

.

That is, the induced surface charge is calculated byfirst taking the partial derivative of V with respect

to the direction that is normal to the conductingsurface and then evaluating the result at the con-ducting surface.

6. The total charge that is induced on the conductingsurface by our charge q can be found by integratingthe surface charge density σ over the entire con-ducting surface

Q =

ˆσ da.

In general, the total charge induced on the conduct-ing surface should be the negative of your chargedoing the inducing.

7. The image method doesn’t work to compute thework because of the Faraday cage effect that occurswith conductors. To calculate the work required tobring the charge q in from infinity to a distance dfrom the infinite conducting surface we have to use

W =

ˆ d

∞~F · d~.

Separation of Variables

To perform separation of variables on Laplace’s equation

∇2V = 0,

follow this general procedure:

1. Study the problem statement. Draw a pictureShould you use cartesian or spherical coordinates?In what region are you asked to find the potential?Is the configuration independent of one or more co-ordinates?

2. Identify and list the boundary conditions.3. Is it a 1D, 2D, or 3D problem. Identify the variables

of which the solution will be independent.4. Write down Laplace’s equation. Assume a separable

solution and separate variables, writing Laplace’sequation as a set of ODEs.

5. Solve the set of ODEs using the boundary condi-tions identified in step 2.

6. Write the general solution as the linear combinationof the products of the solutions to the ODEs.

7. Use the final boundary condition (i.e. the “initialcondition”) and orthogonality to write a formula toobtain the Fourier coefficients.


In 2D Cartesian coordinates, Laplace’s equation is

∂2V

∂x2+∂2V

∂y2= 0.

Depending on how the problem is set up, one term couldbe in terms of z instead of x or y. Assuming a separable


solution of the form

V (x, y) = X(x) · Y (y),


X ′′

X+Y ′′

Y= 0.

Each term in this equation must be constant. One is equalto k2, and the other is equal to −k2. So you get a pair ofordinary second-order ODEs. Notice that in one of them,the constant must be negative and the other positive.

The solution to the ODE with the negative constant(e.g. X ′′ = −k2X or Y ′′ = −k2Y ) will be sines andcosines. The solution to the ODE with the positive con-stant will be exponentials, e±kx.

Use the boundary conditions to determine whichvariable the negative constant goes with. For example,if one of your boundary conditions is V → 0 as y → ∞,then we know the Y ODE needs to have the solutionwith the exponentials which can be made to go to zero asy →∞. In that case, your Y ODE would need to be asso-ciated with the positive constant. Similarly, if you have aboundary condition like V (0, y) = V (a, y) = 0, then youneed the X ODE solution to be sines and cosines sincethose can be forced to be zero at x = 0 and x = a.

Once you have the solutions for your ODEs, you mul-tiply them together (since we originally assumed a sepa-rable solution). Then the general solution will typicallybe an infinite sum. The sum may contain exponentialsand sine (or cosine).

To obtain the unknown coefficients, use the orthog-onality of sine (or cosine) as with Fourier series. To dothat, you need to have the following two orthogonalityconditions memorized or be able to derive them on thespot.ˆ a

0

sin(nπx

a

)sin(mπx

a

)dx =

a

2δnm

ˆ a

0

cos(nπx

a

)cos(mπx

a

)dx =

a n = m = 0a2 n = m 6= 0

0 n 6= m

.


In 3D cartesian coordinates the full Laplace’s equa-tion is

∂2V

∂x2+∂2V

∂y2+∂2V

∂z2= 0.


V (x, y, z) = X(x) · Y (y) · Z(z),


X ′′

X+Y ′′

Y+Z ′′

Z= 0.

Each term of this equation must separately be equal to aconstant, so

C1 + C2 + C3 = 0,

where C1 is the constant associated with the x term, C2

is the constant associated with the y term, and C3 is theconstant associated with the z term. One or two of theconstants must be positive and the rest negative in orderto add to zero. To determine which ones need to be neg-ative and which need to be positive, you have to considerthe boundary conditions. For example, if V = 0 at theboundaries in one direction, then the constant for that di-rection must be negative so that the solution is in termsof sines and cosines rather than exponentials. Once youhave defined two of the constants, you can write the thirdin terms of the first two.

For the 3D Cartesian case, the whole solution pro-cess is similar to the 2D case, but now you end up witha double sum.

3D Spherical Coordinates

The general procedure for solving Laplace’s equationin spherical coordinates is as follows:

1. Use this when you’re trying to find the potentialV (r, θ) when there is spherical symmetry. Typi-cally, you will be given V0, which may be a functioninvolving cos θ, on some surface. Start by writingthe general solution to Laplace’s equation in spher-ical coordinates

V (r, θ) =

∞∑l=0

(Alr

l +Blrl+1

)Pl(cos θ).

This assumes V has no φ dependence. Here,Pl(cos θ) is the l-th Legendre polynomial.

2. Next, consider your region(s). The potential can-not blow up to infinity, so if your region includesr = 0, then Bl = 0 for all l. If your region includesr → ∞, then Al = 0 for all l. You will often haveone or more regions, and you will end up with adifferent solution for each region.

3. If you have two regions, you can eliminate Al orBl by setting your solutions equal to each other atthe boundary between the regions. We know thepotential must be continuous everywhere.

4. If you’re given V0 on a boundary, set your solutionat the boundary equal to V0. Make sure that V0 iswritten in terms of the Legendre polynomials.

5. Multiply both sides by Pm(cos θ) and integrate from0 to π, exploiting the orthogonality and complete-ness of the Legendre polynomials with orthogonalitycondition

ˆ π

0

Pl(cos θ)Pm(cos θ) sin θ dθ =2δlm

2l + 1.

2.5. SUMMARY: SPECIAL TECHNIQUES 53

6. Once you’ve found the potential inside and outsideof a surface, you can find the charge distributionσ(θ) using(

∂Vout∂r

− ∂Vin∂r

) ∣∣∣r=R

= −σ(θ)

ε0.

Notice that the differentiation is with respect to r,and in the end, you evaluate it at the surface of in-terest r = R. Your charge distribution may dependon θ. If the potential on the inside is constant, thenthis reduces to

∂Vout∂r

∣∣∣r=R

= −σ(θ)

ε0.

The first several Legendre polynomials in cosine are

P0(cos θ) = 1

P1(cos θ) = cos θ

P2(cos θ) =1

2(3 cos2 θ − 1).

Instead of being given the potential on a sphericalshell, you may also be given a surface charge density σ(θ)that is glued over a spherical surface at r = R. Now, youwould write σ(θ) in terms of Legendre polynomials andsolve as before, but now you use the boundary conditions

Vin(R, θ) = Vout(R, θ)(∂Vout∂r

− ∂Vin∂r

) ∣∣∣r=R

= −σ(θ)

ε0.

Multipole Expansion

Multipole expansion allows us to approximate potentialsat a distance. From this, we can also calculate the electricfield.

The general formula for multipole expansion is

V (~r) =1

4πε0

∞∑n=0

1

rn+1


The first term is the monopole term, and the second termis the dipole term, so

V ' Vmon + Vdip.

If you’re given a charge distribution and asked tocalculate V at large distance, use multipole expansion.

The Monopole Term

The monopole term is

Vmon =1

4πε0

Q

r,

where Q is the total charge and r is your distance fromthe origin of your coordinate system. If the total charge

of a group of particles is zero, then there is no monopoleterm.

The monopole term is very simple to calculate. Justfind the total charge and plug into the equation above.Don’t make the mistake of thinking the r in the monopoleequation means you need to compute your distance fromeach charge. No matter where the charges are, this r issimply your distance from the origin of your coordinatesystem.

The Dipole Term

The dipole term is

Vdip =1

4πε0

r · ~pr2

.

where ~p is the dipole moment.

For a single dipole, that is, a charge +q and a charge−q separated by a distance d, the dipole moment is

~p = q~d,

where ~d is the vector from the−q charge to the +q charge.The dipole moment of a collection of n point charges is

~p =

n∑i=1

qi~ri′,

where ~ri′ is the vector from the origin of your coordi-

nate system to the ith particle. For a continuous chargedistribution, the dipole moment is

~p =

ˆ~r ′ dq =

ˆ~r ′ρ(~r ′) dτ ′.

Note: For the dipole term, these formulas are for thecase in which the dipole moment ~p is at the origin of ourcoordinate system and pointing in the z direction.

Electric Field

If you’re given a charge distribution and are askedto find the approximate electric field far from the origin,use multipole expansion to calculate the approximate po-tential V . Then use ~E = −~∇V to find the approximateelectric field far from the origin.

For an ideal dipole, for example, we would compute~E = −~∇Vdip(~r).This gives us the following formula forcalculating the electric field in spherical coordinates giventhe dipole moment ~p.

~Edip =p

4πε0r3


).


Miscellaneous

The area element in spherical coordinates is

da = r2 sin θ dθ dφ.

The following trick is sometimes useful when usingthe orthogonality of the Legendre polynomialsˆ π

0

Pm(cos θ) sin θ dθ =

ˆ π

0

P0(cos θ)Pm(cos θ) sin θ dθ

= 2δ0m.

Note: If you have to actually compute Legendre poly-nomials in the form P`(cos θ), switch to x using x = cos θ.Then

ˆ π

0

P`(cos θ)Pm(cos θ) sin θ dθ =

ˆ 1

−1

P`(x)Pm(x) dx.

Then the integrals will just be polynomials in x insteadof in cosines.

Chapter 3

Polarization and Dielectrics

3.1 Polarization

We can classify matter as being of two types:

1. Conductors2. Insulators

Since conductors do not contain electric fields, we will befocusing on insulators (or dielectrics) in this chapter. Ininsulators, we still have electrons, but they are not freeelectrons—they are bound to nuclei.

Induced Dipoles

If we apply an electric field to an atom, the charges will bemicroscopically displaced. For example, the positive nu-clei of the atoms will be displaced a small amount in thedirection of the electric field, and the negatively-chargedelecton clouds will be displaced a little in the directionopposite the electric field. This results in a lot of smalldipole moments in an insulator in an electric field, andwe say the material has become polarized.

If the electric field is extremely strong, the nuclei canbe completely separated from their electrons. This is ion-ization and can cause an insulator to suddenly becomea conductor.

There are really two types of displacements:

1. Stretching occurs when the bond lengths of atomsis increased.

2. Rotating occurs when the electric field exerts atorque on an atom causing it to rotate. This isdiscussed in the next subsection.

For induced dipoles, our dipole moment is

~p = α~E,

where ~E is the applied electric field and α is the atomicpolarizability of the atom. Note that this is a rule-of-thumb that is approximately true for small atoms.

Consider a simple model of an atom with a charge+q at the center and a charge −q (the electron cloud)smeared uniformly throughout a sphere of radius a. If weapply an electric field ~E, then the +q at the center of thesphere will be displaced from the center of the sphere ofnegative charge by some distance d in the direction of theelectric field. What is the atomic polarizability of thisatom?

Assuming that the sphere (i.e. the electron cloud)maintains its shape, which it should, then the electricfield due to the electron cloud at the position of the pos-itive nucleus, which is now a distance d from the centerof the cloud is

E =1

4πε0

qd

a3.

This is just the field at a distance d from the center of auniformly charged sphere. We are considering the systemto be at equilibrium. That is, the nucleus has been dis-placed from the center of the cloud, but the two are nowstable. The force that the electron cloud is exerting onthe nucleus in one direction is exactly equal to the forcethat the external electric field is exerting on the nucleusin the opposite direction. We know that the dipole mo-ment can be calculated as p = qd, so solving the abovefor qd gives us

p = 4πε0a3E.

Comparing this with ~p = α~E, we have that the atomicpolarizability in this case is

α = 4πε0a3 = 3ε0V ,

where V is the volume of our atom. This is a crude butacceptable model for a single atom.

For a molecule, which is asymmetric compared to asingle atom, calculating the polarizability is a little morecomplicated. For a linear molecule, for example, the po-larizability is different depending on whether the externalfield is applied parallel to the molecule’s axis or perpen-dicular. For a linear molecule, the polarizability is

~p = α⊥ ~E⊥ + α|| ~E||,55

56 CHAPTER 3. POLARIZATION AND DIELECTRICS

where α⊥ is the polarizability in the direction perpendic-ular to the molecule’s axis and α|| is the polarizability inthe direction parallel to its axis.

For completely asymmetric molecules, the problembecomes even worse. Instead of a polarizability number(single atom) or a polarizability vector (linear molecule),we need in general a polarizability tensor containing nine

terms.

α =

αxx αxy αxz

αyx αyy αyz

αzx αzy αzz

.Then, for example,

px = αxxEx + αxyEy + αxzEz.

In any case, there are always principal axes that you canchoose such that the polarizability tensor becomes diag-onal, and you only have to deal with three terms.

For a hydrogen atom with atomic polarizabilityα = 7.42× 10−42 C2m/N and radius of 5× 10−11 mwhat is the displacement d between the nucleus and itselectron cloud when it is placed between two parallelplates with a voltage difference of 500 V and a separa-tion of 1mm? What voltage is required to ionize theatom?

We know that the potential difference between thetwo parallel plates is V = Es, where s is the distancebetween the two plates. We know V and s, so the elec-tric field between the plates is

E =V

s=

500 V

10−3 m= 5× 105 V/m.

The dipole moment can be calculated as p = qd =αE. We know that the charge of the proton is e =1.602× 10−19 C, so the displacement d can be calcu-

lated as

d =αE

q=

7.42× 10−41 C2m/N · 5× 105 V/m

1.602× 10−19 C

= 2.32× 10−16 m.

Notice that the radius of the atom is hundreds of thou-sands of times larger than the displacement caused bythis electric field.

To completely ionize the atom, we have to increasethe voltage until the displacement of the nucleus be-comes comparable to the radius of the atom. We needa potential difference of

V =sqd

α

=10−3 m · 1.602× 10−19 C · 5× 10−11 m

7.42× 10−41 C2m/N

= 1.08× 108 V.

Example:

Intrinsic Dipoles

Consider a polar molecule such as water. This moleculeis bent and since the oxygen atom is more electronegativethan the hydrogen atoms, there is a built-in polarity suchthat the oxygen side of water is more negative and thehydrogen side is more positive. Therefore, water has anatural dipole moment that points from the oxygen atomtoward the hydrogen side.

In a glass of water, for example, the dipoles of thewater molecules point in random directions. When anelectric field is applied to the glass of water, the watermolecules will experience a torque (since one side is morenegative and the other side is more positive), and all thewater molecules will rotate until their dipoles align withthe applied electric field.

For a neutral polar molecule, the total force on themolecule is zero since its total charge is zero. How-ever, since the molecule is polar, it has a dipole moment~p = q~d. An external field ~E will exert a force ~F+ = q ~E+

on the positive charge and a force ~F− = −q ~E− on thenegative charge. If the field is uniform then ~E+ = ~E−and the net force on the dipole is ~F++~F− = q ~E−q ~E = 0.However, since ~F+ is in one dirction and ~F− is in the op-posite direction, there will be a net torque on the dipole ifit is not already aligned with the electic field. This torqueis

~N = q~d× ~E = ~p× ~E,

about the center of the dipole (or molecule). About anyother point, this torque is

~N = ~p× ~E + ~r × ~F .

This torque will cause a polar molecule to rotate until itsdipole aligns with the applied electric field.

If the electric field is not uniform, i.e., ~E+ 6= ~E−,then ~F+ 6= ~F−. So in addition to having a net torque,we also have a net force of

~F = ~F+ − ~F− = q(~E+ − ~E−) = q∆~E.

3.1. POLARIZATION 57

on the molecule. If the dipole is short then

∆Ex = (~∇Ex) · ~d,

or more generally,

∆~E = (~d · ~∇)~E.

This implies that

∆~F = (~p · ~∇)~E.

A dipole (e.g. a polar molecule) near a conductingsurface will induce a charge in the conducting surface.The electric field of this induced charge will then exert atorque on the dipole and cause it to rotate.

In general, a material has many dipoles either causedby an applied electric field that slightly displaces the nu-clei of the atoms from their electron clouds or by themolecules themselves if they are polar. Regardless of howthe dipoles came about, if an external electric field is ap-plied, they will all align themselves with the electric fieldcausing the material to become polarized. It is then use-ful to speak of the polarization as the dipole momentper unit volume

~P = dipole moment per unit volume.

Electric Field

For a single dipole ~p, the potential produced by the dipoleis

V (~r) =1

4πε0

u · ~pu2

.

For a polarized material, we sum over all of the dipolesusing ~p = ~P dτ ′. So the potential due to the dipoles in avolume V is

V (~r) =1

4πε0

ˆV

u · ~P (~r ′)

u2dτ ′.

O

~p ~u

~r~r ′

V

By writing ~∇′ 1u = u

u2 and applying the divergencetheorem, we can write this in the much more convenientform

V (~r) =1

4πε0

˛S

~P · d~a ′

u− 1

4πε0

ˆV

~∇′· ~Pu

dτ ′.

The first term looks like the potential of a surface chargedensity with

σb = ~P · n.

We call this the bound surface charge density. Thesecond term looks like the potential of a volume chargedensity with

ρb = −~∇ · ~P .

We call this the bound volume charge density. So thepotential due to a polarized object is just the potentialdue to the bound surface charge plus the potential due tothe bound volume charge.

V (~r) =1

4πε0

˛S

σbud~a+

1

4πε0

ˆV

ρbudτ ′.

One way to find the electric field due to polarizationof a polarized object is to compute σb and ρb and thencompute the potential using the integral above. Finally,we can easily find the electric field due to the polarizedobject by using ~E = −~∇V .

If the polarization is spherically symmetric, it’s mucheasier to find the bound charges and calculate their fieldsusing Gauss’s law.


Find the electric field produced by a uniformly po-larized sphere of radius R.a.

We choose our coordinate system so that the polar-ization vector is parallel to the z-axis. I.e., ~P ‖ z. Then,because it is uniformly polarized, the volume boundcharge is

ρb = −~∇ · ~P = 0.

The surface bound charge is

σb = ~P · n = P cos θ.

We are, therefore, looking for the field of a sphere witha surface charge density P cos θ. We’ve previously cal-culated this as

Vin(r, θ) =P

3ε0r cos θ,

for r ≤ R, and

Vout(r, θ) =P

3ε0

R3

r2cos θ,

for r ≥ R.

We can write r cos θ = z, then

~Ein = −~∇Vin = − P

3ε0z = −

~P

3ε0,

for r < R.

Outside the sphere, V is identical to the potentialfor a perfect dipole at the origin

Vout(r, θ) =1

4πε0

~p · rr2

,

for r ≥ R. Here, the dipole moment vector is

~p =4

3πR3 ~P .

The electric field on the outside is that of a perfectdipole at the origin with the above dipole moment.

aThis is example 4.2 in Griffiths

Example:

Be careful when you try to apply Gauss’s law. For ex-ample, Gauss’s law can be used for a sphere when thepolarization is radial, but not when the polarization isuniform.

Tip:

For a uniformly polarized dielectric, the polarizationeffectively pastes a charge over the surface of the mate-rial. If the polarization is nonuniform, you can also getbound charges inside the material.

Calculate the electric field inside and outside ofa sphere of radius R with polarizationa

~P (~r) = k~r.

The bound surface and volume charge densitiesare

σb = ~P · n = P = kr

ρb = −~∇ · ~P = − 1

r2

∂

∂r

(r2kr

)= −3k.

The bound surface charge density is just P = kr be-cause the polarization is parallel to the normal vectorn.

Since the polarization is radial, we can useGauss’s law to find the electric field. Inside the sphere,only ρb contributes to the charge, so we get

~E = − kε0~r, r < R.

Outside, the sphere, we have the bound surfacecharge plus the bound volume charge. Since these sumto zero, we get

~E = 0, r > R.

aThis is problem 4.10 in Griffiths

Example:

3.2. ELECTRIC DISPLACEMENT 59

3.2 Electric Displacement

In a dielectric, free charge is any charge that is not dueto the polarization of the dielectric. The total volumecharge density of a dielectric is

ρ = ρb + ρf ,

where ρb is the bound charge, and ρf is the free charge.Plugging this into the differential form of Gauss’s lawgives us

ε0~∇ · ~E = ρ = ρb + ρf .

Recall that the volume bound charge density is ρb =−~∇ · ~P , so

ε0~∇ · ~E = −~∇ · ~P + ρf ,

which we can write as

~∇ · ~D = ρf ,

where

~D = ε0 ~E + ~P ,

is called the electric displacement.

Notice that

~∇ · ~D = ρf ,

is the differential form of Gauss’ law in terms of the elec-tric displacement instead of the usual electric field. Wecan write this in integral form as

˛~D · d ~A = Qf,enc. (3.1)

Consider a straight wire carrying a uniform linecharge λ that is surrounded by rubber out to a radiusa. Find the electric displacement.a

We have cylindrical symmetry, so we can useGauss’ law with the electric displacement. We use aGaussian cylinder of length l and radius s. The cylin-drical symmetry implies that ~D ‖ d ~A along the curved

part of the Gaussian cylinder, so | ~D| is constant on thecurved part of the Gaussian surface, and there is no fluxthrough the end caps of the cylinder.

This gives us˛~D · d ~A = Qf,enc

| ~D|2πsl = λl.

This gives us the electric displacement

~D =λ

2πss.

If we knew the polarization ~P of the rubber, we coulduse this to calculate the electric field within the rubberdielectric.

For s > a, we have ~P = 0, so

~E =~D

ε0=

λ

2πε0ss.

Notice that the field outside the rubber dielectricis unaffected by the polarization.


Example:

Don’t be fooled by the apparent analogy between ~Eand ~D. There is no ‘Coulomb’s law’ for ~D. There is alsono corresponding potential for ~D. We know that the curlof ~E is always zero, but the same is not true of ~D. Infact, the curl of ~D is

~∇× ~D = ε0~∇× ~E + ~∇× ~P = ~∇× ~P .

Eq. (3.1) implies the boundary condition

D⊥above −D⊥below = σf ,

for the electric displacement and

~D‖above − ~D

‖below = ~P

‖above − ~P

‖below.

These boundary conditions are often more useful than the

boundary conditions for the electric field at a surface:


ε0

~E‖above − ~E

‖below = 0.

3.3 Linear Dielectrics

Earlier, we said that often

~P = α~E.

Usually, this is written as

~P = ε0χe ~E,

where χe is the electric susceptibility of the material.Materials that obey this equation are called linear di-electrics.


In some materials, like crystals, the material is easierto polarize in one direction than in others. In that case,we use the more general susceptibility tensor

χ =

χexx χexy χexz

χeyx χeyy χeyz

χezx χezy χezz

.Then for example,

Px = ε0 (χexxEx + χexyEy + χexzEz) .

Typically, if we say “linear dielectric”, we mean “isotropiclinear dielectric”, and then we can use the simpler equa-tion above instead of the tensor.

Note that ~E here is the total electric field. Supposewe start with a dielectric with no polarization. If we placethis dielectric in an external electric field ~Eext, this willinduce a polarization in the dielectric. This polarizationthen modifies the electric field, which modifies the polar-ization, which modifies the field, and so on. The field~E in the equation above, is the total electric field wheneverything is in equilibrium. That is, it is the applied

external field plus the field due to the polarization of thedielectric.

For a linear dielectric,

~D = ε0 ~E + ~P

= ε0 ~E + ε0χe ~E

= ε0(1 + χe)~E

= ε~E,

whereε ≡ ε0(1 + χe),

is called the permittivity of the material. In a vacuum,χe = 0, then ε = ε0.

Another useful definition is

εr ≡ε

ε0= 1 + χe,

called the relative permittivity of the material. It isalso sometimes called the dielectric constant of the ma-terial.

A metal sphere of radius a carries a charge Q. It issurrounded by a linear dielectric material out to radiusb. The material has permittivity ε. Find the potentialat the center relative to infinity.a

We have spherical symmetry, so we can find ~D.Using Gauss’ law, we find

~D =Q

4πr2r,

for r > a. Inside the sphere, we know that ~E = 0 sinceit is a metal. So ~E = ~P = ~D = 0 and we get

~Ein =Q

4πεr2,

for a < r < b, and

~Eout =Q

4πε0r2,

for r > b. Notice that we have ε for inside the rubberand ε0 for outside.

The potential at the center is

V = −ˆ 0

∞~E· d~l

= −ˆ b

∞

Q

4πε0r2dr −

ˆ a

b

Q

4πεr2dr

=Q

4π

[1

ε0b+

1

εa− 1

εb

].

We didn’t have to calculate ~P to solve this prob-lem, but now we can.

~P = ε0χe ~E =ε0χeQ

4πεr2r.

In the dielectric material,

ρb = −~∇ · ~P = 0.

Normally, ~∇ · rr2 = 4πδ(~r), but here, the polarization

doesn’t go all the way to r = 0.

To find the surface bound charge, we use σb = ~P ·n.

σb =

ε0χeQ4πεb2 r = b

− ε0χeQ4πεa2 r = a..

Note, at the inner surface, n points toward the centerof the sphere since it always points outward from thedielectric.

aThis is example 4.5 in Griffiths.

Example:

3.3. LINEAR DIELECTRICS 61

Suppose space is filled with a linear dielectric, then

~∇ · ~D = ρf

~∇× ~D = 0.

This implies that~D = ε0 ~Evac,

where ~Evac is the field you would have if there were nodielectric. This then implies that

~E =1

ε~D =

1

εr~Evac.

If all space were filled with a homogeneous linear dielec-tric, the electric field everywhere would be reduced by1εr

.

A dielectric acts like a frustrated conductor. In aconductor, charges move until the field is completely can-celed out. In a dielectric, charges move as much as theycan. The field is diminished, but not completely canceled.In other words, a dielectric ‘screens’ a charge.

So the field of a free charge embedded in a large di-electric, is

~E =1

εr~Evac =

1

εr

1

4πε0

q

r2r =

1

4πε

q

r2r.

Notice that it has ε instead of the usual ε0.

Consider a parallel plate capacitor that is filled withan insulating material with dielectric constant εr. Whateffect does this have on its capacitance? The field is con-fined to the region between the plates, and it is assumedthe plates are much larger than their separation. Thenthe dielectric will reduce the field between the places bya factor of εr. This implies that the potential differencebetween the plates is reduced by the same factor. Sincethe capacitance is inversely proportional to the potential,

this implies that the capacitance is increased by the factorεr.

C = εrCvac,

where Cvac is the capacitance that the capacitor wouldhave if there was vacuum between its plates.

Boundary Value Problems

In a homogeneous linear dielectric, the bound charge den-sity is proportional to the free charge density

ρb = −~∇ · ~P

= −~∇ ·(ε0χe

ε~D)

= − χe1 + χe

ρf .

Here we used the fact that ~P = ε0χe ~E and ~E = 1ε~D.

On the last line, we used the fact that ~∇ · ~D = ρf andε = ε0(1 + χe).

So if no free charge is embedded in the material, thenρ = 0, and any net charge must sit on the surface. Withinsuch a dielectric, the potential obeys Laplace’s equation.

It is helpful to write the boundary conditions as

D⊥above −D⊥below = σf .

For linear dielectrics, we can write

εaboveE⊥above − εbelowE⊥below = σf ,

or

εabove∂

∂nVabove − εbelow

∂

∂nVbelow = −σf ,

and since the potential is always continuous, we also have

Vabove = Vbelow.


A sphere of homogeneous linear dielectric is placedin an otherwise uniform electric field ~E0. What is theelectric field inside the sphere?a

We will solve Laplace’s equation for Vin(r, θ) andVout(r, θ) so we can use the boundary conditions at thesurface. Our boundary conditions are

Vin(R, θ) = Vout(R, θ)

ε∂Vout∂r

∣∣∣r=R

= ε0∂Vin∂r

∣∣∣r=R

Vout → −E0r cos θ as r →∞.

Notice the ε on one side of the second boundary condi-tion and ε0 on the other side.

Our solutions will have the form

Vin(r, θ) =

∞∑l=0

AlrlPl(cos θ)

Vout(r, θ) =

∞∑l=0

(Clr

l +Blrl+1

)Pl(cos θ).

The third boundary condition implies that C1 =−E0 and Cl = 0 for l 6= 1. Then the first boundarycondition gives us

∞∑l=0

AlRlPl(cos θ) = −E0R cos θ +

∞∑l=0

BlRl+1

Pl(cos θ).

This implies that

AlRl =

BlRl+1

,

for l 6= 1 and

A1R = −E0R+B1

R2.

The second boundary condition implies that

εlRl−1Al =−(l + 1)ε0Rl+2

Bl,

for l 6= 1 and

εA1 = −E0ε0 −2B1

R3ε0.

These imply that Al = Bl = 0 for l 6= 1 and that

A1 = − 3

2 + εrE0

B1 =εr − 1

εr + 2R3E0.

This gives us

Vin(r, θ) = − 3

2 + εrE0r cos θ = − 3

2 + εrE0z.

Taking the negative of the gradient gives us the field

~E =3E0

εr + 2~E0.


Example:

3.3. LINEAR DIELECTRICS 63

Suppose the entire region below the z = 0 plane isfilled with a uniform dielectric material of susceptibilityχe. Calculate the force on a point charge a distance dabove the origin.a

Recall that ρb ∝ ρf . Here we have no free charge,so ρb = 0. The surface bound charge density is

σb = ~P · n = Pz = ε0χeEz.

Here, Ez is the electric field just inside the dielectric.

The contribution from the point charge at(x, y, z) = (0, 0, d) is

− 1

4πε0

q

r2 + d2cos θ = − 1

4πε0

qd

(r2 + d2)32

.

The z-component of the field due to the bound chargeis −σb/2ε0. Then

σb = ε0χeEz = ε0χe

[− 1

4πε0

qd

(r2 + d2)32

− σb2ε0

].

Solving for σb gives us

σb = − 1

2π

χe2 + χe

qd

(r2 + d2)32

.

If we had a conductor instead of a dielectric, the onlydifference would be that χe

2+χewouldn’t be there.

We can write the total induced charge on the sur-face of the dielectric as

qb = − χe2 + χ2

q.

We can then find the field due to σb by integrating

~E =1

4πε0

û

u2σb da.

The easier way to do it is to solve it using image charges.But we have to be careful. We have to consider thefield inside and outside the dielectric. We’ll solve theproblem piecewise and then put them together so theboundary conditions are satisfied.

To find the potential in the region z > 0, we replacethe dielectric by a point charge q′ sitting at the imagepoint (0, 0,−d). Then

V+(r) =1

4πε0

q√x2 + y2 + (z − d)2

+1

4πε0

q′√x2 + y2 + (z + d)2

.

For the potential in the region z < 0, we replace q bya point charge q′′ at z = d, then the potential in theregion z < 0 is

V−(r) =1

4πε

q√x2 + y2 + (z − d)2

.

Note that we have ε here instead of ε0.

Our boundary conditions are

V+ = V− at z = 0

ε0∂V+

∂z

∣∣∣z=0

= ε∂V−∂z

∣∣∣z=0

E⊥above − E⊥below =σbε0.

From the first boundary condition, we get

q + q′

ε0=q′′

ε.

From the second boundary condition, we get

q − q′ = q′′.

Solving this system for the two unknowns gives us

q′ =ε0 − εε0 + ε

q = − χe2 + χe

q

q′′ =2ε

ε0 + εq.

For the first one, we simplified using ε = ε0(1 + χe).

For the third boundary condition, if we plug every-thing in and solve for σb, we get

σb = − 1

2π

χe2 + χe

dq

(r2 + d2)32

,

as expected.

Since our potential satisfies the boundary condi-tions, by the uniqueness theorem, it must be the solu-tion.

The force on the original charge is

~F =1

4πε0

qq′

(2d)2z = − 1

4πε0

χe2 + χe

q2

4d2z.

Notice the ε0 since we’re outside the dielectric here.


Example:


Energy in Linear Dielectrics

Recall that the work required to charge a capacitor is

W =1

2CV 2.

A capacitor with a linear dielectric between the plates hascapacitance

C = εrCvac.

Recall that

W =1

2ε0

Ê2 dτ. (3.2)

The capacitor case suggests that with a linear dielectric,this should become

W =1

2ε0

ˆεrE

2dτ =1

2

ˆ~D · ~E dτ.

As we increase ρf by an amount ∆ρf , the polarization ofthe dielectric will change, and so will the bound charges,but we’re only interested in the work required to bring inthe free charge

∆W =

ˆ∆ρfV dτ.

The solid material objects we deal with on a day-to-day basis are either conductors or dielectrics.

Since ~∇ · ~D = ρf , we have that

∆ρf = ∆(~∇ · ~D) = ~∇ · (∆ ~D).

Then

∆W =

ˆ [~∇ · (∆ ~D)

]V dτ.

We can expand

~∇ · (∆ ~DV ) = (~∇ ·∆ ~D)V + ∆ ~D · ~∇V.

This implies that

∆W =

ˆ(~∇ ·∆ ~DV ) dτ +

ˆ∆ ~D · ~E dτ.

If we make the region large, the surface term vanishes, sowe get

∆W =

ˆ∆ ~D · ~E dτ.

This holds for any material.

For a linear dielectric, ~D = ε~E. This implies

1

2∆( ~D · ~E) =

1

2∆(εE2) = ε(∆~E) · ~E = (∆ ~D) · ~E.

Then

∆W = ∆1

2

ˆ~D · ~E dτ,

which gives us

W =1

2

ˆ~D · ~E dτ. (3.3)

What’s the difference between Eq. (3.2) and Eq. (3.3)? InEq. (3.2), all charges including bound charges are broughtin from infinity. It does not include the bending andstretching of atoms in the dielectric as you bring thecharges in. This bending and stretching is included inEq. (3.3).

We can also use

W =ε02

ˆεrE

2 dτ,

but this is only valid for linear dielectrics.

Forces on Linear Dielectrics

Like conductors, dielectric materials are also attracted toelectric fields. For an example of calculating this force,see section 4.4.4 in Griffiths.

3.4. SUMMARY: POLARIZATION AND DIELECTRICS 65

3.4 Summary: Polarization and Dielectrics

Polarization

Recall that dipole moments ~p = q~d are formed when apair of opposite charges with magnitude q are displacedby ~d from each other. Electric fields in a material causethe nuclei of atoms and their electron clouds to be dis-placed from each other, creating induced dipoles.

For induced dipoles, our dipole moment is typicallyabout

~p = α~E,

where ~E is the applied electric field and α is the atomicpolarizability of the atom. More generally, α is a rank-2tensor.

A dipole in a uniform electric field experiences atorque

~N = q~d× ~E = ~p× ~E,about its center.

An electric field applied to a material with manydipoles causes the dipoles to align. The material is thenpolarized, and its polarization is given by

~P = dipole moment per unit volume.

A polarized object has bound surface and volumecharge densities

σb = ~P · nρb = −~∇ · ~P .

The potential V due to a polarized object is just the po-tential due to these bound charge densities

V (~r) =1

4πε0

˛S

σbud~a+

1

4πε0

ˆV

ρbudτ ′.

One way to find the electric field of a polarized objectis to compute σb and ρb and then compute the potentialusing the integral above. Finally, we can find the electricfield due to the polarized object by using ~E = −~∇V .

Remember, on a surface, r has some specific valuelike r = R or r = a. The σb should depend on this spe-cific value and not on r.

If the polarization is spherically symmetric, it’s mucheasier to find the bound charges and calculate their fieldsdirectly by using Gauss’s law. Be careful when you tryto apply Gauss’s law. For example, Gauss’s law can beused for a sphere when the polarization is radial, but notwhen the polarization is uniform.

Remember that n points outward from the surfaceof a polarized object. So for example, if you have a po-larized spherical shell, then on the outer surface it points

outward, n = r and on the inner surface it points inward,n = −r.

You need to remember the divergence (at least theradial component)

~∇ · ~P =1

r2

∂

∂r

(r2Pr

).

Also, recall the following result which you may have touse at times

~∇ · uu2

= 4πδ3(~u).

Electric Displacement

In a dielectric, the total volume charge density is the sumof the bound charge density (due to polarization) and thefree charge density (everything else)

ρ = ρb + ρf .

The differential form of Gauss’s law can be writtenas

~∇ · ~D = ρf ,

and the integral form as

˛~D · d ~A = Qf,enc,

where

~D = ε0 ~E + ~P ,

is the electric displacement. Notice that the electric dis-placement is in the same direction as the electric field andthe polarization. If you’re given ~P , you can use this formof Gauss’s law to calculate the electric displacement dueto a polarized object, and then calculate the electric field.

The boundary conditions for the electric field at asurface are


ε0

~E‖above − ~E

‖below = 0.

For the electric displacement, the boundary conditionsare

D⊥above −D⊥below = σf

~D‖above − ~D

‖below = ~P

‖above − ~P

‖below.


Linear Dielectrics

A linear dielectric is a material that obeys

~P = ε0χe ~E,

where χe is the electric susceptibility of the material.Note that ~E here is the total electric field–the appliedfield causing the polarization plus the field produced bythe polarization.

For a linear dielectric,

~D = ε0(1 + χe)~E = ε~E,

whereε ≡ ε0(1 + χe),

is the permittivity of the material. In a vacuum, χe = 0,then ε = ε0. The relative permittivity or dielectric con-stant of the material is

εr ≡ε

ε0= 1 + χe.

If a space is filled with a linear dielectric, then

~D = ε0 ~Evac,

and~E =

1

ε~D =

1

εr~Evac.

where ~Evac is the electric field that would be there in theabsence of the dielectric. In other words, in a dielectric,the electric field is reduced by 1

εr.

If a parallel plate capacitor has a linear dielectricbetween its plates, its capacitance is

C = εrCvac,

where Cvac is the capacitance that the capacitor wouldhave if there was vacuum between its plates.

In a homogeneous linear dielectric, the bound chargedensity is proportional to the free charge density

ρb = − χe1 + χe

ρf .

This implies that if an object contains no free charge em-bedded in it, then ρ = 0, and any net charge sits on thesurface.

Boundary conditions at a surface for linear dielectricsare

D⊥above −D⊥below = σf

εaboveE⊥above − εbelowE⊥below = σf

εabove∂

∂nVabove − εbelow

∂

∂nVbelow = −σf

Vabove − Vbelow = 0.

We can use separation of variables to solve some problemsinvolving linear dielectrics. The process is the same aswith conductors, but now the boundary conditions mayinclude ε and ε0 as shown above. May also be able tosolve linear dielectrics problems using the image method.

For linear dielectrics,

W =1

2

ˆ~D · ~E dτ =

ε02

ˆεrE

2 dτ.

Chapter 4

Magnetostatics

4.1 Lorentz Force Law

A charge Q moving with velocity ~v in a magnetic field ~Bexperiences a force

~F = Q~v × ~B.

If it is moving in both magnetic and electric fields, thenthe force is

~F = Q~E +Q~v × ~B.

These relations are completely empirical.

The typical motion of a charged particle in a mag-netic field is circular, with the magnetic force providingthe acceleration toward the center of the circle. Imag-ine a uniform field ~B pointing into the page. The chargemoves counterclockwise with speed v, around a circle ofradius R. The magnetic force points inward and has afixed magnitude of QvB. Then

QvB =mv2

R2.

This gives usp = QBR,

where p = mv is the momentum of the particle. This isthe cyclotron formula.

If the only force on a charged particle is due to a mag-netic field, and the magnetic field is perpendicular tothe velocity of the particle, then the particle will exe-cute uniform circular motion. Any segment of its pathwill be a portion of the arc of a circle. Generally, you

can simplify these problems using qvB = mv2

R .

Tip:

Suppose a charge Q moves an amount d~l = ~v dt.Then the magnitude of the work done is

dW = ~F · d~l = Q(~v × ~B

)· ~v dt = 0.

This shows us that magnetic forces do no work.

Magnetic forces do no work.

Tip:

67

68 CHAPTER 4. MAGNETOSTATICS

Suppose we have a particle moving in uniform elec-tric and magnetic fields. The fields are perpendicularwith ~B pointing in the x direction and ~E pointing inthe z direction. If a particle of charge q is initially atrest at the origin, what is its motion?a

The particle is initially at rest. Then the electricfield causes it to move in the z direction. But as soonas it gains velocity, the magnetic field begins exerting aforce on it. By the right hand rule, this force initiallypushes the particle in the y direction. Since magneticforce is always at right angles to the direction of motion,the magnetic force makes particle want to move in a cir-cle, but depending on where it is along the curve, theelectric field either amplifies or diminishes this circularmotion.

Since there is no motion in the x-direction, we canwrite the particle’s trajectory as

~r(t) = (0, y(t), z(t)),

then its velocity and acceleration are

~v = (0, y, z)

~a = (0, y, z).

Taking the cross product of the velocity with themagnetic field ~B = (B, 0, 0) gives us

~v × ~B = Bz y −By z.Then Newton’s law combined with the Lorentz force lawgives us

~F = q(~E + ~v × ~B

)= ma

q (Ez +Bz y −By z) = m(yy + zz)

We can separate the x and y components to get thepair of coupled ODEs

my = qBz

mz = q(E −By).

Substituting the cyclotron frequency ω = qBm , lets us

write these as

y = ωz

z = ω

(E

B− y).

These coupled ODEs can be solved by differenti-ating one and plugging it into the other. The generalsolutions are

y(t) = C1 cosωt+ C2 sinωt+E

Bt+ C3

z(t) = C2 cosωt− C1 sinωt+ C4.

Applying the initial conditions (0, y(0), z(0)) = (0, 0, 0)and (0, y(0), z(0)) = (0, 0, 0), gives us

y(t) =E

ωB(ωt− sinωt)

z(t) =E

ωB(1− cosωt)

Setting R = EωB allows us to write y − ωRt =

−R sinωt and z − R = −R cosωt. Squaring both andadding them gives us

(y −Rωt)2 + (z −R)2 = R2.

This is a circle of radius R centered at (0, Rωt,R). Thatis, the center of the circle is moving in the y directionwith speed Rω = E

B . The particle itself is moving asif on the rim of a wheel rolling in the y direction. Theparticle’s trajectory is then a series of cycloids.


Example:

The current in a wire is the charge per unit timethat passes a given point along the wire. From ~F =Q~v× ~B, we see that if we change the signs of both Q and~v, we get the same thing. So a positive charge moving tothe right is equivalent to a negative charge moving to theleft.

The unit of current is the Ampere defined as 1Coulomb per second

1 A =1 C

1 s.

A line charge traveling at speed v, as in a wire, is a

current I = λv, where λ is the linear charge density. Butsince current is actually a vector, we write this as

~I = λ~v.

This equation refers only to the moving negative chargesas in a wire. If you had both positive and negative chargesmoving (e.g. ions in a tube instead of a metal wire), then

you would write ~I = λ+~v+ + λ−~v−.

The magnetic force on a small segment of wire car-rying a current is

d~F = ~v × ~Bdq.

4.1. LORENTZ FORCE LAW 69

If we integrate, then the magnitude of the magnetic forceon a general segment of wire carrying a current is

~F =

ˆ~v × ~B dq =

ˆ (~v × ~B

)λ dl.

Substituting ~I = λ~v gives us

~F =

ˆ (~I × ~B

)dl.

Since ~I and d~l are in the same direction for a wire, wecan write this as

~F =

Î(d~l× ~B

).

Finally, if the current is constant, which it is in magneto-statics, we can pull the I outside the integral as

~FB = I

ˆ (d~l× ~B

).

For a straight wire, this is just

~FB = I~L× ~B.

Suppose you have a rectangular loop of wire (hang-ing vertically) of width a with the top part of the loopin a uniform magnetic field that points into the page,as shown below. The bottom part of the loop is outsidethe region of the magnetic field. What is the currentI in the loop such that the upward magnetic force onthe loop balances the downward gravitational force onthe mass m attached to the bottom of the rectangularloop? a

m

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

a

The magnitude of the upward magnetic force onthe horizontal segment of wire that is in the magneticfield is F = IBa. There is also a magnetic force on eachvertical segment of the wire—the upper parts of whichextend into the magnetic field. However, the force onthe left wire is to the left, and the force on the right wireis to the right, so they cancel each other. The down-ward gravitational force on the mass is mg, so the loopis suspended motionless when IBa = mg or when thecurrent is

I =mg

Ba.

What if we increase the current in the wire? If thecurrent is increased, the upward magnetic force will ex-ceed the downward gravitational force, and the loop willrise. This makes it look like the magnetic force is doingwork. But we know that the magnetic force never doeswork, so what is happening?

When the loop starts to rise, the charges movingin the upper horizontal wire are no longer moving hor-izontally. They now have a vertical component to theirtotal velocity ~v. If the horizontal velocity of the chargesis ~w and the vertical velocity of the charges is ~u, then~v = ~u+ ~w. The magnetic force is always perpendicularto the total velocity. In this case, the magnitude of themagnetic force due to the horizontal component of thevelocity is qwB = λawB = IBa. The magnitude ofthe magnetic force due to the vertical component of thevelocity is quB.

The net force on the wire is still perpendicular to ~v,so there’s no work being done. Where is the energy thatcauses the loop to rise coming from then? Notice thatthe horizontal component of the magnetic force, quB,is directed opposite the flow of the current. That is,it opposes the flow of the current. So the battery thatis producing the current must work harder to maintainthe current. That is, the missing work is actually beingdone by the battery.

The horizontal force on the top wire is F = λauB.In an infinitesimal time dt, the charges in the wire movea horizontal distance w dt, so the work done by the bat-tery is

Wbattery = λaB

ûw dt = IBah.


Example:


The surface current density ~K is used to describethe flow of charge over a surface. Suppose you have somesurface on which charge is flowing. Consider an infinites-imal strip of width dl⊥, along the direction in which thecharge is flowing. If the current flowing in this strip is d~I,then the surface current density is

~K =d~I

dl⊥.

We can also write this in terms of the surface charge den-sity. If we have a surface charge density σ moving withvelocity ~v, then

~K = σ~v.

In general, ~K varies over a surface. The magnetic forceon the surface is

~FB =

ˆ (~v × ~B

)σ da =

ˆ (~K × ~B

)da.

For charge flowing in a volume, we talk about thevolume current density ~J . Consider a tube orientedparallel to the flow of charge with infinitesimal cross-section da⊥. Then if the current in the tube is d~I, thevolume current density is

~J =d~I

da⊥.

We can also write this in terms of the volume charge den-sity. If we have a volume charge density ρ moving withvelocity ~v, then

~J = ρ~v.

In general, ~J varies in a volume. The magnetic force onthe volume is

~FB =

ˆ (~v × ~B

)ρ dV =

ˆ (~J × ~B

)dV.

Suppose we have a current I that is uniformlydistributed in a cylindrical wire of radius a. What isthe volume current density? a

The area perpendicular to the flow is the wire’scross-sectional area πa2, so the volume current densityis just

J =I

πa2.

aThis is example 5.4.a in Griffiths.

Example:

Consider the same example with the cylindricalwire but now let the volume current density be J = ks.What is the total current? a

We know that J = dIda⊥

, so

dI = J da⊥.

Since the volume charge density is not uniform, wehave to integrate. What is da⊥? We start by lookingat the whole area a⊥ perpendicular to the flow of cur-rent. This is just the circular cross-section of the wire.In polar coordinates, a small element of this disk hasarea da⊥ = s ds dφ. We now integrate over the entiredisk to get the total current I.

I =

ˆJ da⊥

=

ˆ(ks)s ds dφ

= 2πk

ˆ a

0

s2 ds

=2πka3

3.

aThis is example 5.4.b in Griffiths.

Example:

In the last example, we calculated the current cross-ing a surface S by integrating

I =

ˆS

J da⊥ =

ˆS

~J · d~a.

The total charge that leaves a volume V per unit time is

˛S

~J · d~a =

ˆV

~∇ · ~J dV.

Because charge is conserved, we know that the chargeleaving a volume reduces the charge left in the volume,and we get

ˆV

~∇ · ~J dV = − d

dt

ˆV

ρ dV = −ˆV

∂ρ

∂tdV.

The negative sign is there because the charge in V is beingdiminished.

This local charge conservation implies the continu-ity equation

~∇ · ~J = −∂ρ∂t.

4.2 Biot-Savart Law

Magnetostatics deals with steady currents. For steadycurrents, the charge is not piling up anywhere, and so

4.2. BIOT-SAVART LAW 71

ρ is constant. That is, ∂ρ∂t = 0, and so the continuity

equation becomes~∇ · ~J = 0.

The Biot-Savart law gives the magnetic field of aline current

~B(~r) =µ0

4π

ˆ ~I × u dl′

u2,

whereµ0 = 4π × 10−7 N/A2,

is the permeability of free space. The SI unit of mag-netic field is the Tesla (T) defined as

1 T =N

Am.

The cgs unit is the “gauss” equal to 10−4 Tesla. Earth’smagnetic field is about half a gauss.

Id~l ′

P

~u

In magnetostatics, the current is steady, so we canwrite the Biot-Savart law as

~B(~r) =µ0I

4π

ˆd~l ′ × uu2

.

Use the Biot-Savart law to calculate the magneticfield at the center of a circular loop of current I.

From the center of a circle, or any portion thereof,the line element d~l ′ is always perpendicular to u. Soif our circle of current is lying in the xy-plane andcentered at the origin, then

d~l ′ × u = dl′ z,

and the Biot-Savart law becomes

~B(~r) =µ0I

4π

ˆ1

u2dl′ z.

But the distance from the line element to the originis a constant u = R, so u2 = R2 can be pulled outsidethe integral

~B(~r) =µ0I

4πR2

ˆdl′ z.

But this integral just gives the circumference of thecircle, so

~B(~r) =µ0I

4πR22πR z =

µ0I

2Rz.

Example:


Find the magnetic field a distance s from a longstraight wire carrying a steady current I. a

z

x

y

I

d~l ′

~uθ

P

We start by putting the wire in the z direction asshown above. Because we will be doing a cross product,we need to follow a meticulous process to ensure thatwe have the right vectors.

Our field point P is at

~r = (s, 0, 0).

Our current element is at ~r ′ = (0, 0, z). Since zs = tan θ,

we can write this as

~r ′ = (0, 0, s tan θ).

This gives us

~u = ~r − ~r ′ = (s, 0,−s tan θ).

Then

u2 = s2 sec2 θ =s2

cos2 θ,

and our unit vector is

u =(s, 0,−s tan θ√s2(1 + tan2 θ)

= (cos θ, 0,− sin θ).

Our differential vector is

d~l ′ = dz z = d~r ′ = (0, 0, s sec2 θ)dθ.

We are now ready to take the cross product. Whenwe do so, we get

d~l ′ × u =

∣∣∣∣∣∣∣∣x y z

0 0 sec2 θ dθ

cos θ 0 − sin θ

∣∣∣∣∣∣∣∣ =s

cos θdθ y.

Plugging this result into the Biot-Savart law givesus

~B(~r) =µ0I

4π

ˆ θ2

θ1

cos2 θ

s2

s

cos θdθ y

=µ0I

4πs

ˆ θ2

θ1

cos θ dθ y

=µ0I

4πs(sin θ2 − sin θ1) y.

For an infinite line, we have θ1 = −π2 and θ2 = π2 , then

~B =µ0I

2πsy =

µ0I

2πsφ.

We can replace y by φ because of the cylindrical sym-metry.


Example:

In the example above, we found that the magneticfield a distance s from an infinite wire carrying a steadycurrent I is

~B =µ0I

2πsφ.

Consider now the electric field of an infinite line of charge.We get

~E =1

4πε0

ˆλu

u2dz

=λ

4πε0

ˆ θ2

θ1

(cos θ, 0,− sin θ)s dθ

cos2 θ

cos2 θ

s2

=λ

4πε0s

ˆ θ2

θ1

(cos θ, 0,− sin θ) dθ

=λ

4πε0s(sin θ2 − sin θ1, 0, cos θ2 − cos θ1) .

So for the electric field, we generally have components intwo directions. For an infinite line, we have θ1 = −π2 andθ2 = π

2 , then

~E = − λ

4πε0s(2, 0, 0) = − λ

2πε0sx.

4.2. BIOT-SAVART LAW 73

What is the force between two parallel wires sepa-rated by a distance d and carrying currents I1 and I2?a

We know that the field at wire 2 due to wire 1 is

B2 =µ0I12πd

.

The direction can be found using the right hand rule.

Then the force on wire 2 due to wire 1 is

F2 = I2

ˆd~l× ~B2 = I2

µ0I12πd

ˆdl.

But this force is infinite for wires of infinite length, solet’s consider the force per unit length instead. Then

F

l=µ0I1I2

2πd.

By the right-hand rule, the force is attractive if I1and I2 are in the same direction.


Example:

What is the magnetic field a distance s above thecenter of a circular loop of radius R carrying a currentI? a.

y

z

xd~l ′

P

~u

φ θ

d ~Bθ

I

θ

θP

~u

d ~B

Here, d ~B sweeps out a cone, so by symmetry, theonly component of ~B remaining will be in the z direc-

tion. Our field point P is at

~r = (0, 0, z).

Our current element is at

~r ′ = (R cosφ,R sinφ, 0) .

Then our differential element is

d~l ′ = d~r ′ = (−R sinφ,R cosφ, 0) dφ.

Then

~u = ~r − ~r ′ = (−R cosφ,−R sinφ, z) .

The unit vector is

u =(−R cosφ,−R sinφ, z)√R2 cos2 φ+R2 sin2 φ+ z2

=(−R cosφ,−R sinφ, z)√

R2 + z2.

Taking the cross product gives us

d~l ′ × u =Rz cosφdφ x+Rz sinφdφ y +R2 dφ z√

R2 + z2.

Plugging this into the Biot-Savart law gives us

~B =µ0I

4π

ˆRz cosφ x+Rz sinφ y +R2 z

(R2 + z2)32

dφ

=µ0IR

2

2(R2 + z2)32

z.


Example:

For surface currents, the Biot-Savart law is

~B(~r) =µ0

4π

ˆ ~K(~r ′)× uu2

da′.

For volume currents, it is

~B(~r) =µ0

4π

ˆ ~J(~r ′)× uu2

dV ′.


There is no Biot-Savart law for a point charge since asingle moving charge is not a steady current.

4.3 Ampere’s Law

Recall that the magnetic field a distance s from wire car-rying a current I is

B =µ0I

2πs.

The magnetic field lines are circles around the wire. Sup-pose we integrate ~B over a circle of radius s about thewire,

¸~B · d~l. Then since the field along the loop is par-

allel to the loop, we know that ~B · d~l = B dl. So ourintegral is˛

~B · d~l = B

˛dl =

µ0I

2πs2πs. = µ0I.

So integrating the magnetic field over a circular loop en-closing a wire gives us µ0 times the current in the wire.

We can generalize this result to loops of any shape.The magnetic field of a current carrying wire in vectorform is

~B =µ0I

2πsφ.

The line element in cylindrical coordinates is

d~l = ds s+ s dφ φ+ dz z.

Then~B · d~l =

µ0I

2πss dφ =

µ0I

2πdφ.

So our integral becomes˛~B · d~l =

µ0I

2π

˛dφ.

This result is independent of the shape of the loop. Thisassumes that the wire is at the origin. If we integratearound a loop, of any shape, that completely encloses theorigin, then φ goes from 0 to 2π. Then our result is˛

~B · d~l = µ0I.

If the current is not enclosed by the loop, i.e., we inte-grate over a loop that does not include the origin, then¸dφ = 0.

This result also holds for bundles of wires. In general,˛~B · d~l = µ0Ienc,

where Ienc is the current enclosed by the closed “Am-perian” loop that we are integrating over. This is theintegral form of Ampere’s law. It is valid for arbitrarybut steady currents.

If we are given a current density, we can calculatethe current enclosed in an Amperian loop encircling thecurrent density with

Ienc =

ˆ~J · d~a.

Here, we integrate over the surface that is bounded bythe Amperian loop.

From Stokes’ theorem, we get

˛~B · d~l =

ˆ (~∇× ~B

)· d~a = µ0

ˆ~J · d~a.

This implies the differential form of Ampere’s law

~∇× ~B = µ0~J .

The Biot-Savart law

~B(~r) =µ0

4π

ˆ ~J(~r ′)× uu2

dV ′,

gives us the magnetic field at a point ~r due to a volumecurrent density ~J . Taking the divergence of both sidesgives us

~∇ · ~B =µ0

4π

ˆ~∇ ·(~J × u

u2

)dV ′.

Next, we apply the vector product rule

~∇ ·(~A× ~B

)= ~B ·

(~∇× ~A

)− ~A ·

(~∇× ~B

),

to get

~∇ ·(~J × u

u2

)=u

u2·(~∇× ~J

)− ~J ·

(~∇× u

u2

).

But we know that ~∇× ~J = 0 because ~∇ is of the x, y, zcoordinates, but ~J depends only on the x′, y′, z′ coor-dinates. We also know that the curl of u

u2 is zero, soapparently the entire quantity above is zero. That is, thedivergence of the magnetic field is zero

~∇ · ~B = 0.

Ampere’s law in integral form can be used to find themagnetic field of certain current distributions in much thesame way that Gauss’s law in integral form can be usedto find the electric field of certain charge distributions. InGauss’s law, we have a closed surface integral, so we usea Gaussian surface. In Ampere’s law, we have a closedline integral, so we use an Amperian loop.

Ampere’s law is always true for steady currents, butit is not always useful. There are four symmetric currentconfigurations that can be solved by Ampere’s law

• Infinite straight lines• Infinite planes• Infinite solenoids• Toroids

4.3. AMPERE’S LAW 75

Find the magnetic field due to an infinite straightline current I using Ampere’s law.

For an infinite line current, our Amperian loop is acircle of radius r about the infinite line current. By theright-hand-rule, we know that the magnetic field of aline current is circumferential. Therefore, the magneticfield is parallel to our Amperian loop everywhere alongthe loop. In other words, ~B ‖ d~l, so ~B · d~l = B dl. Thecurrent enclosed by our loop is I, so Ampere’s law gives

us˛~B · d~l = µ0Ienc

B

˛dl = µ0I

B(2πs) = µ0I

B =µ0I

2πs.

So our magnetic field is

~B =µ0I

2πsφ.

Example:

Find the magnetic field due to an infinite planecurrent ~K using Ampere’s law.

Suppose the xy-plane is filled with a sheet of cur-rent ~K = Kx flowing in the x direction. We use anAmperian rectangle of width l and place it perpendic-ular to the current flow (so that current flows throughthe loop). We place the loop so that half of it is abovethe plane and half of it is below the plane.

Our rectangular loop now encloses a current

Ienc = Kl.

What is the direction of the magnetic field due toa plane current? Consider, briefly, only the strip of cur-rent that passes through your loop. We can think ofthis strip as a flat wire and use the right-hand rule todeduce the field due to it. We note that above the plane,the field points in the −y direction and below the planeit points in the y direction. We can ignore the mag-netic field to either side because they are “cancelled”

out by the infinite other strips to either side. We con-clude that the magnetic field is parallel to the plane, inthe −y direction for z > 0, and in the y direction forz < 0.

Since the magnetic field is parallel to the plane, itis parallel to the top and bottom of our rectangular Am-perian loop and perpendicular to the sides of our loop.Then Ampere’s law gives us

˛~B · d~l = µ0Ienc

B

˛dl = µ0Kl

B2l = µ0Kl

B =µ0K

2.

So, after adding the direction vectors,

~B =

µ0K

2 y for z < 0

−µ0K2 y for z > 0.

Example:


Find the magnetic field due to an infinite solenoidusing Ampere’s law.

b

a

l

Suppose we have a cylinder of radius R. A wire istightly wrapped around the cylinder so that there are nturns of wire per unit length of the cylinder. When cur-rent flows through the wire, it flows around the cylinder,and you have a solenoid.

Alternatively, you can think of there being a surfacecurrent K = nI flowing through a sheet of aluminumthat has been wrapped around the cylinder. Either way,the result is the same.

For a solenoidal current, the magnetic field is lon-gitudinal, i.e., parallel to the solenoid. This conclusioncan be reached via a variety of arguments. To find themagnitude of the field, we use two Amperian rectanglesof height l. Notice in the image how these rectanglesare oriented so that any current through them is per-pendicular to the plane of the rectangle.

For the outermost loop, there is no current en-closed. We know the magnetic field is longitudinal, so

the vertical sides of our loop are parallel to the field, andthe horizontal sides are perpendicular. So Ampere’s lawgives us

˛~B · d~l = lBa − lBb = µ0Ienc = 0.

When we integrated clockwise around the loop, we gotlBa from the left side of the loop, 0 from the top −lBbfrom the right side, and 0 from the bottom side. So

Ba = Bb.

This implies that the magnetic field outside the solenoiddoes not depend on the distance from the solenoid.Since we know the field goes to zero at infinite distancefrom the solenoid, this implies that the field is zero ev-erywhere outside the solenoid

~B = 0, s > R.

Our second Amperian loop is half inside and halfoutside the solenoid. Suppose the left side of the loopis at a distance s from the center of the solenoid, andthe right side is at a distance s + k. It doesn’t matterwhat k is. Since the field outside the solenoid is zero,when we integrate the field over the right side of theloop, which is outside, there will be no contribution. Infact, the only contribution comes from the left side ofthe loop

˛~B · d~l = Bl = µ0Ienc = µ0nIl.

So~B = µ0nI z, s < R.

inside the solenoid.

Example:

4.4. THE VECTOR POTENTIAL 77

Find the magnetic field due to a toroid using Am-pere’s law. The toroid consists of a circular ring ofarbitrary but consistent cross-section around which Ntotal turns of wire carrying a current I are wrapped.

A toroid is essentially just a solenoid bent into acircle. Like the magnetic field of a solenoid is longitu-dinal, the magnetic field of a toroid is circumferentialboth inside and outside the coil.

The magnitude of the field is obtained using Am-pere’s law with an Amperian circle of radius s that is

in the same plane as the toroid. That is, the Amperianloop and the toroid share the same axis. If s is smallenough so that it does not include any part of the toroid,then the enclosed current is zero and the magnetic fieldis zero. If s is large enough that the Amperian loop liesentirely within the toroid, then the enclosed current isNI, and Ampere’s law gives us

˛~B · d~l = B(2πs) = µ0Ienc = µ0NI.

If s is larger than the outer radius of the toroid, thenthe net current enclosed is zero.

So inside the toroid, the magnetic field is

~B =µ0NI

2πsφ,

and everywhere outside the toroid, the field is zero.

Example:

Be careful when applying Ampere’s law to cylindricalthings because cylinders can carry longitudinal (e.g.a wire) or solenoidal currents. The magnetic fields ofthese two are very different. For longitudinal currents,the magnetic field lines are circles wrapped around thecylinder. Your Amperian loop needs to be a loop aboutthe same axis. For solenoidal currents (i.e. the currentflows circularly over the cylinder’s surface), the mag-netic field lines are straight and parallel to the cylin-der. Then you use a pair of Amperian rectangles—onecompletely outside the solenoid and one half inside andhalf outside the solenoid.

Tip:

4.4 The Vector Potential

In electrostatics, ~∇× ~E = 0 implies that the electric fieldcan be written as the derivative of a scalar potential, asin, ~E = −~∇V . Similarly, ~∇ · ~B = 0 implies that

~B = ~∇× ~A,

where ~A is the magnetic vector potential.

This means we can rewrite Ampere’s law using

~∇× ~B = ~∇×(~∇× ~A

)= ~∇

(~∇ · ~A

)−∇2 ~A = µ0

~I.

We can add any curl-less function to ~A, and we getthe same magnetic field ~B when we take its curl. That is,taking ~A to ~A ′+∇f gives us the same ~B. This gives usthe freedom to set ~∇ · ~A = 0. For example, suppose wehave a vector potential ~A ′ such that ~∇ · ~A 6= 0. Then we

can set ~A = ~A ′+ ~∇λ and we get ~∇ · ~A = ~∇ · ~A ′+∇2λ.We can satisfy ~∇ · ~A = 0 by setting ∇2λ = −~∇ · ~A. Butthis is just Poisson’s equation.

Recall Poisson’s equation from electrostatics

∇2V = − ρ

ε0.

If ρ = 0 at infinity, then the solution is

V =1

4πε0

ˆρ

udV.

Similarly, if ~∇ · ~A ′ goes to zero at infinity, then

λ =1

4π

ˆ ~∇ · ~A ′

udV ′.

If ~∇ · ~A ′ = 0, then

∇2 ~A = −µ0~J .

This is again just Poisson’s equation, so if ~J → 0 at ∞then

~A(~r) =µ0

4π

ˆ ~J(~r ′)

udV ′.

Remember that ∇2 ~A is really three equations—onein each component of ~A.

For line current ~I, we have

~A(~r) =µ0

4π

ˆ ~I

udl′.

For steady line current, we have

~A(~r) =µ0I

4π

ˆ1

ud~l ′.


For surface current, we have

~A(~r) =µ0

4π

ˆ ~K

uda′.

The vector potential ~A is not as useful as the scalarpotential V is in electrostatics. First, it is a vector, so itis harder to work with. Second, it is not easy to interpretit physically. It can be interpreted as momentum per unitcharge.

A spherical shell of radius R with a uniform sur-face charge density σ is spinning with angular velocity~ω. What is the vector potential ~A inside and outsidethe shell?a

The integration is easiest if we set our measuringpoint ~r to be on the z-axis such that ~ω is in the xzplane and makes an angle ψ with the z-axis.

The vector potential is given by

~A(~r) =µ0

4π

ˆ ~K

uda′,

where ~K = σ~v.

By the law of cosines, the magnitude of the vector~u from the integration point

~r ′ = (R sin θ′ cosφ′, R sin θ′ sinφ′, R cos θ′),

to the field point

~r = (0, 0, r),

isu =

√R2 + r2 − 2Rr cos θ′.

Our surface element is

da′ = R2 sin θ′ dθ′ dφ′.

The velocity of a point in a rotating rigid body is

~v = ~ω × ~r ′.

If we expand this cross product, we get a bunch of termsand many of them contain sinφ′ or cosφ′. Since we willbe integrating this from 0 to 2π, those terms with sinφ′

or cosφ′ will end up being zero so we can ignore themfrom the start and say that

~v = −ωR sinψ sin θ′ y.

Then plugging in ~K = σ~v and da′ gives us

~A(~r) = −µ0ωR3σ sinψ

2

ˆ π

0

sin θ′ cos θ′√R2 + r2 − 2Rr cos θ′

dθ′ y.

To evaluate this integral, we can make the substitutiont = cos θ′, then we end up with

~A(~r) = W (R2 + r2 +Rr)|R− r|y−W (R2 + r2 −Rr)(R+ r)y.

where

W =µ0ωR

3σ sinψ

6R2r2.

If ~r is inside the sphere, then R > r, and

~A(~r) = −µ0ωRσ sinψ

3y, for r ≤ R.

If ~r is on the outside, then

~A(~r) = −µ0ωR4σ sinψ

2r2y, for r ≥ R.

We can write this a little more compactly by noting that

~A(~r) =µ0Rσ

3~ω × ~r, r ≤ R,

=µ0R

4σ

3r3~ω × ~r, r ≥ R.

We now rotate the solution back to the more naturalcase where ~w is along the z axis and ~r = (r, θ, φ). Then

~A(~r) =µ0Rσωr sin θ

3φ, r ≤ R,

=µ0R

4σω sin θ

3r2φ, r ≥ R.

We can calculate ~B by taking the curl of ~A. Insidethe spherical shell, we get

~B =µ0Rωσ

3

[2 cos θ r − 2 sin θ θ

]=

2µ0Rωσ

3z.

Notice that this is a uniform field in the direction of ~ω.


Example:

4.4. THE VECTOR POTENTIAL 79

What is the vector potential inside and outside aninfinite solenoid of radius R constructed with n turns ofwire per unit length each carrying current I.a

To solve this, we’ll use the fact that

˛~A · d~l =

ˆ(~∇× ~A) · d~a =

ˆ~B · d~a = ΦB ,

is the magnetic flux through the surface bounded by theloop we are integrating over. Recall Ampere’s law

˛~B · d~l = µ0Ienc.

If we replace ~B by ~A and replace µ0Ienc by ΦB , thenthe two are the same.

For the solenoid, we know that ~Bin = µ0nIz forr < R and ~Bout = 0 for r > R. Using an Amperiancircle of radius s < R centered at the center of thesolenoid, we get

˛~A · d~l = | ~A|2πs =

ˆ~B · d~a = µ0nIπs

2.

This gives us

~A =1

2µ0nIs φ, s < R,

inside the solenoid. Repeating for s > R, we get

~A =µ0nIR

2

2sφ, s > R.


Example:

The vector potential ~A is usually in the same directionas the current ~I.

Tip:

Magnetostatic Boundary Conditions

Recall that the divergence of the magnetic field is zero,i.e., ~∇ · ~B = 0. The integral form of this is˛

~B · d~a = 0.

By considering a surface with surface current ~K and us-ing a small Gaussian box, we get the boundary condition

B⊥above = B⊥below.

Using the integral form of Ampere’s law and an Ampe-rian loop that crosses through the surface, we can showthat

B‖above −B

‖below = µ0K.

We can combine these boundary conditions as

~Babove − ~Bbelow = µ0

(~K × n

),

where n is the unit normal vector pointing in the direc-tion from below to above.

Recall that the potential V is always continuous.Similarly, the vector potential ~A is always continuous

~Aabove = ~Abelow.

The derivative of the vector potential is discontinuousacross a surface current

∂

∂n~Aabove −

∂

∂n~Abelow = −µ0

~K.

Multipole Expansion of Vector Potential

Recall the multipole expansion of V ~r)

V (~r) =1

4πε0

∞∑n=0

1

rn+1


Using the same approach as we used with V , we can ob-tain a multipole expansion of the vector potential

~A(~r) =µ0I

4π

∞∑n=0

1

rn+1

˛(r′)nPn(cos θ′) d~l ′.

This gives the vector potential at ~r due to a current loop(which is being integrated over) carrying current I.

The first term of the multipole expansion is themonopole term

~Amon(~r) = 0.

It is zero because¸d~l = 0 is the total displacement

around a closed loop.

The second term in the multipole expansion is thedipole term

~Adip(~r) =µ0I

4πr2

˛r′ cos θ′ d~l ′ =

µ0I

4πr2

˛(r · ~r ′) d~l ′.

We can write this in the form

~Adip(~r) =µ0

4π

~m× rr2

,

where

~m = I

ˆd~a = I~a,


is the magnetic dipole moment, and ~a is the vectorarea of the loop.

For a flat loop, the vector area a has a magnitudeequal to the ordinary area enclosed by the loop, andits direction is perpendicular to the plane of the loop.

Tip:

The actual magnetic field of something typically con-tains a dipole term, quadrupole term, octopole term, andso on. Provided that it is not zero, the dipole term typ-ically dominates and gives a good approximation of the

field at distances much larger than the diameter of thecurrent loop.

Often, the easiest way to calculate the vector poten-tial and field of a pure dipole is to put the dipole momentvector ~m at the origin and pointing in the z direction.Then in spherical coordinates, the vector potential is

~Adip(~r) =µ0

4π

m sin θ

r2φ.

and the magnetic field is then ~Bdip(~r) = ~∇× ~A or

~Bdip(~r) =µ0m

4πr3


).

Find the magnetic dipole moment of the bent wireloop shown below. The wire is carrying a current I.Assume each straight segment has length w.a

y

z

x

To solve this, we’ll use the principle of superposi-tion by breaking the loop, along the bend, and creat-ing two square loops. Notice that along the two new

lengths that have been added, the currents flow in op-posite direction, so when it’s all added together, thesecontributions cancel, and we get the original result.

Our total dipole moment is now easily calculatedas

~m = Iw2y + Iw2z.

It has magnitude

m =√

2Iw2,

and points along the line z = y.


Example:

4.5 Magnetic Fields in Matter

Magnetization

Consider a magnetic dipole formed by a rectangular loopof current. Suppose the magnetic field ~B is in the z direc-tion, and the current loop, with sides a and b is centeredat the origin with the sides of length b parallel to the xaxis. The current loop is tilted such that its plane makesan angle θ with the xy plane. Suppose the current in theloop is counterclockwise if viewed from above.

Remember the formula ~F = q~v × ~B. If the currentloop was laying flat in the xy plane, the force exertedon each segment of wire would be directed outward, and

since each side has a paired opposite side, the total forceon the current loop is zero. However, if the current loop istilted as described, then the outward forces on the sides oflength b are still parallel to the xy plane but they are nolonger parallel to the plane of the loop. These two forcescreate a torque on the current loop, rotating it until it isagain parallel to the xy plane.

Here is a top-down view of the current loop:

4.5. MAGNETIC FIELDS IN MATTER 81

x

y

I b

b

aa

Below is a side-view of this current loop.

y

z

~F

~F

~m

θ

θ

This torque is ~N = IabB sin θ x, where the magni-tude of the force on the arms of length b is F = IbB. Thisof course comes from the formula ~F = I~L× ~B. Note thatwe can write the torque on this magnetic dipole as

~N = ~m× ~B,

where ~m = I~a is the magnetic dipole moment. Recallthat ~a is the normal vector with a magnitude ab equalto the area of the loop. This equation for the torque ona magnetic dipole is true for current loops of any shape.However, it is only valid for uniform fields, although evenin a non-uniform field, it is true for sufficiently small cur-rent loops.

Electrons are moving charges. Think of the electronsin a material as tiny currents about the atoms. Sincethese “orbiting” electrons form small current loops, theyare magnetic dipoles. In a typical material, since electronorbits are oriented in random directions, these magneticdipoles cancel each other out. However, if you apply amagnetic field to the material, the magnetic dipoles allalign themselves with the field. The material is then mag-netically polarized or magnetized.

For some materials, the magnetization is parallel tothe applied magnetic field ~B. Such materials are param-agnetic. In other materials, the magnetization is in thedirection opposite the field. Such materials are diamag-netic. In still other materials, the magnetization persistsafter the external field is removed. Such materials arecalled ferromagnetic.

In many materials, the magnetization cancels be-cause the Pauli exclusion principle locks electrons intopairs with opposite spin. However, paramagnetism occursin materials composed of atoms having an extra unpairedelectron.

If the magnetic field ~B is uniform, then the net forceon a current loop is zero. This can be seen in

~F = I

˛d~l× ~B.

Since ~B is constant, it can be pulled outside the integral

~F = I

(˛d~l

)× ~B.

But¸d~l is just the net displacement, which is zero around

a closed loop, and so ~F = 0. If the field is not uniform,then a current loop may experience a net force. For ex-ample, if you have a current loop just outside the end ofa solenoid, where the magnetic field is not at all uniform,the current loop will be attracted to or repelled from thesolenoid depending on the direction of the current.

For an infinitesimal current loop with magneticdipole moment ~m, the force on the loop due to a magneticfield ~B is

~F = ~∇(~m · ~B

).

Notice that this quantity is zero if ~B is uniform.

When electromagnetism was first being developed,the parallels between magnetostatics and electrostaticssuggested that the magnetic dipole was formed by twomagnetic monopoles (a north monopole and a southmonopole) much like in electrostatics a dipole is a pairof opposite charges. This is the Gilbert model of mag-netostatics, and it gives good results in the far-field. Touse it, just treat magnetostatics like electrostatics, butreplace ~P with ~M , 1

ε0with µ0, and ~E with ~B. We now

know that the Gilbert model is wrong. The correct model,the Ampere model treats magnetic dipoles as currentloops instead of a pair of magnetic monopoles.

Consider an electron orbiting a nucleus. Suppose itorbits classically with a radius R and a period T = 2πR

v .This is not a steady current, but the period is shortenough that it can be modeled as such

I =e

T=

ev

2πR.

Then the orbital dipole moment is

~v = −1

2evRz.

If this system is placed in a magnetic field, it experiencesa torque ~N = ~m× ~B.


Without the external magnetic field, the electricforce equals the centripetal force needed to keep the elec-tron in orbit

1

4πε0

e2

R2=mev

2

R.

If an external magnetic field is applied perpendicular tothe electron’s orbital plane, then there is an additionalforce on the electron so that

1

4πε0

e2

R2+ evB =

mev2

R,

where v is the new speed of the electron. Subtracting thefirst equation from the second gives us

evB =me

R(v − v)(v + v).

If the change in speed ∆v = v − v is small, then

evB =me

R2v∆v,

which gives us

∆v =eBR

2me.

This is how much the electron speeds up when the exter-nal magnetic field is turned on.

This orbiting electron is a very classical view. Inreality, we use statistical mechanics.

The change in the magnetic dipole moment when youturn the magnetic field on is

∆~m = −1

2e∆vRz = −e

2R2

4me

~B.

We see that the change in the magnetic dipole moment isopposite the direction of the magnetic field. This effectoccurs in all atoms in an external magnetic field. Thisis the phenomenon of diamagnetism—i.e. when an ex-ternal magnetic field is turned on, magnetic dipoles areinduced, and these induced dipoles point antiparallel tothe magnetic field. Although this occurs in all atoms, itis very small. It is negligible compared to the strength ofparamagnetism, so for atoms with unpaired electrons, youdon’t notice the diamagnetism because it is completelyovershadowed by the paramagnetization.

To quantify the strength of the magnetic polariza-tion, we define the magnetization ~M to be the inducedmagnetic dipole moment per unit volume. This is similarto the ~P from electrostatics.

Magnetic Fields of Magnetized Materials

For a single dipole ~m, the vector potential is

~A(~r) =µ0

4π

~m× nu2

.

For a lump of magnetized material, we can break it intomany dipoles and integrate to get

~A(~r) =µ0

4π

ˆ~m× nu2

dV ′.

Recall that ~∇ ′ ( 1u

)= u

u2 , so we can write this as

~A(~r) =µ0

4π

ˆ~m× ~∇ ′

(1

u

)dV ′.

If we expand this, we can write it in the form

~A(~r) =µ0

4π

ˆV

~Jb(~r′)

udV ′ +

µ0

4π

˛S

~Kb(~r′)

uda′,

where

~Jb = ~∇× ~M ,

is the volume bound current, and

~Kb = ~M × n,

is the surface bound current. The first integral is a vol-ume integral over the volume of the magnetized object,and the second integral is a surface integral over the sur-face of the object. Recall that n is the unit normal vectorso d~a ′ = n da′.

Once you have the bound currents, don’t calculatethe field using the integral above to find ~A and thencomputing ~B = ~∇ × ~A if you can at all help it. If yourproblem has the appropriate symmetry, just use Ampere’slaw.


Find the magnetic field of a sphere with a uniformmagnetization ~M = M z.a

Since ~M is uniform, its curl is zero, i.e., ~Jb =~∇ × ~M = 0. Calculating the surface bound current,we find

~Kb = ~M × n = M sin θ φ.

Recall from an earlier example that a rotatingspherical shell of uniform surface charge σ correspondsto a surface current density of

~K = σ~v = σωR sin θ φ.

So if we set M = σωR, we can think of a uniformlymagnetized sphere as a rotating spherical shell with asurface current. Using the same earlier example, wehave that the magnetic field on the inside is

~B =2

3µ0~M .

On the outside, the field is the same as a pure dipolewith moment

~m =4

3πR3 ~M .


Example:

The Auxiliary Field

The effect of magnetization is to create bound volumeand surface currents which produce magnetic fields. It ishelpful to separate these bound current densities from thefree current densities by writing the total current densityas

~J = ~Jb + ~Jf .

The bound current comes from the magnetization, andthe free current is all the other sources–like current dueto a battery.

Recalling that ~Jb = ~∇× ~M , we can write Ampere’slaw in differential form as

1

µ0

~∇× ~B = ~J = ~Jb + ~Jf = ~∇× ~M + ~Jf .

Rearranging, we get

~∇× ~H = ~Jf ,

where

~H ≡ 1

µ0

~B − ~M ,

is called the auxiliary field. In integral form, we canwrite ˛

~H · d~l = If,enc,

where If,enc is the free current enclosed. We now haveAmpere’s law for magnetized materials.

You cannot assume from the integral form of Ampere’slaw that ~H = 0 if If,enc = 0. This is only true if~∇ · ~M = 0 everywhere. Your first step should alwaysbe to look at the symmetry of the problem and makesure that it can actually be solved using Ampere’s law.Ampere’s law, here, is only useful when ~H is alwaysparallel or perpendicular to the Amperian loop.

Tip:

The auxiliary field ~H is similar to the magnetic field~B, however, its source is ~Jb instead of ~J . We knowthat ~∇ · ~B = 0, but inside of a magnetized object,~∇ · ~H = −~∇ · ~M .

If ~M = 0 in some region, then the auxiliary field is thesame as the magnetic field in that region except for afactor. That is, ~H = 1

µ0

~B.

Tip:

The auxiliary field ~H is like the ~D from electrostat-ics. Some authors actually call ~H the magnetic field.


Consider a long copper rod of radius R that car-ries a uniformly distributed free current I. What isthe auxiliary field ~H inside and outside the rod? a

We start with an Amperian loop of radius s strad-dling the axis of the rod. For s ≤ R, we get

˛~H · d~l = 2πsH = If,enc =

Iπs2

πR2,

or~H =

Is

2πR2φ, s ≤ R.

For s ≥ R, we use the same equation, but now ourfree current enclosed is the total free current in therod. We get

~H =I

2πsφ, s ≥ R.

Since ~H ≡ 1µ0

~B − ~M and ~M = 0 outside therod, we can easily calculate the magnetic field outsidethe rod

~B = µ0~H =

µ0I

2πsφ, s ≥ R.

Inside the rod, we are not yet able to calculate ~B,because we don’t know enough about ~M for copper.However, since copper is so weakly diamagnetic, it isgenerally okay to approximate it as ~M = 0.


Example:

Recall the boundary conditions for the magnetic field

B⊥above −B⊥below = 0

~B‖above − ~B

‖below = µ0

~K × n.

The corresponding boundary conditions for the auxiliaryfield are

H⊥above −H⊥below = −(M⊥above −M⊥below

)~H‖above − ~H

‖below = ~Kf × n.

Linear and Nonlinear Materials

For a linear material, we might be tempted to write themagnetization as proportional to the magnetic field ~B.In reality, it is customarily written as being proportionalto ~H

~M = χm ~H,

where χm is the magnetic susceptibility of the mate-rial. The magnetic susceptibility is dimensionless. Forparamagnetic materials, it is positive, and for diamag-netic materials, it is negative. Typical values are ∼ 10−5.

Materials that obey this equation are said to be lin-ear materials. Recall the definition of the auxiliary field

is ~H = 1µ0

~B− ~M . This implies that ~B = µ0

(~H + ~M

).

For linear materials, then

~B = µ0 (1 + χm) ~H.

In other words, ~B is proportional to ~H,

~B = µ ~H,

where

µ = µ0 (1 + χm) ,

is permeability of the material. Then

µ

µ0= 1 + χm,

is the relative permeability.


Consider an infinite solenoid straddling the z axis.The solenoid is wrapped with n turns of wire per unitlength carrying current I in the φ direction. Thesolenoid is filled with a linear material of magnetic sus-ceptibility χm. What is the magnetic field inside thesolenoid? a

Since we have solenoidal symmetry, we use Am-pere’s law like we did before with a solenoid. To calcu-late the field inside the solenoid, we use a rectangularAmperian loop in the yz-plane that is half inside thesolendoid and half outside the solenoid. There could bebound volume and bound surface currents, but we don’tneed to know them. All we know is the free current inthe wires, and we can use˛

~H · d~l = If,enc.

to find ~H from the free current alone. We can then get~B.

The free current enclosed in our loop is If,enc =nIL, where L is the height of our rectangular loop. Weknow that ~B is in the z direction and ~H is proportionalto ~B, so it is in the same direction. On the outside,~H = 0, so only the vertical side of our loop on theinside of the solenoid contributes, and we get

HL = nIL,

or~H = nIz.

Then the magnetic field on the inside is

~B = µ0 (1 + χm) ~H = µ0 (1 + χm)nIz.

Since χm is positive for paramagnetic materials,the magnetic field is increased if the material withinthe solenoid is paramagnetic. If the material is diamag-netic, the magnetic field is decreased.


Example:

For linear materials, we can calculate the bound sur-face current from ~H as

~Kb = ~M × n = χm ~H × n,

and the bound volume current as

~Jb = ~∇× ~M = ~∇×(χm ~H

)= χm ~Jf .

A ferromagnet is an example of a nonlinear ma-terial. In a ferromagnetic material, a field is not requiredto maintain magnetization.


4.6 Summary: Magnetostatics

Lorentz Force Law

A charge q moving with velocity ~v in a magnetic field ~Bexperiences a magnetic force

~F = q~v × ~B.

If it is moving in both magnetic and electric fields, thenit experiences an electromagnetic force

~F = q ~E + q~v × ~B.

If the only force on a charged particle is due to amagnetic field, and the magnetic field is perpendicular tothe velocity of the particle, then the particle will executeuniform circular motion. Any segment of its path willbe a portion of the arc of a circle. Generally, you can

simplify these problems using qvB = mv2

R .

Remember, magnetic forces do no work.

Currents

The current in a wire is the charge per unit time thatpasses a given point along the wire. A line charge travel-ing at speed v, as in a wire, is a current I = λv, where λis the linear charge density. But since current is actuallya vector, we write this as

~I = λ~v.

The magnetic force on a wire carrying a constant currentI while in a magnetic field ~B is

~FB = I

ˆ (d~l× ~B

).

For a straight wire and a uniform field, this simplifies to

~FB = I~L× ~B = ILB sin θ.

On a surface, the flow of charges is given by the sur-face current density ~K

~K =d~I

dl⊥= σ~v,

where dl⊥ is the width of a ribbon carrying current d~I,and σ is the surface charge density moving with velocity~v. The magnetic force on the surface is

~FB =

ˆ (~v × ~B

)σ da =

ˆ (~K × ~B

)da.

In a volume, the flow of charges is given by the vol-ume current density ~J

~J =d~I

da⊥= ρ~v,

where da⊥ is the cross-sectional area of a tube carryingcurrent d~I, and ρ is the volume charge density movingwith velocity ~v. If the current I in a volume with cross-sectional area A is uniformly distributed, then the volumecurrent density is just J = I

A . To calculate the total cur-rent given the volume charge density, just separate vari-

ables in ~J = d~Ida⊥

and integrate,

I =

ˆS

J da⊥ =

ˆS

~J · d~a.

The magnetic force on the volume is

~FB =

ˆ (~v × ~B

)ρ dV =

ˆ (~J × ~B

)dV.

Biot-Savart Law

The magnetic field at ~r due to a steady current I in awire is given by the Biot-Savart law

~B(~r) =µ0I

4π

ˆd~l ′ × uu2

,

where the integral is done along the wire.

To solve a Biot-Savart law problem, use the followingprocedure:

1. Write down the field point ~r (i.e. where you aremeasuring the magnetic field) in vector form.

2. Write down the location of the current element ~r ′

in vector form.3. Calculate ~u = ~r − ~r ′.4. Calculate u2 for the denominator in the Biot-Savart

law.5. Calculate u = ~u

|~u| .

6. Differentiate ~r ′ to get d~l ′ = d~r ′.7. Calculate the cross product d~l ′ × u for the numer-

ator in the Biot-Savart law.8. Plug everything into the Biot-Savart law and calcu-

late the integral.

For surface currents, the Biot-Savart law is

~B(~r) =µ0

4π

ˆ ~K(~r ′)× uu2

da′.

For volume currents, it is

~B(~r) =µ0

4π

ˆ ~J(~r ′)× uu2

dV ′.

4.6. SUMMARY: MAGNETOSTATICS 87

Ampere’s Law

The divergence of the magnetic field is always zero

~∇ · ~B = 0.

The curl of the magnetic field (in magnetostatics) is

~∇× ~B = µ0~J .

This is the differential form of Ampere’s law.

Ampere’s law in integral form is more useful, and itis ˛

~B · d~l = µ0Ienc,

where Ienc is the current enclosed by the closed “Ampe-rian” loop that we are integrating over. It is valid forarbitrary but steady currents. Ampere’s law is used tofind the magnetic field due to a configuration of current.

Ampere’s law is used just like Gauss’s law in elec-trostatics, but instead of finding the charge enclosed in aGaussian surface, we find the current enclosed in an Am-perian loop. For Ampere’s law to be useful, we need ~Bto be perpendicular or parallel to each part of the Am-perian loop. Only then does the ~B · d~l in the integralsimplify. As such, before we can apply Ampere’s law, wehave to deduce the direction of the magnetic field pro-duced by the configuration of current. You also need tonote the direction of the current flow, which needs to beperpendicular to the plane of your Amperian loop.

There are four symmetric current configurations thatcan be solved by Ampere’s law:

• Infinite straight lines: In this case, the magneticfield is circumferential, so your Amperian loop is acircle with the same axis as the line of current.

• Infinite planes: The magnetic field is parallel aboveand below the infinite plane, so your Amperian loopstraddles the plane (i.e. half the loop is above theplane and half is below) and is oriented so that thecurrent flows through the loop.

• Infinite solenoids: The current is circumferential,so the magnetic field is parallel to the axis of thesolenoid. Here, you use two Amperian loops–onecompletely outside, and one straddling the surfaceof the solenoid.

• Toroids: A toroid is just a solenoid bent into a cir-cle. The magnetic field is circumferential, and yourAmperian loop should be a circle of radius s withthe same axis as the toroid.

Be careful when applying Ampere’s law to cylindri-cal things because cylinders can carry longitudinal (e.g. awire) or solenoidal currents. The magnetic fields of thesetwo are very different.

If we are given a current density, we can calculatethe current enclosed in an Amperian loop encircling thecurrent density with

Ienc =

ˆ~J · d~a.

Here, we integrate over the surface that is bounded bythe Amperian loop.

Vector Potential

The magnetic vector potential ~A is related to the mag-netic field by

~B = ~∇× ~A.

The vector potential ~A is usually in the same directionas the current ~I that generates it.

For line current ~I,

~A(~r) =µ0

4π

ˆ ~I

udl′ =

µ0I

4π

ˆ1

ud~l ′.

For surface current,

~A(~r) =µ0

4π

ˆ ~K

uda′.

For volume current,

~A(~r) =µ0

4π

ˆ ~J(~r ′)

udV ′.

One way to calculate ~B is to calculate ~A using one ofthese integrals and then computing ~B = ~∇ × ~A. Avoidthis if at all possible.

To find the vector potential, we can use˛~A · d~l = ΦB .

This is just like Ampere’s law but with ~B replaced by ~Aand instead of the current Ienc through the loop, we usethe magnetic flux ΦB through the loop.

The boundary conditions in magnetostatics are

B⊥above −B⊥below = 0

~B‖above − ~B

‖below = µ0

(~K × n

)~Aabove − ~Abelow = 0

∂

∂n~Aabove −

∂

∂n~Abelow = −µ0

~K.

The magnetic dipole moment of a current loop is

~m = I

ˆd~a = I~a,

where ~a is the vector area of the loop. For a flat loop, thevector area a has a magnitude equal to the ordinary area


enclosed by the loop, and its direction is perpendicular tothe plane of the loop.

Like with the electric potential V , we can do a multi-pole expansion of the vector potential. In this expansion,the monopole term is always zero for the vector potential.This means the dipole term

~Adip(~r) =µ0

4π

~m× rr2

,

provided it is not zero, gives a good approximation of themagnetic field at distances much larger than the diameterof the current loop.

Magnetic Fields in Matter

In a uniform magnetic field, a magnetic dipole (i.e. cur-rent loop) experiences a torque

~N = ~m× ~B.

For an infinitesimal current loop with magneticdipole moment ~m, the force on the loop due to a magneticfield ~B is

~F = ~∇(~m · ~B

).

This force is zero if ~B is uniform.

If a material containing magnetic dipoles is subjectto an external magnetic field, it becomes magneticallypolarized or magnetized. The material’s magnetization is

~M ≡ magnetic dipole moment per unit volume.

This is like the ~P from electrostatics.

A magnetized object produces its own small mag-netic field. A magnetized material contains a volumebound current

~Jb = ~∇× ~M ,

and a surface bound current

~Kb = ~M × n.

To calculate the magnetic field due to the magnetizationof an object, first calculate these bound currents, andthen calculate the magnetic field due to these currents,for example, by applying Ampere’s law.

We can separate the bound current densities from thefree current densities by writing the total current densityas

~J = ~Jb + ~Jf .

The bound current comes from the magnetization, andthe free current is all the other sources–like current dueto a battery.

We can rewrite the differential form of Ampere’s lawas

~∇× ~H = ~Jf ,

and the integral form as

˛~H · d~l = If,enc,

where

~H ≡ 1

µ0

~B − ~M ,

is called the auxiliary field. Note that If,enc is the free

current enclosed. The auxiliary field ~H is similar to themagnetic field ~B, however, its source is ~Jb instead of ~J .

We now have Ampere’s law for magnetized materials,so if you’re asked to calculate the magnetic field inside ofa linear material, use

¸~H · d~l = If,enc.

Note: We cannot assume from the integral form ofAmpere’s law that ~H = 0 if If,enc = 0. This is only true

if ~∇ · ~M = 0 everywhere. Your first step should alwaysbe to look at the symmetry of the problem and make surethat it can actually be solved using Ampere’s law. Am-pere’s law, here, is only useful when ~H is always parallelor perpendicular to the Amperian loop.

The boundary conditions for the auxiliary field are

H⊥above −H⊥below = −(M⊥above −M⊥below

)~H‖above − ~H

‖below = ~Kf × n.

Typically in a material,

~M = χm ~H,

where χm is the magnetic susceptibility of the ma-terial. Materials that obey this equation are said to belinear materials. From the definition of the auxiliary field,we can now write ~B as

~B = µ0 (1 + χm) ~H = µ ~H,

where

µ = µ0 (1 + χm) ,

is the permeability of the material.

Chapter 5

Electrodynamics

5.1 Electromotive Force

Ohm’s Law

In most substances, the current density ~J is

~J = σ ~f ,

where σ is the conductivity of the material, and ~f isthe force per unit charge. The conductivity σ is relatedto the resistivity of the material ρ via

ρ =1

σ.

Note that ρ and σ here have nothing to do with the chargedensity. For a perfect conductor, σ →∞.

In our case, the force pushing the charges is the elec-tromagnetic force

~J = σ(~E + ~v × ~B

).

Here, q has been divided out since we are looking at theforce per unit charge. When ~v is small, which it usuallyis, we get Ohm’s law

~J = σ ~E.

Inside a conductor, we have ~E = 0 only for stationarycharges. We are now dealing with moving charges. For aperfect conductor

~E =~J

σ' 0.

Consider a cylindrical resistor of length L with aconstant, but not necessarily circular, cross-section ofarea A and conductivity σ. If the potential is constantover each end, and the potential difference betweenthe ends is V , what is the current that flows? a

In the resistor, the electric field is uniform. Sofrom J = I

A , J = σE, and V = EL, we get

I = JA = σEA = σAV

L=V

R,

with

R =L

σA.

This is an example of the form of Ohm’s law, V = IR,that we are used to.


Example:

89

90 CHAPTER 5. ELECTRODYNAMICS

Suppose you have a pair of long concentric cylin-ders. The outer cylinder has radius b and the innercylinder has radius a. Between the cylinders, thereis a material with conductivity σ. If the two cylin-ders are maintained at a potential difference V , whatcurrent flows from one to the other in a length L? a

From previous problems, we know the field be-tween the cylinders is

~E =λ

2πε0ss,

where λ is the charge density on the inner cylinder.The current must be flowing radially outward (or in-ward), and

I =

ˆ~J · d~a = σ

ˆ~E · d~a.

Using a Gaussian cylinder of radius s gives us

I = σE2πsL =σλ

2πε0s2πsL =

σλL

ε0.

But the voltage difference is

V = −ˆ a

b

~E · d~l =λ

2πε0ln

(b

a

).

Solving this for λ and plugging it into I gives us

I =2πσL

ln(ba

)V.aThis is example 7.2 in Griffiths

Example:

As the previous two examples show, the current flow-ing between two electrodes is

V = IR,

where R, measured in ohms (Ω), is the resistance of thematerial between the electrodes. This is the more familiarform of Ohm’s law.

For steady currents and uniform conductivity,

~∇ · ~E =1

σ~∇ · ~J = 0.

This implies that the charge density inside the material iszero, and that Laplace’s equation holds within a uniformohmic material carrying steady current. Previously, weknew these to be true for electrostatics, where the chargeis not moving. Here, we see it is also true in electrody-namics where we have moving charges. So we can use ourtechniques from electrostatics to calculate the potential.

In a real material, the electrons undergo a lot of colli-sions. As a result, much of the work done by the electrical

force is converted to heat in the resistor. The work doneper unit charge is V , and the charge flowing per unit timeis I, and the power delivered is

P = V I = I2R.

Electromotive Force

In an electric circuit individual electrons move slowly, butthe current in the circuit occurs almost instantaneouslyand is the same everywhere. This is because of a feed-back mechanism. If charge builds up somewhere alongthe wire, the electric field there becomes stronger. Thisfield causes incoming electrons to slow down (reducingthe rate of charge buildup) and causes the outgoing elec-trons to speed up (reducing the charge build up). Inthis way, the electric field that is created nearly instanta-neously throughout a circuit causes the current to quicklybecome the same everywhere.

There are really two forces that drive current arounda circuit—the source (e.g. a battery) and the electrostaticforce from the electric field

~f = ~fs + ~E.

We define a quantity called the electromotive force oremf as

E =

˛~f · d~l =

˛~fs · d~l,

where the line integral is taken around the circuit. Itdoesn’t matter if we use ~f here or ~fs since

¸~E · d~l = 0.

The emf can interpreted as the work done per unit chargeby the battery.

In an ideal source (e.g. an ideal battery), we have~E = ~fs, then the voltage between the terminals is

V = −ˆ~E · d~l =

ˆ~fs · d~l =

˛~fs · d~l = E.

Here, the first two integrals are from one battery terminalto the other. The line integral is taken around the cir-cuit. The last two integrals are the same because ~fs = 0outside the battery. So for an ideal battery, the voltageV across the terminals is equal to the emf

V = E.

In a non-ideal battery,

V = E − Ir,

where I is the current in the circuit, and r is the internalresistance of the battery.

If a wire is moved through a magnetic field, the mag-netic field acts on the free charges in the wire, and acurrent begins flowing. This motional emf is how gen-erators work. In a generator, a wire coil is rotated in amagnetic field to produce the motional emf that drivesthe current.

5.1. ELECTROMOTIVE FORCE 91

The motional emf produced in a circuit is the neg-ative of the rate of change of the magnetic flux throughthe circuit

E = −dΦ

dt.

This is a completely general result that holds for loops ofany shape—even loops that are changing in time. Gener-ally, to find the emf produced in a circuit, this equationis much easier to use than finding the emf by integratingthe force over the circuit.

To determine the direction in which a current due toan emf flows in a circuit, you must start with the direc-tion of the flux. Pick a positive direction for the circuit,say clockwise. Then, the area vector ~a for the circuitpoints into the page. This can be determined by theright-hand rule in which your fingers wrap around thecircuit in the positive direction, then you thumb pointsin the direction of ~a. Once you have ~a. You can computeE = −dΦ

dt = − ddt~B×~a. If E turns out to be positive, then

the current will flow in the positive (clockwise) direction.Otherwise, if E is negative, then the current flows in thenegative direction.

Suppose a metal disk of radius a is rotating withangular velocity ω about a vertical metal axis. It is ina uniform magnetic field ~B pointing upward. A circuitis made by running a wire such that one end touchesthe rotating axis and the other end slides along theouter edge of the rotating disk. On the wire is a re-sistor R. Find the current that flows.a

This is a problem of motional emf. We have freecharges (on the disk) rotating through a magneticfield. The magnetic field will exert a force on thesecharges and cause a current to flow in the circuit.

In this case, we cannot use the shortcut E = −dΦdt

to calculate the emf since we don’t know how to cal-culate the flux through the circuit. Instead, we mustgo back to

E =

˛~f · d~l.

The tangential speed of a point on the disk a distancer from the axis of rotation is v = ωr. So the force perunit charge is

~fB = ~v × ~B = ωBr r.

Then

E =

ˆ a

0

~fB · d~r = ωB

ˆ a

0

r dr =ωBa2

2.

Then from Ohm’s law V = IR, the current is

I =V

R=

E

R=ωBa2

2R.


Example:


Suppose you have a square loop cut out of a thicksheet of aluminum. The top part of the loop is in amagnetic field pointing into the page, and the bottompart is hanging down outside the magnetic field. Whatis the terminal velocity of the loop as it falls down? a

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x y

Recall that terminal velocity occurs when the up-ward force equals the downward force. In our case, theterminal velocity occurs when

FB = mg,

where FB is the upward magnetic force and mg is thedownward gravitational force.

When the loop falls, an emf will be generated creat-ing a current in the loop. This current, in the presenceof the magnetic field, will create the upward magneticforce on the loop.

Suppose our loop has side lengths L, and y is thedistance that the loop extends into the magnetic field.Then the magnetic flux through the loop is

Φ = ~B · ~A = BA = BLy.

Here, we are taking ~A to point into the page parallel to~B. Then the emf that is generated is

E = −dΦ

dt= −BLdy

dt= BLv.

Note that dydt = −v since y is decreasing.

From Ohm’s law V = IR, the current in the loopis

I =V

R=

E

R=BLv

R.

We took ~A to point into the page. From the right-handrule, that implies the positive direction for our loop isclockwise. Since I is positive, this means our currentflows clockwise in the loop.

If a wire carrying a current moves through a mag-netic field, a magnetic force acts on the wire ~FB =I~L × ~B. From the righthand rule, with a clockwisecurrent, we see that an upward force occurs on the topwire. An outward force occurs on each of the verticalsegments that is in the field. These cancel each otherout, so we only care about the top wire.

FB = ILB =vL2B2

R.

Equating with mg and solving for v to get the ter-minal velocity,

v =mgR

L2B2.

Suppose the “wire” forming the loop has cross-sectional area a. Then the volume of the loop is4aL. Since density is mass divided by volume, we havem = 4aLρ, where ρ is the density of aluminum. Recallfrom an earlier example that the resistance of a pieceof wire is R = `

σa , where σ is the conductivity of thematerial, and ` is the length of the wire. For our circuit,` = 4L. In our case, then,

v =g

L2B2·m ·R =

g

L2B2· 4aLρ · 4L

σa=

16gρ

σB2.

The density of aluminum is ρ = 2700 kg/m3, andits conductivity is σ = 3.77× 107 1/Ω ·m. Plugging ev-erything in and taking care of units give our terminalvelocity

v = 0.0112 m/s.

aThis is part of problem 7.11 in Griffiths.

Example:

5.2 Electromagnetic Induction

Faraday’s Law

If a loop of wire is pulled through a magnetic field, a cur-rent flows in the loop. This is just motional emf, and weknow that

E = −dΦ

dt.

If a loop of wire is held stationary and a magnetic

field is carried past it, a current again flows in the loop.In fact, the exact same emf is produced, but it must beby a different mechanism. Since the loop is stationary,the charges in the loop are stationary, and we know thatstationary charges are not affected by a magnetic field.What’s happening is that the changing magnetic field in-duces an electric field, and it is this induced electric fieldthat causes the charges in the loop to move.

5.2. ELECTROMAGNETIC INDUCTION 93

If the emf is equal to the rate of change of flux

E =

˛~E · d~l = −dΦ

dt,

then the relation between a changing magnetic field andthe induced electric field is given by Faraday’s law˛

~E · d~l = −ˆ∂ ~B

∂t· d~a.

If the magnetic field is not changing, we get the old result¸~E ·d~l = 0. Stokes’ theorem allows us to write Faraday’s

law in differential form as

~∇× ~E = −∂~B

∂t.

If both the loop and the magnet are stationary, butthe strength of the magnetic field is varied, a current againflows in the loop.

In all three cases,

E = −dΦ

dt.

In other words, whenever the magnetic flux through acircuit changes, an emf is produced in the circuit.

When a changing magnetic flux through a circuit in-duces an emf in the circuit, the emf drives a current in thecircuit. To figure out which direction this current flowsin the circuit, we use Lenz’s law, which is essentiallythat nature opposes a change in flux. That is, whenevera changing magnetic flux induces a current in a circuit,the current will flow in the direction, such that the mag-netic field it produces opposes the change in flux thatoriginally induced the current flow.

Suppose you have a cylindrical magnet of length Land radius a moving in the z direction with constantspeed v. The magnetic is uniformly magnetized with~M = M z. The magnet passes through a circular con-

ducting ring with a diameter slightly larger than that ofthe magnet. We know the magnetic field of the magnet isthat of a long solenoid with a surface current ~Kb = M φ,which has a magnetic field ~B = µ0

~M . As the magnetapproaches the ring from far away, the magnetic fluxthrough the ring is initially zero. As the front of themagnet passes through the ring, the flux through the ringspikes up. This abrupt change in flux causes an emf spikein the wire ring. The direction of the induced currentmust be clockwise (if viewed from above) to oppose thechange in flux from the magnet. After the front end of themagnet has passed through the ring, the flux stays con-stant, so the emf drops to zero. As the back end of themagnet passes through the ring, the flux abruptly dropsto zero. Again, this change in flux produces an emf spikein the ring, albeit with the opposite sign.

Induced Electric Fields

Recall that to find the magnetic field due to current, weuse Ampere’s law

˛~B · d~l = µ0Ienc,

where the integral is over an Amperian loop enclosing thecurrent Ienc. We can use the same techniques to calcu-late the electric field induced by a changing magnetic flux.The flux form of Faraday’s law is

˛~E · d~l = −dΦ

dt.

This is the same kind of integral as Ampere’s law. Soto calculate the induced electric field, we integrate over aloop enclosing the change in flux dΦ

dt . Essentially, we cal-

culate the induced field ~E using Ampere’s law and thencalculate the rate of change of the enclosed flux throughthe loop.

Consider a circular region centered on the originand lying in the xy-plane. In this region, a uniformbut time varying magnetic field ~B(t) points straightup. What is the electric field? a

If ~B is changing in time, we know by Faraday’slaw that an electric field is produced. Since the mag-netic field lines are straight, we expect the electricfield lines to be circular. That is, we expect the elec-tric field to be circumferential.

With an Amperian loop of radius s, smaller thanthe radius of our region, and lying flat within the cir-cular region, we can apply Faraday’s law to calculatethe exact field.˛

~E · d~l = E · 2πs = −dΦ

dt.

The rate of change of flux is

dΦ

dt=

d

dt~B(t) · ~A = πs2 dB

dt,

So the electric field is

~E = −s2

dB

dtφ.


Example:


Consider an infinite straight wire carrying a timevarying current I(t) in the z direction. What is theinduced electric field near the wire? a

We start with an Amperian loop of length L andwidth s− s0 placed as in the image below.

I

L

s0

s

We know the magnetic field curls around the wire,so we expect the electric field to be parallel to the wire.So ~E is parallel to the top and bottom of our Amperianloop and perpendicular to the sides of our Amperianloop. However, the electric field is not the same at thetop and bottom of the loop. So˛

~E · d~l = E(s0)L− E(s)L.

The negative sign comes because as we integrate coun-terclockwise around the loop, we take the bottom part

in the positive direction and the top part in the negativedirection.

The magnetic field of an infinite current at a mo-ment in time is

~B(s) =µ0I

2πsφ.

The flux through the loop is then

Φ =

ˆ~B · d~a =

µ0LI

2π

ˆ s

s0

1

s′ds =

µ0LI

2πln

(s

s0

).

The rate of change of the flux is

dΦ

dt=µ0L

2π

dI

dtln

(s

s0

).

So by Faraday’s law

E(s0)L− E(s)L = −µ0L

2π

dI

dtln

(s

s0

)E(s) =

µ0

2π

dI

dtln

(s

s0

)− E(s0)

~E(s) =

[µ0

2π

dI

dtln s+K

]z.

where K is a constant with respect to s.


Example:

Inductance

Suppose you have two loops of wire that are close to eachother. If you pass a current through one loop, then thecurrent will produce a magnetic flux through the otherloop. From the Biot-Savart law, we know that the fluxthrough loop 2 is proportional to the current in loop 1

Φ2 = MI1,

where M is the mutual inductance of the two loops. Itturns out that the flux through loop 2 is

Φ2 =µ0I14π

˛ ˛d~l1 · d~l2

u,

where I1 is the current in loop one. This implies that themutual inductance is

M =µ0

4π

˛ ˛d~l1 · d~l2

u.

This is called the Neumann formula. One of the lineintegrals goes around loop 1 and the other line integralgoes around loop 2. It is called mutual inductance be-cause the same M is used whether the current is in loop1 and we are interested in the flux through loop 2 or viceversa.

If we vary the current in loop 1, then the changingflux through loop 2 induces an emf in loop 2

E2 = −dΦ2

dt= −M dI1

dt.

Interestingly, the magnetic field produced by a cur-rent in a loop creates a flux of

Φ = LI,

through itself, where L is the self-inductance of theloop. This self-flux and self-inductance means that if thecurrent is varied, an emf

E = −LdIdt.

is produced in a circuit due to its own self-flux.

Note, self-inductance is always positive. It has unitsof henries (H).

In general, any circuit has some self-inductance, sowhen the current in a circuit is varied, its self-flux andself-inductance produces an emf in the circuit. This emfacts to oppose the change in current, so it is often called

5.2. ELECTROMAGNETIC INDUCTION 95

back emf. If the current is reduced, the back emf thatis produced will act to increase the current, and if thecurrent is increased, the back emf works to reduce it. As

such, self-inductance is kind of like an inertia of the cur-rent in a circuit.

Consider a long solenoid with radius b and N turnsof wire of wire per unit length. A short solenoid oflength l, radius a, and n turns of wire per unit lengthis inside (and coaxial) the long solenoid. If there is acurrent I in the short solenoid, what is the total fluxthrough the long solenoid? a

We want Φlong = MIshort. However, we don’tknow what the field is of a short solenoid. Instead,we use the fact that the inductance is mutual. So in-stead, we will find Φshort = MIlong. If we use the samecurrent then Φlong = Φshort.

We know the field inside of a long solenoid is

B = µ0NI.

The flux through a single turn of short solenoid is then

Φ = ~B · ~A = Bπa2.

The total flux through the short solenoid is then

Φshort = πa2Bnl = µ0πa2nNlI.

This is the same as the flux through the long solenoid,Φlong, that we wanted. This implies that the mutualinductance of the two solenoids is

M = µ0πa2nNl.


Example:

Consider an RL circuit containing a battery withemf E0, a resistor R and an inductor L. What is thecurrent in the circuit? a

The total emf in the circuit is the emf of the batteryplus the back emf produced by the inductor

E = E0 − LdI

dt.

Ohm’s law E = IR then tells us that

E0 − LdI

dt= IR.

We can solve this differential equation by separatingvariables and integrating. If our initial condition is thatthe current is turned on at t = 0, then we get

I(t) =E0

R

(1− e− t

τ

),

where

τ =L

R,

is the circuit’s time constant.


Example:


Consider a toroid with outer radius b, inner radiusa, N turns of wire, and a rectangular cross-section withheight h. What is the self-inductance of the toroid?a

Within the coils of a toroid, we know the field is

B =µ0NI

2πs.

The flux through a single loop is

ˆ~B · d~a =

µ0NIh

2π

ˆ b

a

1

sds =

µ0NIh

2πln

(b

a

).

a

s

b

ds

axis

Multiplying this by N gives us the total flux

Φ =µ0N

2Ih

2πln

(b

a

).

Then from Φ = LI, the inductance of the toroid is

L =µ0N

2h

2πln

(b

a

).


Example:

Energy of Magnetic Fields

To run a current in a circuit, we need to supply energy toovercome the back emf generated in the circuit. The workdone on a unit charge against this back emf is −E whereE is the back emf. Since the charge flowing per unit timeis I = dQ

dt , the total work done in time t is W = −EIt, so

dW

dt= −EI = LI

dI

dt.

Integrating this from I = 0 to I = I gives us

W =1

2LI2.

This is the work that must be performed against the backemf to establish a line current I in a circuit. If the cur-rent is turned off, this energy is returned to us as the backemf works to maintain the current for a little longer. Thiswork is also the energy stored in an inductor. It is theenergy stored in a magnetic field.

A slightly more general way of writing this is

W =1

2

˛ (~A · ~I

)dl.

For volume current,

W =1

2

ˆV

(~A · ~J

)dτ.

With a little work, we can show that

W =1

2µ0

âllspace

B2 dτ.

The energy stored per unit volume is

B2

2µ0.

Recall that the energy stored in an electric field is

WE =1

2

ˆ(V ρ) dτ =

ε02

Ê2 dτ.

The energy stored in a magnetic field has the same form

WB =1

2

ˆ (~A · ~J

)dτ =

1

2µ0

ˆB2 dτ.

5.3. MAXWELL’S EQUATIONS 97

Consider a long coaxial cable (i.e. nested cylin-ders). The inner surface of radius a carries a currentI. The current I is carried back along the outer surfaceof radius b. What is the magnetic energy stored in alength l? a

By Ampere’s law, the field between the cylinders is

~B =µ0I

2πsφ.

Elsewhere, there is no magnetic field. The energy perunit volume is

B2

2µ0=

µ0I2

8π2s2.

Next, consider an elemental cylindrical shell of length l,radius a < s < b, and thickness ds. This has a volume

2πls ds, so it contains energy

µ0I2

8π2s2· 2πls ds =

µ0I2l

4π

1

sds.

Outside of a ≤ s ≤ b, the magnetic field is zero, so thetotal energy in a length l of the coaxial cable is

W =µ0I

2l

4π

ˆ b

a

1

sds =

µ0I2l

4πln

(b

a

).

Since W = 12LI

2, we see that the self-inductance is

L =µ0l

2πln

(b

a

).

This illustrates a new way of calculating the self-inductance. We first calculate the energy stored in themagnetic field, and then use W = 1

2LI2 to calculate L.


Example:

5.3 Maxwell’s Equations

Before Maxwell, the equations of electromagnetism were

~∇ · ~E =ρ

ε0~∇ · ~B = 0

~∇× ~E = −∂~B

∂t~∇× ~B = µ0

~J .

The first is Gauss’s law. The third is Faraday’s law. Thelast is Ampere’s law.

There is a problem with Faraday’s law. Recall thatthe divergence of any curl is zero. However, if we take thedivergence

~∇ ·(~∇× ~B

)= µ0

~∇ · ~J ,

we don’t generally get zero on the right side. We do getzero when the current is steady, but if the current is vary-ing, we generally do not.

From the continuity equation, we have

~∇ · ~J = −∂ρ∂t

= − ∂

∂t

(ε0~∇ · ~E

)= −~∇ ·

(ε0∂ ~E

∂t

).

So if we write Ampere’s law as

~∇× ~B = µ0

(~J + ~Jd

),

where

~Jd = ε0∂ ~E

∂t,

called the displacement current to the right side ofAmpere’s law, we fix it. When we take the divergence ofthe modified form of Ampere’s law, we get zero on theright side as we should. This doesn’t change anything in

electrostatics because in electrostatics, ∂ ~E∂t = 0.

Our final laws of magnetism, Maxwell’s equations,are

~∇ · ~E =ρ

ε0~∇ · ~B = 0

~∇× ~E = −∂~B

∂t

~∇× ~B = µ0~J + µ0ε0

∂ ~E

∂t.

Just as Faraday’s law (third equation) implies that achanging magnetic field creates an electric field, Ampere’scorrected law (fourth equation) implies that a changingelectric field creates a magnetic field.

Maxwell’s Equations in Matter

Recall that an electric polarization produces a boundcharge density, and a magnetic polarization (i.e. mag-netization) produces a bound current density

ρb = −~∇ · ~P~Jb = ~∇× ~M .

A change in electric polarization results in a flow of boundcharge. If ~P increases, the bound charge increases, lead-ing to a net current

dI =∂σa⊥∂t

da⊥ =∂ρ

∂tda⊥.


This implies a polarization current density

~JP =∂ ~P

∂t.

We want to include this effect in Maxwell’s equations.

We can write the total charge density as

ρ = ρf + ρb = ρf − ~∇ · ~P ,

and the total current density as

~J = ~Jf + ~Jb + ~JP = ~Jf + ~∇× ~M +∂ ~P

∂t.

Now Gauss’s law can be written as

~∇ · ~E =1

ε0

(ρf − ~∇ · ~P

),

or in terms of the electric displace ~D = ε0 ~E + ~P as

~∇ · ~D = ρf .

In Ampere’s law (with Maxwell’s correction), we nowhave

~∇× ~B = µ0

(~Jf + ~∇× ~M +

∂ ~P

∂t

)+ µ0ε0

∂ ~E

∂t.

Or in terms of the auxiliary field µ ~H = ~B − µ0~M , we

can write it as

~∇× ~H = ~Jf +∂ ~D

∂t.

So in terms of free currents and charges, Maxwell’sequations can be written as

~∇ · ~D = ρf

~∇ · ~B = 0

~∇× ~E = −∂~B

∂t

~∇× ~H = ~Jf +∂ ~D

∂t.

In integral form, they are

˛~D · d~a = Qf,enc˛~B · d~a = 0

˛~E · d~l = − d

dt

ˆ~B · d~a

˛~H · d~l = If,enc +

d

dt

ˆ~D · d~a

This leads us to the boundary conditions

D⊥1 −D⊥2 = σf

B⊥1 −B⊥2 = 0

~E‖1 − ~E

‖2 = 0

~H‖1 − ~H

‖2 = ~Kf × n.

5.4. SUMMARY: ELECTRODYNAMICS 99

5.4 Summary: Electrodynamics

Electromotive Force

In most substances, the current density is ~J = σ ~f , whereσ is the conductivity of the material, and ~f is the forceper unit charge. The conductivity σ is related to the re-sistivity of the material ρ via ρ = 1

σ . In electrodynam-

ics, the force ~f that drives charges is the electromagneticforce. When ~v is small, which it usually is, we get Ohm’slaw

~J = σ ~E.

Recall that ~J = d~Ida⊥

. With this, we can write the currentthrough a closed surface as

I =

˛~J · d~a = σ

˛~E · d~a =

σQencε0

.

The current flowing between two electrodes is

V = IR,

where R is the resistance of the material between theelectrodes. This is the more familiar form of Ohm’s law.

The power delivered to a resistor is

P = V I = I2R.

The force that drives current around a circuit isthe electromagnetic force plus the force provided by thesource ~f = ~fs + ~E. We define the electromotive force,which is not actually a force, as

E =

˛~f · d~l =

˛~fs · d~l,

where the line integral is taken around the circuit.

In a circuit powered by an ideal battery, V = E.With non-ideal battery, V = E − Ir, where I is the cur-rent in the circuit, and r is the internal resistance of thebattery.

Whenever there is a change in magnetic flux througha circuit, there is an emf produced in the circuit

E = −dΦBdt

,

and this emf drives a current in the circuit. This is a gen-eral result that holds for loops of any shape. The changein flux can occur in different ways. If the change in fluxoccurs because the circuit is moved through a magneticfield, then the emf is called “motional emf”. If a loopof wire is held stationary and a magnetic field is carriedpast it, the same emf is produced. The changing magneticfield induces an electric field, and it is this induced electricfield that causes the charges in the loop to move. This

is “electromagnetic induction”. A third way in which theflux in the loop changes is when both the loop and themagnet are stationary, but the strength of the magneticfield is varied.

To figure out which direction the current (that isdriven by an emf) flows in the circuit, we use Lenz’s law,which is essentially that nature opposes a change in flux.Whenever a changing magnetic flux induces a current ina circuit, the current will flow in the direction, such thatthe magnetic field it produces opposes the change in fluxthat originally induced the current flow.

Faraday’s Law

Faraday’s law relates changing magnetic fields to the elec-tric fields they induce.

Faraday’s law in differential form is

~∇× ~E = −∂~B

∂t.

The integral form of Faraday’s law is more useful˛~E · d~l = −dΦ

dt.

This is the same kind of integral as Ampere’s law. Tocalculate the induced electric field, we integrate over thecircuit enclosing the change in flux dΦ

dt . Essentially, we

calculate the induced field ~E using Ampere’s law and thencalculate the rate of change of the enclosed flux throughthe loop.

A typical electrodynamics problem is to find the cur-rent induced in one circuit by a changing current in an-other circuit. Suppose you have a time varying currentI(t) in circuit 1. This implies a changing magnetic fielddue to circuit 1. You can often calculate this field usingAmpere’s law ˛

~B · d~l = µ0Ienc.

The changing magnetic field implies a changing fluxthrough circuit 2. The flux through circuit 2 is calcu-lated using

Φ =

ˆ~B · d~a.

The changing flux through circuit 2 means an emf

E = −dΦ

dt,

is generated in circuit 2. This emf drives a current incircuit 2, and this current can be calculated using Ohm’slaw

E = IR.


Inductance

Suppose you have two loops of wire that are close to eachother. If you pass a current through one loop, then thecurrent will produce a magnetic flux through the otherloop. From the Biot-Savart law, we know that the fluxthrough loop 2 is proportional to the current in loop 1

If you have two circuits near each other and youpass current through one circuit, then a current will beinduced in the second circuit. The current in the firstcircuit creates a magnetic field. This means a changingmagnetic flux in the second circuit, and therefore, an emfthat drives a current in the second circuit. The magneticflux through loop 2, is related to the current in loop 1 by

Φ2 = MI1,

where M is the mutual inductance of the two circuits.Then

E2 = −dΦ2

dt= −M dI1

dt.

A current in a circuit will also create a magnetic fluxthrough itself

Φ = LI,

where L is the self-inductance of the circuit. When thecurrent in a circuit changes, the self-flux through the cir-cuit changes, producing a “back emf”

E = −LdIdt.

that opposes the original change in the current. Self-inductance is always positive.

Energy

To establish a current I in a circuit, work

W =1

2LI2.

must be performed against the back emf. This energy isreturned when the current is turned off. In other words,the circuit stores this energy. This is the energy that isstored in the magnetic field of an inductor.

The energy stored per unit volume in a magnetic fieldis

B2

2µ0.

The total energy stored in a magnetic field is

W =1

2µ0

âllspace

B2 dτ.

This gives us a second way of calculating the self-inductance L of a circuit. Start by calculating the energystored in the field, and then use W = 1

2LI2 to calculate

L.

Maxwell’s Equations

Maxwell modified Ampere’s law as

~∇× ~B = µ0

(~J + ~Jd

),

where

~Jd = ε0∂ ~E

∂t,

is called the “displacement current”.

Maxwell’s equations, are

~∇ · ~E =ρ

ε0~∇ · ~B = 0

~∇× ~E = −∂~B

∂t

~∇× ~B = µ0~J + µ0ε0

∂ ~E

∂t.

Chapter 6

Conservation Laws

6.1 Charge

Recall that the charge in a volume is the integral of thecharge density over the volume. If the charge density istime-dependent, then the charge in a volume V is

Q(t) =

ˆV

ρ(~r, t) dτ.

The current flowing through a boundary is the integral ofthe current density over the boundary. So, for a closedboundary S enclosing our surface V, local charge conser-vation implies that

dQ

dt= −

˛S

~J · d~a.

Plugging the definition of Q(t) given above into this equa-tion gives us

d

dt

ˆV

ρ(~r, t) dτ = −˛S

~J · d~a.

From the divergence theorem, we can rewrite the rightside to get ˆ

V

∂ρ

∂tdτ = −

˛V

~∇ · ~Jdτ.

This implies the continuity equation

∂ρ

∂t= −~∇ · ~J .

This is the statement of (local) charge conservation.

6.2 Energy

Recall that the work performed against the Coulomb forceto assemble a static charge distribution is

We =ε02

Ê2 dτ.

Similarly, the work performed against the back emf toinitiate currents is

Wm =1

2µ0

ˆB2 dτ.

So the total energy stored in electromagnetic fields is

Uem =1

2

ˆ (ε0E

2 +1

µ0B2

)dτ.

A more general relation can be derived1, and it is calledPoynting’s theorem,

dW

dt= − d

dt

ˆV

1

2

(ε0E

2 +1

µ0B2

)dτ − 1

µ0

˛S

(~E × ~B

)· d~a. (6.1)

The first integral is the energy stored in the fields,and the second integral is the rate that the energy is car-ried across the boundary S of the volume V. Poynting’stheorem is the work-energy theorem of electrodynamics.

The differential form of Poynting’s theorem is

∂

∂t(umech + uem) = −~∇ · ~S, (6.2)

1See Griffiths, Introduction to Electrodynamics101

102 CHAPTER 6. CONSERVATION LAWS

where the energy density of the electromagnetic fields is

uem =1

2

(ε0E

2 +1

µ0B2

),

the mechanical energy density, umech, of the particles sat-isfies

dW

dt=

d

dt

ˆV

umech dτ,

and

~S =1

µ0

(~E × ~B

),

is called the Poynting vector. The Poynting vectorgives the energy per unit time per unit area transportedby the electromagnetic fields.

Notice that the differential form of Poynting’s the-orem has the same form as the continuity equation but

with charge density replaced by the total energy densityand current density replaced by the Poynting vector. Sojust like the continuity equation is the statement of (local)charge conservation, the Poynting theorem is the state-ment of energy conservation.

To calculate the total energy in a volume V stored inthe fields, we just integrate the energy density over thevolume

Uem =

ˆV

uem dτ =

ˆV

1

2

(ε0E

2 +1

µ0B2

)dτ.

The power, or energy per unit time, flowing into a regionwith boundary S can be calculated as

P =dU

dt=

ˆ~S · d~a.

Suppose you have a parallel-plate capacitor withplates of radius a and separated by a gap of width w.The current flowing “through” the capacitor is I. Inthe gap between the plates, the electric and magneticfields are

~E(s, t) =It

ε0πa2z

~B(s, t) =µ0Is

2πa2φ,

where s is the distance from the axis. What is the me-chanical energy density umech in the gap, the energydensity uem of the electromagnetic fields in the gap, andwhat is the Poynting vector? Use these and Eq. (6.2)to verify that energy is conserved.a

There are no particles in the gap, so

umech = 0.

The energy density is

uem =1

2

(ε0E

2 +1

µ0B2

)=

1

2

(ε0

[It

ε0πa2

]2

+1

µ0

[µ0Is

2πa2

]2)

=I2

2π2a4

(t2

ε0+µ0s

2

4

).

The magnitude of the Poynting vector is

S =1

ε0EB sin θ

=1

ε0EB

=I2st

2ε0π2a4.

The electric field is in the z direction, and the mag-netic field is circumferential and in the φ direction. Sothe right-hand-rule tells us that ~E × ~B is in the −sdirection. Therefore, the Poynting vector is

~S = − I2st

2ε0π2a4s.

The divergence of ~S is

~∇ · ~S =1

s

∂

∂t(sSs)

=1

s

∂

∂t

(−s I2st

2ε0π2a4

)= − I2t

ε0π2a4.

Plugging these in, we find that Eq. (6.2) is satisfied, andso energy is conserved.

aThis is part (b) of Problem 8.2 in Griffiths.

Example:

6.3. MOMENTUM 103

Continuing with the same capacitor example, cal-culate the total energy in the gap as a function of time,and calculate the total power flowing into the gap. Thenverify that energy is conserved by verifying Eq. (6.1).a

Note thatdW

dt= 0,

in the gap since W = 0 due to there being no chargesin the gap.

The total energy stored in the fields in the gap is

Uem =

ˆV

uem dτ

=

ˆV

I2

2π2a4

(t2

ε0+µ0s

2

4

)dτ

=I2

2π2a4

(2πw

ˆ a

0

s ds+ 2πw

ˆ a

0

s3 ds

)=

I2w

2πa2

(t2

ε0+µ0a

2

8

).

Note that

− d

dtUem = − I2wt

ε0πa2.

This is the first term on the right in Eq. (6.1).

The total power flowing into the gap is

P = −ˆ~S · d~a = − 1

µ0

ˆ (~E × ~B

)· d~a

=I2t

2ε0π2a4

ˆ â2 dφ dz

=I2wt

ε0πa2.

This is the second term on the right in Eq. (6.1).

Plugging these three results into Eq. (6.1), we seethat it is verified.

aThis is part (c) of Problem 8.2 in Griffiths.

Example:

6.3 Momentum

For a moving point charge, we cannot calculate the elec-tric field using Coulomb’s law. The electric field of a mov-ing point charge is radial, but it’s not symmetric like it isfor a stationary charge. Instead, it is sort of flattened inthe forward and backward direction. Where the surface ofconstant field for a stationary charge forms a sphere, thesurface of constant field for a moving charge looks morelike a pancake. Similarly, a moving point charge is nota current, so its magnetic field cannot be calculated bythe Biot-Savart law. Nevertheless, a moving point chargedoes have a magnetic field. Where the surface of constantfield for a line current forms a cylinder, the surface of con-stant magnetic field of a single moving particle looks morelike a sphere. Like with a line current, the direction of thefield lines can be determined using the right-hand-rule.

Suppose you have two charged particles moving atright angles to each other toward a common point. Theelectric forces that one exerts on the other are equal andopposite—in agreement with Newton’s third law. Themagnetic forces that one exerts on the other are equal inmagnitude, but not opposite. The magnetic forces be-tween the two moving particles are not central forces,so they do not cancel out—violating Newton’s third law.This is a problem because conservation of momentum re-quires that the internal forces cancel, which they do inelectrostatics and magnetostatics, but apparently not inelectrodynamics. Conservation of momentum can only berestored in electrodynamics by considering not just themomenta of the moving particles but also the momentaof the moving fields.

The total electromagnetic force on charges in a vol-ume V is

~F =

ˆV

(~E + ~v × ~B

)ρ dτ =

ˆV

(ρ~E + ~J × ~B

)dτ.

The force per unit volume is

~f = ρ~E + ~J × ~B.

Using Maxwell’s equations, we can eliminate ρ and ~J andwrite this in terms of the fields alone as

~f = ε0

(~∇ · ~E

)~E +

(1

µ0

~∇× ~B − ε0∂ ~E

∂t

)× ~B.

With some further manipulation, we can write this as

~f = ε0

[(~∇ · ~E

)~E +

(~E · ~∇

)~E]

+1

µ0

[(~∇ · ~B

)~B +

(~B · ~∇

)~B]

−1

2~∇(ε0E

2 +1

µ0B2

)− ε0

∂

∂t

(~E × ~B

).

We can write this as

~f = ~∇ ·T− ε0µ0∂~S

∂t,

where T is the Maxwell stress tensor. Then the totalforce on the charges in the volume V is

~F =

˛S

T · d~a− ε0µ0d

dt

ˆV

~S dτ. (6.3)

If´~S dτ is independent of time, or for a static charge dis-

tribution, the second term is zero, then to calculate the


total force, we integrate the stress tensor over the surfaceS of the volume V.

The Maxwell stress tensor is a rank-2 tensor withnine-elements

Tij = ε0

(EiEj −

1

2δijE

2

)+

1

µ0

(BiBi −

1

2δijB

2

),

where δij is the Kronecker delta, and the i and j indicescan be x, y, or z. This is a symmetric tensor, and theoff-diagonal elements (i 6= j so δij = 0) are, for example,

Txy = Tyx = ε0ExEy +1

µ0BxBy.

After simplification, the diagonal elements have the form

Txx =ε02

(E2x − E2

y − E2z

)+

1

2µ0

(B2x −B2

y −B2z

).

Physically, the element Tij is the force per unit area inthe ith direction acting on a surface element oriented inthe jth direction. The diagonal elements are pressuresand the off-diagonal elements are shears.

The jth element of the “dot product” of T with avector ~a is

(~a ·T)j =∑

i=x,y,z

aiTij .

The result is a vector. Since T is a rank-2 tensor with ninecomponents, it can be thought of as a 3×3 matrix. Thenthe dot product can be treated as matrix multiplication

T · ~a =

Txx Txy Txz

Tyx Tyy Tyz

Tzx Tzy Tzz

ax

ay

az

= (Txxax + Txyay + Txzaz) x

+ (Tyxax + Tyyay + Tyzaz) y

+ (Tzxax + Tzyay + Tzzaz) z.

We can also take the “divergence” of T. The jthelement is(~∇ ·T

)j

= ε0

[(~∇ · ~E

)Ej +

(~E · ~∇

)Ej −

1

2∇jE2

]+

1

µ0

[(~∇ · ~B

)Bj +

(~B · ~∇

)Bj −

1

2∇jB2

].

6.3. MOMENTUM 105

Calculate the net force exerted on the northernhemisphere of a uniformly charged sphere of radius Rby the southern hemisphere. Previously, we might havesolved this problem by doing a volume integral, butnow, we do it using Eq. (6.3).a

Since this is a static example, the second integral inEq. (6.3) is zero. Our surface, of the upper hemisphere,is a hemispherical bowl of radius R and a disk of radiusR at θ = π

2 . So the total force on the upper hemisphereis

~F =

ˆbowl

T · d~a+

ˆdisk

T · d~a.

We’ll start by computing the contribution from thebowl. The area element is

d~a = R2 sin θ dθ dφ r,

where

r = sin θ cosφ x+ sin θ sinφ y + cos θ z.

The electric field at the surface of the hemisphere is

~E =1

4πε0

Q

R2r = Kr,

where for brevity in the rest of this example, we havedefined K = Q/(4πε0R

2). Plugging in r, we see that

E2 = K2

Ex = K sin θ cosφ

Ey = K sin θ sinφ

Ez = K cos θ.

We are now ready to calculate the components of thestress tensor T. In our case, there is no magnetic fieldpart, so all terms with B are zero, and the componentsare

Tij = ε0

(EiEj −

1

2δijE

2

).

Since T is symmetric, we only need six independentquantities:

Txx =1

2ε0K

2(2 sin2 θ cos2 φ− 1

)Tyy =

1

2ε0K

2(2 sin2 θ sin2 φ− 1

)Tzz =

1

2ε0K

2(2 cos2 θ − 1

)Txy = Tyx = ε0K

2 sin2 θ sinφ cosφ

Txz = Tzx = ε0K2 sin θ cos θ cosφ

Tyz = Tzy = ε0K2 sin θ cos θ sinφ.

We want to findˆbowl

T · d~a =

ˆbowl

(T · d~a)x x

+

ˆbowl

(T · d~a)y y

+

ˆbowl

(T · d~a)z z.

The components of the dot product, after simpli-cation, turn out to be

(T · d~a)x =1

2ε0K

2R2 sin2 θ cosφdθ dφ

(T · d~a)y =1

2ε0K

2R2 sin2 θ sinφdθ dφ

(T · d~a)z =1

4ε0K

2R2 sin(2θ) dθ dφ

These three are integrated over the hemispherical bowl.The first two will give us zero when we integrate overφ, so all that remains is

ˆbowl

T · d~a =

ˆbowl

(T · d~a)z z

=

ˆbowl

1

4ε0K

2R2 sin(2θ) dθ dφz

=Q2

64π2ε0R2

ˆ 2π

0

ˆ π2

0

sin(2θ) dθ dφz

=Q2

32πε0R2z.

That is the contribution from the bowl. Next, wecalculate the contribution from the disk. Again, by thesymmetry, the x and y components vanish in the surfaceintegral, so we only have to calculate the z component.On the disk, the field is

~E = Mr r, M =1

4πε0

Q

R3.

The surface element on the disk is d~a = −r dr dφ z. Itis negative so that d~a points outward. The directionvector is r = cosφ x+ sinφ y. For the disk, we get

ˆbowl

T · d~a =

ˆbowl

(T · d~a)z z =Q2

64πε0R2z.

Adding this to the contribution from the bowl gives usthe total force exerted on the upper hemisphere

~F =Q2

32πε0R2z +

Q2

64πε0R2z =

3Q2

64πε0R2z.

aThis is Example 8.2 in Griffiths.

Example:


Repeat the previous example, but now instead ofintegrating over the surface of the upper hemisphere, in-tegrate over a larger hemisphere of radius ρ > R that isconcentric with the upper hemisphere of the uniformlycharged sphere.

Again, the force is given by

~F =

ˆbowl

T · d~a+

ˆdisk

T · d~a.

Again, by symmetry, the net force on the upper hemi-sphere of the charged sphere is in the z direction, so wecan neglect the x and y components of T ·d~a right fromthe start and just calculate

~F =

ˆbowl

(T · d~a)z z +

ˆdisk

(T · d~a)z z.

On the hemispherical surface of integration, theelectric field is

~E =1

4πε0

Q

ρ2r.

The area vector is

d~a = ρ2 sin θ dθ dφ r,

and the direction vector is

r = sin θ cosφ x+ sin θ sinφ y + cos θ z.

This is the same as in the previous example, except Rhas been replaced with ρ > R. So, our result for thebowl is the same as in the previous example except withR replaced by ρ

ˆbowl

T · d~a =Q2

32πε0ρ2z.

For the disk, we now have two pieces. We have thedisk r ≤ R and the annulus R ≤ r ≤ ρ. For r < R, the

result is exactly the same as in the previous example,soˆdisk

(T · d~a)z z =Q2

64πε0R2z +

ˆR≤r≤ρ

(T · d~a)z z.

In the annulus R ≤ r ≤ ρ, the electric field is

~E =1

4πε0

Q

r2r,

the area element is

d~a = −r dr dφ z,

and the direction vector is

r = 〈cosφ, sinφ, 0〉.

The z-component of the dot product is

(T · d~a)z = Tzxdax + Tzyday + Tzzdaz

=1

2ε0(E2z − E2

x − E2y

)=

Q2

32π2ε0

1

r3dr dφ.

Then

ˆR≤r≤ρ

(T · d~a)z z =Q2

32π2ε0

ˆ 2π

0

dφ

ˆ ρ

R

1

r3drz

=Q2

32πε0

(1

R2− 1

ρ2

)z.

Combining everything gives us the same result as before

~F =Q2

32πε0ρ2z +

Q2

64πε0R2z +

Q2

32πε0

(1

R2− 1

ρ2

)z

=3Q2

64πε0R2z.

Example:

6.3. MOMENTUM 107

Suppose you have an infinite parallel plate capaci-tor with the upper plate at z = d/2 with charge densityσ and the lower plate at z = −d/2 with charge density+σ. Calculate the stress tensor and force on the upperplate due to the lower plate.a

The electric field between the plates is

~E = −σε 0z.

Since Ex = Ey = 0, all of the off-diagonal elements arezero. The diagonal elements are

Txx = Tyy = − σ2

2ε0, Tzz =

σ2

2ε0,

so the stress tensor is

T =σ2

2ε0

−1 0 0

0 −1 0

0 0 1

.The force is given by Eq. (6.3). Since the charges

are static, the second integral is zero, and we have

~F =

˛S

T · d~a.

The area element for the upper plate is

d~a = −dx dy z.

It is negative because −z is the outward direction forthe closed surface (think hemisphere with infinite ra-dius) enclosing the upper plate. Our dot product isthen

T · d~a =σ2

2ε0

−1 0 0

0 −1 0

0 0 1

0

0

−dx dy

= − σ2

2ε0dx dy z.

The total force on the upper plate is then

~F =

˛S

T · d~a = − σ2

2ε0

ˆ ∞0

ˆ ∞0

dx dy z.

This is of course infinite, but it shows that the force onthe upper plate per unit area is

~f = − σ2

2ε0z.

aThis is parts (a) and (b) of Problem 8.5 in Griffiths.

Example:

We’ve looked at the mathematical statements ofcharge and energy conservation, and we will now do thesame for linear momentum. From Newton’s second law,we know that force is equal to the rate of change of mo-mentum, so we can write Eq. (6.3) as

d~pmechdt

= −ε0µ0d

dt

ˆV

~S dτ +

˛S

T · d~a,

where ~pmech is the total (mechanical) momentum of theparticles in the volume V. The first integral gives themomentum stored in the electromagnetic fields. That is,

~pem = ε0µ0

ˆV

~S dτ.

The second integral gives the momentum per unit timethat is flowing in through the surface S of the volume V.Overall, the equation is the general statement of momen-tum conservation in electromagnetism.

If ~pmech is the density of mechanical momentum,then

~pmech =

ˆV

~pmech dτ,

and

~pem = µ0ε0~S,

is the density of the momentum in the fields, then thedifferential form of the statement of momentum conser-vation is

∂

∂t

(~pmech +~pem

)= ~∇ ·T.

This implies that −T is the momentum flux den-sity. This means that −Tij is the momentum in the idirection crossing a surface oriented in the j direction,per unit area per unit time.


For the parallel capacitor in the previous exam-ple, what is the momentum per unit area per unit timecrossing the xy-plane?a

Since −Tij is the momentum in the i direction

crossing a surface oriented in the j direction, per unitarea per unit time, the momentum crossing the xy-planeper unit area per unit time must be

−Tzz = − σ2

2ε0.

aThis is part (c) of Example 8.5 in Griffiths.

Example:

A parallel plate capacitor has the upper plate atz = d/2 and the lower plate at z = −d/2. The area ofeach plate is A. The electric field between the plates is~E = Ez, and the whole system is in a magnetic field~B = Bx. Calculate the electromagnetic momentumbetween the plates. a

The electromagnetic momentum between theplates is

~pem = µ0ε0

ˆV

~S dτ.

The Poynting vector, in this case, is

~S =~E × ~Bµ0

=EB sin θ

µ0y =

EB

µ0y.

So

~pem = ε0EB

ˆV

dτ y = ε0EBAd y,

since the volume of the gap is Ad.

aThis is part (a) of Problem 8.6 in Griffiths.

Example:

6.4 Angular Momentum

We can also define angular momentum density of elec-tromagnetic fields in terms of the linear momentum defi-nition

~lem = ~r ×~pem = ε0

[~r ×

(~E × ~B

)].

Provided that ~E× ~B 6= 0, even static fields can hold angu-lar momentum. In general, classical angular momentumconservation does not hold unless this contribution fromthe fields is also taken into account.

To find the total angular momentum of the fields, weof course just integrate the angular momentum density

~Lem =

ˆV

~lem dτ.

Dealing with angular momentum in the fields is eas-ier than with linear momentum in the fields because withangular momentum, there is never any hidden momen-tum.

Suppose we have a very long solenoid2 concentricwith the z axis, with radius R, n turns of wire per unitlength, and carrying current I. Inside and coaxial tothe solenoid is a long cylindrical shell of radius r andlength l (long but much shorter than the length of thesolenoid) carrying a uniformly distributed surface charge

+Q. Outside and coaxial to the solenoid is a second cylin-drical shell of radius b and length l carrying a uniformlydistributed surface charge −Q. The system is initiallystatic—nothing is rotating, so there is no obvious angu-lar momentum in the system.

Due to the charged cylinders, there is an electric fieldbetween the cylinders

~E =Q

2πε0lss, a < s < b.

Due to the solenoid, there is a magnetic field inside thesolenoid

~B = µ0nI z s < R.

If the current is slowly turned off, the changing mag-netic field induces a circumferential electric field. Insidethe solenoid, the induced field is

~Eind = −1

2µ0n

dI

dts φ, s < R,

and outside the solenoid, the induced field is

~Eind = −1

2µ0n

dI

dt

R2

sφ, s > R.

The circumferential electric field exerts a torque onthe inner cylinder of

~Na = ~r ×(Q~E

)= −1

2µ0nQa

2 dI

dtz,

2This is example 8.4 in Griffiths

6.4. ANGULAR MOMENTUM 109

and a torque on the outer cylinder of

~N b =1

2µ0nQR

2 dI

dtz.

These torques cause the two charged cylinders to rotate—the inner cylinder in the φ direction, and the outer cylin-der in the −φ direction. The angular momentum of eachcylinder can be calculated by integrating the torque fromthe initial current I to the final current 0:

~La = −1

2µ0nQa

2

ˆ 0

I

dI

dtdt z =

1

2µ0nIQa

2 z

~Lb =1

2µ0nQR

2

ˆ 0

I

dI

dtdt z = −1

2µ0nIQR

2 z.

The total angular momentum of the system is

~La + ~Lb = −1

2µ0nIQ

(R2 − a2

)z.

Given that angular momentum is conserved, wheredid this angular momentum come from? Recall that in

the beginning, everything was static—there was no obvi-ous angular momentum. The answer is that, initially, theangular momentum was in the fields. Returning to the ~Eand ~B that existed before we turned the current down,we calculate the linear momentum density of the fields is

~pem = −µ0nIQ

2πlsφ,

in a < s < R. The angular momentum density is then

~lem = ~r ×~pem = −µ0nIQ

2πlz.

To get the total angular momentum of the fields, we justmultiply this by the volume πl(R2−a2) of the region con-taining angular momentum density, and we find that itprecisely accounts for the angular momentum measuredwhen the current is turned off

~Lem = −1

2µ0nIQ

(R2 − a2

)z.


6.5 Summary: Conservation Laws

Charge

The continuity equation is

∂ρ

∂t= −~∇ · ~J .

This is the statement of (local) charge conservation.

Energy

The Poynting vector

~S =1

µ0

(~E × ~B

),

gives the energy per unit time per unit area transportedby the electromagnetic fields.

The energy density of the electromagnetic fields is

uem =1

2

(ε0E

2 +1

µ0B2

).

The total energy of the electromagnetic fields is

Uem =

ˆV

uem dτ.

The mechanical energy density, umech, of the parti-cles satisfies

dW

dt=

d

dt

ˆV

umech dτ.

If there are no particles in the region, then umech = 0.

The differential form of Poynting’s theorem is

∂

∂t(umech + uem) = −~∇ · ~S.

This is the statement of energy conservation. If this equa-tion is satisfied, then energy is conserved.

The power, or energy per unit time, flowing into aregion with boundary S can be calculated as

P =dU

dt=

ˆ~S · d~a.

Momentum

Where the surface of constant field for a stationary chargeforms a spherical shell, the surface of constant field for amoving charge looks more like a pancake. Where the sur-face of constant field for a line current forms a cylindricalshell, the surface of constant magnetic field of a singlemoving particle looks more like a sphere. Like with a linecurrent, the direction of the field lines can be determinedusing the right-hand-rule.

Momentum in electrodynamics is conserved by con-sidering not just the momenta of the moving particles butalso the momenta of the fields.

The force on charges in a volume V is

~F =

˛S

T · d~a− ε0µ0d

dt

ˆV

~S dτ,

where T is the Maxwell stress tensor with elements

Tij = ε0

(EiEj −

1

2δijE

2

)+

1

µ0

(BiBj −

1

2δijB

2

).

The Maxwell stress tensor is a symmetric rank-2 tensorwith nine elements—six of them independent. It can bethought of as a 3× 3 matrix, then the dot product T · d~ais a vector obtained by multiplying the 3 × 3 matrix Tby the 3-element vector d~a. Physically, the element Tijis the force per unit area in the ith direction acting on asurface element oriented in the jth direction. The diago-nal elements are pressures and the off-diagonal elementsare shears.

If´~S dτ is independent of time, or if the charge dis-

tribution is static, then the second term is zero. Thento calculate the total force, we integrate the stress tensorover the surface S of the volume V.

To calculate the total electromagnetic force on a vol-ume of charge by integrating the Maxwell stress tensor,use the following procedure:

1. Identify the volume of interest. If´~S dτ is inde-

pendent of time, or if the charge distribution isstatic, which it probably is, then the second inte-gral is zero, and the force acting on the volume ofcharge is just

~F =

˛S

T · d~a.

This is integrated over the closed surface of the vol-ume of interest. You may have to break this intomultiple surfaces. For example, if your volume isthe upper half of a sphere, then one surface wouldbe the upper hemisphere, and the second surfacewould be the disk capping the bottom of the hemi-sphere.

2. Determine the area element d~a. This vector shouldbe perpendicular to the surface and point outwardfrom the volume of interest.

3. Determine the electric and magnetic fields at thesurfaces of integration.

4. Compute the elements of the Maxwell stress tensor.T is symmetric, so there are only six independentelements.

5. Compute the elements of the dot product (T · d~a)x,(T · d~a)y, and (T · d~a)z by multiplying the 3 × 3matrix T by the vector d~a.

6.5. SUMMARY: CONSERVATION LAWS 111

6. Now, for each surface, computeˆT · d~a =

ˆ(T · d~a)x x

+

ˆ(T · d~a)y y

+

ˆ(T · d~a)z z.

If, in the beginning, you identify by symmetry thedirection of the resultant force, then you may onlyhave to compute one of the three elements of T ·d~a.

The momentum density of the electromagnetic fieldsis

~pem = µ0ε0~S.

Note that ~S = 0 if either ~E = 0 or ~B = 0, so ~pem is

nonzero only in regions containing both ~E and ~B. Thetotal momentum stored in the electromagnetic fields isthen

~pem =

ˆ~pem dτ = ε0µ0

ˆV

~S dτ.

The mechanical energy of the particles is

~pmech =

ˆV

~pmech dτ,

where ~pmech is the density of mechanical momentum.

The differential form of the statement of momentumconservation is

∂

∂t

(~pmech +~pem

)= ~∇ ·T.

This implies that −T is the momentum flux density.This means that −Tij is the momentum in the i directioncrossing a surface oriented in the j direction, per unitarea per unit time.

Angular Momentum

The angular momentum density of electromagnetic fieldsin terms of the linear momentum definition is

~lem = ~r ×~pem = ε0

[~r ×

(~E × ~B

)].

Even static fields can hold angular momentum. In gen-eral, classical angular momentum conservation does nothold unless this contribution from the fields is also takeninto account.

The total angular momentum of the fields is foundby integrating the angular momentum density

~Lem =

ˆV

~lem dτ.

To calculate the total angular momentum of elec-tromagnetic fields, first calculate the linear momentumdensity, then the angular momentum density, and fromthat the total angular momentum. All three of these arezero except in regions with both ~E and ~B.

Chapter 7

Electromagnetic Waves

7.1 One-dimensional Waves

Griffiths defines a wave as “a disturbance in a continuousmedium that propagates with a fixed shape at constantvelocity.” This is of course an ideal wave. Absorption bythe medium and dispersion will typically cause a wave tochange shape as it propagates.

Suppose we have some time-dependent function onthe z-axis f(z, t). We don’t want just any function, butrather, a function that describes a propagating waveform.Initially, the wave has shape g(z) = f(z, 0). If the wave-form is moving to the right with speed v, then the dis-placement at some point z at time t is the same as theearlier displacement at time 0 at a location vt to the leftof z. That is

f(z, t) = f(z − vt, 0) = g(z − vt).

Any function of the form

f(z, t) = g(z − vt),

represents a propagating wave form.

A wave on a string is a good model for one-dimensional wave motion. Suppose we have a very longstring under tension T that is displaced from equilibrium.We will consider a small piece of the string in the interval[z, z + ∆z]. On the left side, the string makes an angleθ with the horizontal, and at z + ∆z, it makes an angleθ′ with the horizontal. The net transverse force on thispiece of string is

∆F = T sin θ′ − T sin θ

' T (tan θ′ − tan θ)

= T

(∂f

∂z

∣∣∣z+∆z

− ∂f

∂z

∣∣∣z

)' T

∂2f

∂z2∆z.

Here we assumed that the angles are small so sin θ 'tan θ. In the last step, we used the second derivative ap-

proximation f ′(z+∆z)−f ′(z)∆z = f ′′(z).If the string’s linear

mass density is µ, then Newton’s second law gives us

∆F = µ∆z∂2f

∂t2.

Putting the two together gives us

∂2f

∂z2=µ

T

∂2f

∂t2.

This is a wave equation that describes a wave traveling

with speed v =√

Tµ .

In general, the one-dimensional wave equation is

∂2f

∂z2=

1

v2

∂2f

∂t2, (7.1)

where v is the speed of the wave.

All functions of the form f(z, t) = g(z − vt) satisfythe wave equation. To show that, we let u = z − vt thenf(z, t) = g(u). Taking first derivatives gives us

∂f

∂z=

∂g

∂u

∂u

∂z=∂g

∂u∂f

∂t=

∂g

∂u

∂u

∂t= −v ∂g

∂u.

Taking second derivatives gives us

∂2f

∂z2=

∂

∂z

(∂g

∂u

)=∂2g

∂u2

∂u

∂z=∂2g

∂u2

∂2f

∂t2= −v ∂

∂t

(∂g

∂u

)= −v ∂

2g

∂u2

∂u

∂t= v2 ∂

2g

∂u2.

So we have that∂2f

∂z2=

1

v2

∂2f

∂t2.

We can also show that f(z, t) = h(z + vt) satisfiesthe wave equation. In fact, the most general solution tothe wave equation is

f(z, t) = g(z − vt) + h(z + vt).

Every solution of the wave equation, i.e. every classicalwave function, can be put in this form.

112

7.1. ONE-DIMENSIONAL WAVES 113

Write the standing wave

f(z, t) = A sin(kz) cos(kvt),

in the form f(z, t) = g(z − vt) + h(z + vt).a

Using the trig identity

sin(u± v) = sinu cos v ± cosu sin v,

we find that

f(z, t) =A

2sin(kz + kvt) +

A

2sin(kz − kvt).

aThis is problem 9.2 in Griffiths.

Example:

Sinusoidal Waves

A sinusoidal wave is a wave of the form

f(z, t) = A cos (k[z − vt] + δ) .

The argument of the cosine is called the phase and δ thephase constant. The wave number is

k =2π

λ,

where λ is the wavelength. The period, i.e., the timeto complete one full cycle is

T =2π

kv.

The frequency, or the number of complete oscillationsper unit time is

ν =1

T=kv

2π=v

λ.

The wave may also be characterized by its angular fre-quency

ω = 2πν = kv,

which gives the radians swept out per unit time.

In terms of the angular frequency,

f(z, t) = A cos (kz − ωt+ δ) .

This is for a wave traveling to the right. For a wave trav-eling to the left, we just have to change the sign of thewave number

f(z, t) = A cos (−kz − ωt+ δ) .

Using Euler’s formula,

eiθ = cos θ + i sin θ,

we can write the wave as

f(z, t) = Re[Aei(kz−ωt+δ)

].

For calculations, it is often easier to deal with wavesin complex form, so we introduce the function

f(z, t) = Aei(kz−ωt),

whereA = Aeiδ,

Then the real wave is

f(z, t) = Re[f(z, t)

].

For example, to combine two sinusoidal waves, yousimply take the real part of the sum of their complexforms since

f1 + f2 = Re(f1) + Re(f2) = Re(f1 + f2).

In particular, if the two wave numbers and angular fre-quencies are the same, then

f1 + f2 = Re(f1 + f2

)= Re

(A1e

i(kz−ωt) + A2ei(kz−ωt)

)= Re

([A1 + A2

]ei(kz−ωt)

)= Re

([A1 +A2] ei(kz−ωt+δ)

)= (A1 +A2) cos (kz − ωt+ δ)

A sinusoidal wave is a specific kind of waveform, butit turns out that any wave can be written as a linearcombination of sinusoidal waves. This is why the studyof sinusoidal waves is so important and why we will fo-cus entirely on sinusoidal waves and their linear combi-nations.

To solve Eq. (7.1), we assume a separable solution ofthe form

f(z, t) = Z(z)T (t),

then the wave equation is

∂2Z

∂z2T =

1

v2Z∂2T

∂t2.

Dividing through by ZT gives us

1

Z

∂2Z

∂z2=

1

v2

1

T

∂2T

∂t2= −k2.

Since both sides are in terms of one variable only, bothsides must be equal to a constant. For convenience, wechoose that constant to be −k2. Then we get the twosolutions

Z(z) = Aeikz +Be−ikz

T (t) = Ceikvt +De−ikvt.

114 CHAPTER 7. ELECTROMAGNETIC WAVES

Then our general solution is

fk(z, t) =(Aeikz +Be−ikz

) (Ceikvt +De−ikvt

)= ACeik(z+vt) +ADeik(z−vt)

+BCe−ik(z−vt) +BDe−ik(z+vt).

By letting k be negative as well as positive, we can shortenthis to

fk(z, t) = Jeik(z+vt) +Keik(z−vt).

where J and K are new coefficients. Only real terms areallowed, so

fk(z, t) = J cos(kz + ωt) +K cos(kz − ωt),

where ω = kv. Since all negative values of k are allowedand since cosine is an even function, we can actually writethis as

fk(z, t) = Ak cos(kz − ωt),

where Ak is a new constant. When k is negative, wecan factor out a negative from the argument, to getcos(kz + ωt). More generally, we can write this as

fk(z, t) = Akei(kz−ωt).

The general linear combination is then

f(z, t) =

ˆ ∞−∞

A(k)ei(kz−ωt) dk.

This is just a Fourier transform, and the coefficients A(k)can be obtained from the theory of Fourier transforms.

ForA3e

iδ3 = A1eiδ1 +A2e

iδ2 ,

express A3 and δ3 in terms of A1, A2, δ1, and δ2.a

Using Euler’s formula, we can expand both sidesas

A3 (cos δ3 + i sin δ3) = A1 (cos δ1 + i sin δ1)

+A2 (cos δ2 + i sin δ2)

(A3 cos δ3) + i (A3 sin δ3) = (A1 cos δ1 +A2 cos δ2)

+i (A1 sin δ1 +A2 sin δ2) .

Equating real and imaginary parts gives us

A3 cos δ3 = A1 cos δ1 +A2 cos δ2

A3 sin δ3 = A1 sin δ1 +A2 sin δ2.

If we divide the second by the first, the A3’s cancelout, and we can take the inverse tangent to get

δ3 = tan−1

(A1 sin δ1 +A2 sin δ2A1 cos δ1 +A2 cos δ2

).

On the other hand, if we square both equations andadd them together, we get

A23 = A2

1 +A22 + 2A1A2 cos(δ1 − δ2),

after making use of two trigonometric identities.

aThis is Problem 9.3 in Griffiths.

Example:

Reflection and Transmission

Suppose you have one string attached to a second string.We are interested in the mathematical expression of wavespropagating on these joined strings.

The tension in both strings is the same, but the lin-ear mass density µ is not necessarily the same. If theyare different, then the speed of a wave will change as it

travels from one string to the other since v =√

Tµ .

We’ll define our coordinate system such that thestrings lie along the z-axis, and the knot joining the twostrings is at z = 0. Suppose we have an incident wavecoming in from the left (i.e. z < 0) with equation

fI(z, t) = AIei(k1z−ωt).

When the wave hits the knot, there will be some reflectedwave reflected back toward the left in string 1

fR(z, t) = ARei(−k1z−ωt),

and some transmitted wave that travels to the right instring 2

fT (z, t) = AT ei(k2z−ωt).

7.1. ONE-DIMENSIONAL WAVES 115

In both strings and for all three waves, the angu-lar frequency ω is the same. However, the wave speeds,wavelengths, and wave numbers are different. They arerelated via

λ1

λ2=k2

k1=v1

v2.

The total wave function is

f(z, t) =

AIe

i(k1z−ωt) + ARei(−k1z−ωt) for z < 0

AT ei(k2z−ωt) for z > 0.

.

To solve this problem, we need to consider the condi-tions at the boundary between the two strings. Assumingthe combined string is continuous and smooth at the knot,

then f(0−, t) = f(0+, t) and ∂f∂z

∣∣∣0−

= ∂f∂z

∣∣∣0+

. We can gen-

eralize this to the complex form of the wave function, sowe have the conditions

f(0−, t) = f(0+, t),

and∂f

∂z

∣∣∣0−

=∂f

∂z

∣∣∣0+.

These conditions imply that

AI + AR = AT

k1

(AI − AR

)= k2AT .

These allow us to write the reflection and transmissioncoefficients in terms of the incident coefficient as

AR =k1 − k2

k1 + k2AI

AT =2k1

k1 + k2AI .

Alternatively, using the relation k2k1

= v1v2

, we can writethem in terms of the wave speeds as

AR =v2 − v1

v1 + v2AI

AT =2v2

v1 + v2AI .

The real wave functions are then related as

AReiδR =

v2 − v1

v2 + v1AIe

iδI

AT eiδT =

2v2

v2 + v1AIe

iδI .

To obtain the real magnitudes, we simply calculatethe magnitude of both sides and use the fact that, for ex-ample,

∣∣eiδ∣∣ = 1, to get The real wave functions are thenrelated as

AR =

∣∣∣∣v2 − v1

v2 + v1

∣∣∣∣AIAT =

2v2

v2 + v1AI .

If µ2 < µ1, we see that v2 > v1, then

AR =v2 − v1

v2 + v1AI .

Then returning to the complex form, we get

AReiδR =

v2 − v1

v2 + v1AIe

iδI

v2 − v1

v2 + v1AIe

iδR =v2 − v1

v2 + v1AIe

iδI

δR = δI .

Similarly, we find that δT = δI , so all three waves are inphase.

On the other hand, if µ2 > µ1, then v2 < v1, then

AR = −v2 − v1

v2 + v1AI .

Then returning to the complex form, we get

AReiδR =

v2 − v1

v2 + v1AIe

iδI

−v2 − v1

v2 + v1AIe

iδR =v2 − v1

v2 + v1AIe

iδI

−eiδR = eiδI

eiπeiδR = eiδI

δR = δI − π.

So the reflected and incident waves are 180 out of phase.

If the second string is infinitely massive, i.e., the endof the first string is anchored, then v2 = 0, and AR = AIand AT = 0. There is still the 180 phase change.

Polarization

A wave propagating on a string is a transverse wave.For transverse waves, there are two directions that areorthogonal to the direction of propagation and to eachother. In other words, there are two independent polar-izations of the wave.

If our wave is propagating in the z direction, thenwe can define the two polarization directions as being inthe x direction and the y direction. We can have verticalpolarization, in which case

~fv(z, t) = Aei(kz−ωt)x,

or horizontal polarization, in which case

~fh(z, t) = Aei(kz−ωt)y.

For a wave propagating in the z direction, the polariza-tion could be in any other direction in the xy-plane

~f(z, t) = Aei(kz−ωt)n,


where n is the polarization vector and defines the planein which the wave is oscillating. Since n is orthogonal toz, we know that n · z = 0.

The polarization angle is the angle θ for which

n = cos θ x+ sin θ y.

Thus, polarization in the n direction can be thought of asa superposition of vertical and horizontal polarizations.

~f(z, t) = A cos θ ei(kz−ωt)x+ A sin θ ei(kz−ωt)y.

7.2 EM Waves in a Vacuum

Away from sources, Maxwell’s equations are

~∇ · ~E = 0

~∇ · ~B = 0

~∇× ~E = −∂~B

∂t

~∇× ~B = µ0ε0∂ ~E

∂t.

If we take the curl of the third equation and plug in thefirst, we get

~∇×(~∇× ~E

)= ~∇×

(−∂

~B

∂t

)~∇(~∇ · ~E

)−∇2 ~E = − ∂

∂t

(~∇× ~B

)~∇ (0)−∇2 ~E = − ∂

∂t

(µ0ε0

∂ ~E

∂t

).

This implies that

∇2 ~E = µ0ε0∂2 ~E

∂t2. (7.2)

Similarly, if we take the curl of the fourth Maxwell equa-tion and then plug in ~∇ · ~B and ~∇× ~E, we get

∇2 ~B = µ0ε0∂2 ~B

∂t2. (7.3)

We’ve now decoupled ~E and ~B into a pair of waveequations. Recall that the wave equation has the form

∇2f =1

v2

∂2f

∂t2.

This implies that the speed of an electromagnetic wave is

v =1

√ε0µ0

= c.

For electromagnetism, each wave equation is a vectorequation—each of the three components are waves.

We are interested in monochromatic plane waveswhose solutions have the form

~E(z, t) = ~E0ei(kz−ωt)

~B(z, t) = ~B0ei(kz−ωt).

Monochromatic waves are those with a set frequency ω.In our case, we assume they are propagating in the z di-rection. Since they are plane waves, they have no xor y dependence. Note: Our wave equations (7.2) and(7.3) are too general in that Maxwell’s equations add fur-ther constraints. So we now apply Maxwell’s equations

to ~E(z, t) and ~B(z, t).

Taking the divergence of the electric field, we get

~∇ · ~E = ~∇ · ~E0 cos (kz − ωt+ δ)

= ~∇ · 〈E0x, E0y, E0z〉 cos (kz − ωt+ δ)

= −kE0z sin (kz − ωt+ δ) .

However, since ~∇· ~E = 0, this implies that E0z = 0. Since

the real part of ~E0 differs from the imaginary part onlyby sine versus cosine, it must also be true that E0z = 0.We can do the same with the magnetic field and the factthat ~∇ · ~B = 0. So,

E0z = B0z = 0.

This implies that electromagnetic waves are transverse.That is, the electric and magnetic fields have zero com-ponent in the direction in which the wave is propagating.

Similarly, Faraday’s law ~∇× ~E = −∂ ~B∂t implies that

−kE0y = ωB0x

kE0x = ωB0y.

This implies that ~E and ~B are in phase and perpendicularto each other. It also implies that

B0 =k

ωE0 =

1

cE0.

Because of the division by the speed of light, we can seethat | ~B| << |~E|.

Putting this all together, if ~E is in the x-direction,then ~B must be in the y direction if the wave is propa-gating in the z direction. Then

~E(z, t) = E0ei(kz−ωt)x

~B(z, t) =1

cE0e

i(kz−ωt)y.

Taking the real parts gives us

~E(z, t) = E0 cos (kz − ωt+ δ) x

~B(z, t) =1

cE0 cos (kz − ωt+ δ) y.

This is an electromagnetic wave that is polarized in the xdirection. By convention, we look at ~E when classifyingthe polarization.

7.2. EM WAVES IN A VACUUM 117

More generally, for a wave propagating in an arbi-trary direction with wave vector ~k, wave number |~k|,and polarization n, the wave functions are

~E(~r, t) = E0ei(~k·~r−ωt)n

~B(~r, t) =1

cE0e

i(~k·~r−ωt)(k × n

)=

1

ck × ~E.

The real ~E and ~B fields are

~E(~r, t) = E0 cos(~k · ~r − ωt+ δ

)n

~B(~r, t) =1

cE0 cos

(~k · ~r − ωt+ δ

)(k × n

)=

1

ck × ~E.

Since, ~E is transverse, we have that

k · n = 0.

What is the direction of the Poynting vector forthe fields given immediately above?

Recall that

~S =1

µ0

~E × ~B.

In our case, the direction will be

~S ∝ n×(k × n

).

Using the BAC-CAB identity,

~A×(~B × ~C

)= ~B

(~A · ~C

)− ~C

(~A · ~B

),

we get

n×(k × n

)= k (n · n)− n

(n · k

),

Since the direction of polarization, n, is alwaysperpendicular to the direction of propagation, k, weknow that n · k = 0. And since n ·n = 1, the directionof the Poynting vector is

~S ∝ n×(k × n

)= k.

Example:

Recall that the energy density of the electromag-netic fields is

u =1

2

(ε0E

2 +1

µ0B2

).

For a monochromatic plane wave,

B2 =1

c2E2 = µ0ε0E

2,

so the energy density becomes

u = ε0E2 = ε0E

20 cos2 (kz − ωt+ δ) ,

if the wave is propagating in the z direction.

The energy per unit area per unit time, or the en-ergy flux density, transported per unit time by the elec-tromagnetic wave, is given by the Poynting vector~S = 1

µ0

~E × ~B, which becomes

~S = cε0E20 cos2 (kz − ωt+ δ) z = cu z,

for a monochromatic wave traveling in the z direction.

Recall that the momentum density of electromag-netic fields is ~p = µ0ε0~S. But since c = 1/

√µ0ε0, we can

write this as

~p =1

c2~S.

For a monochromatic wave traveling in the z direction,this becomes

~p =1

cε0E

20 cos2 (kz − ωt+ δ) z =

1

cu z.

Usually we only care about average values since typ-ical measurements of light will include many cycles. Theaverage values of the energy density, the Poynting vector,and the momentum density are

〈u〉 =1

2ε0E

20

〈~S〉 =1

2cε0E

20 z

〈~p〉 =1

2cε0E

20 z.

In general, the direction of the Poynting vector is thesame as the direction of wave propagation.

Tip:

The intensity, or average power per unit area trans-ported by the wave, is

I = 〈S〉 =1

2cε0E

20 .

When light hits an ideal absorber, all of the wave’smomentum is delivered to the surface. This results in aradiation pressure, or average force per unit area onthe surface. In a time ∆t, the radiation pressure on anideal absorber of surface area A due to an electromagneticwave is

P =1

A

∆p

∆t=

1

A

〈p〉Ac∆t∆t

=1

2ε0E

20 =

I

c,

where P is the radiation pressure, and p is momentum.For an ideal reflector, the radiation pressure is twice thissince the momentum not only stops, but changes direc-tion and reflects backward.


Calculate the physical ~E and ~B fields, in Carte-sian coordinates, for a monochromatic plane wave ofamplitude E0, phase constant δ = 0, angular frequencyω, polarization in the xz-plane, and traveling from~a = 〈1, 2, 3〉 to ~b = 〈4, 5, 6〉.

We know that ~E and ~B are

~E = E0 cos(~k · ~r − ωt

)n

~B =1

cE0 cos

(~k · ~r − ωt

)(k × ~n

).

We just need to find k, ~k, n, and then ~k · ~r and k× n.

The direction of propagation is

k =~b− ~a|~b− ~a|

=〈3, 3, 3〉|〈3, 3, 3〉|

=〈1, 1, 1〉√

3.

So the wave vector is

~k = k k =ω

c√

3〈1, 1, 1〉.

We know the polarization is in the xz-plane, so

n = 〈A, 0, B〉,

for some unknown A and B. We know this must beperpendicular to the wave vector, so n · k = 0. Taking

this dot product and setting it equal to zero, tells usthat A = −B. So

n = 〈A, 0,−A〉.

But since this is a unit vector, we know it must be nor-malized. This implies that A = 1√

2, so

n =〈1, 0,−1〉√

2.

The other dot product is

~k · ~r =ω

c√

3〈1, 1, 1〉 · 〈x, y, z =

ω

c√

3(x+ y + z).

The cross product is

k × n =1√6〈1, 1, 1〉 × 〈1, 0,−1〉

=1√6

∣∣∣∣∣∣∣∣x y z

1 1 1

1 0 −1

∣∣∣∣∣∣∣∣=

1√6〈−1, 2,−1〉.

Example:

7.3 EM Waves in Linear Materials

In matter with no free charge or free current, recall thatMaxwell’s equations are

~∇ · ~D = 0

~∇ · ~B = 0

~∇× ~E = −∂~B

∂t

~∇× ~H =∂ ~D

∂t.

If the material is linear, then ~D = ε~E, and ~H = 1µ~B, so

these equations become

~∇ · ~E = 0

~∇ · ~B = 0

~∇× ~E = −∂~B

∂t

~∇× ~B = µε∂ ~E

∂t.

These differ from Maxwell’s equations in a vacuum only inthat the µ0ε0 is now µε. Apparently, an electromagneticwave in a linear material travels with speed

v =1√µε

=c

n,

where

n =

√εµ

ε0µ0,

is a material-dependent constant called the index of re-fraction. For most materials, µ ≈ µ0, so n ≈ √εr.

The energy density, Poynting vector, and intensityare still the same, but now with ε and µ instead of ε0 andµ0.

u =1

2

(εE2 +

1

µB2

)~S =

1

µ

(~E × ~B

)I =

1

2εvE2

0 ,

where for monochromatic plane waves, v = ωk .

For a wave that passes from one transparent materialto another, the boundary conditions are:

ε1E⊥1 = ε2E

⊥2

~E‖1 = ~E

‖2

B⊥1 = B⊥21

µ1

~B‖1 =

1

µ2

~B‖2.

7.3. EM WAVES IN LINEAR MATERIALS 119

Reflection and Transmission at NormalIncidence

Suppose an electromagnetic wave of frequency ω passesfrom one linear material with k = k1 to another wherek = k2, where the xy-plane forms the boundary betweenthe two materials, and the incident electromagnetic waveis normal to the boundary. The incident wave is

~EI(z, t) = E0Iei(k1z−ωt)x

~BI(z, t) =1

v1E0Ie

i(k1z−ωt)y.

When the wave hits the boundary, a reflected wave travelsbackward in the first material

~ER(z, t) = E0Rei(−k1z−ωt)x

~BR(z, t) = − 1

v1E0Re

i(−k1z−ωt)y.

The negative sign for the reflected magnetic wave is re-

quired because ~B = 1c k ×

~E. Remember that k pointsin the direction of propagation, in this case, in the −zdirection, and ~E points in the x direction in this case. So

k × ~E points in the −y direction.

The transmitted wave that travels on in the z direc-tion through material 2 is

~ET (z, t) = E0T ei(k2z−ωt)x

~BT (z, t) =1

v2E0T e

i(k2z−ωt)y.

Next, we apply the boundary conditions.

ε1E⊥1 = ε2E

⊥2

~E‖1 = ~E

‖2

B⊥1 = B⊥21

µ1

~B‖1 =

1

µ2

~B‖2.

In this case, ~EI and ~BI are parallel to the surface. Sincethere are no perpendicular components, we don’t careabout the first or third boundary conditions. For the sec-ond condition, we require that the combined electric fieldson the left joins to the electric field on the right at theboundary where z = 0. This gives us

~EI(0, t) + ~ER(0, t) = ~ET (0, t)

E0Ie−iωtx+ E0Re

−iωtx = E0T e−iωtx

E0I + E0R = E0T .

For the magnetic field, we get

1

µ1

(~BI(0, t) + ~BR(0, t)

)=

1

µ2

~BT (0, t)

1

µ1

(1

v1E0I −

1

v1E0R

)=

1

µ2

1

v2E0T

E0I − E0R = βE0T ,

whereβ =

µ1v1

µ2v2=µ1n2

µ2n1.

Adding/subtracting the two results gives us

E0T =2

1 + βE0I

E0R =1− β1 + β

E0I .

For most materials, µ ≈ µ0. In that case, β ≈ v1v2

,then

E0T =2v2

v2 + v1E0I

E0R =v2 − v1

v2 + v1E0I .

Notice that these are identical to the results for the waveon a string. The magnitudes of the real amplitudes are

E0T =2v2

v2 + v1E0I =

2n1

n1 + n2E0I

E0R =

∣∣∣∣v2 − v1

v2 + v1

∣∣∣∣E0I =

∣∣∣∣n1 − n2

n1 + n2

∣∣∣∣E0I .

Recall that to find the real amplitude when given some-thing like

A = kB,

you can first write the complex numbers in polar form as

AeeδA = kBeiδB .

Then you just take the magnitude (complex) of both sides∣∣AeiδA∣∣ =∣∣kBeiδB ∣∣ .

This simplifies toA = kB,

since |eix| = 1.

Recall that the intensity is given by

I =1

2εvE2

0 .

If we want to know the fraction of energy that is reflectedand the fraction that is transmitted, we compute the re-flection coefficient

R =IRII,

and the transmission coefficient

T =ITII.

Note thatR+ T = 1.


In our case, assuming µ1 = µ2 = µ, the reflectioncoefficient is

R =IRII

=

(E0R

E0I

)2

=

(n1 − n2

n1 + n2

)2

,

and the transmission coefficient is

T =ITII

=ε2v2

ε1v1

(E0T

E0I

)2

=4n1n2

(n1 + n2)2.

Reflection and Transmission at ObliqueIncidence

Now we repeat the previous analysis of a monochromaticelectromagnetic wave hitting the surface between two ma-terials, but now we will not assume that the incident waveis normal to the surface. In general, we have the incidentwave

~EI(~r, t) = ~E0Iei(~kI ·~r−ωt)

~BI(~r, t) =1

v1kI × ~EI .

This gives us a reflected wave

~ER(~r, t) = ~E0Rei(~kR·~r−ωt)

~BR(~r, t) =1

v1kR × ~ER.

and a transmitted wave

~ET (~r, t) = ~E0T ei(~kT ·~r−ωt)

~BT (~r, t) =1

v2kT × ~ET .

Since the incident wave is no longer in the z direc-tion, we no longer have k = kz. We now have threedifferent angles to consider. The incident wave hits thesurface at some angle θI , called the angle of incidence,from the normal. The reflected wave leaves the surfaceat some angle θR, called the angle of reflection. fromthe normal. The transmitted wave leaves the surface atsome angle θT , called the angle of refraction, from thenormal.

The geometry of our setup implies three fundamentallaws of geometrical optics. These laws hold for all kindsof waves incident on surfaces—not just electromagneticwaves.

Law 1: The incident, reflected, and transmitted wavevectors are all in the same plane. This plane in-cludes the normal to the surface.

Law 2: The incident angle is equal to the reflected an-gle

θI = θR.

This is called the law of reflection.

Law 3: The third law is also called the law of refrac-tion or Snell’s law

sin θTsin θI

=n1

n2.

Recall the boundary conditions

ε1E⊥1 = ε2E

⊥2

B⊥1 = B⊥2

~E‖1 = ~E

‖2

1

µ1

~B‖1 =

1

µ2

~B‖2.

From the first law, we know that ~kI , ~kR, and ~kT areall in the same plane—the incident plane. If the boundaryof the surface between the two materials is the xy-plane,then we can take the plane of incidence to be the xz-plane. That is, the wave vectors ~kI , ~kR, and ~kT are alllying in the xz-plane. The z-component of the fields isthe component that is perpendicular to the the surface,and the part of the fields that is parallel to the surfaceare the x and y components of the fields.

At the boundary, z = 0, and the exponential factorsof the fields cancel. That is,

~kI · ~r = ~kR · ~r = ~kT · ~r.

So our boundary conditions become

ε1

(~E0I + ~E0R

)z

= ε2

(~E0T

)z(

~B0I + ~B0R

)z

=(~B0T

)z(

~E0I + ~E0R

)x,y

=(~E0T

)x,y

1

µ1

(~B0I + ~B0R

)x,y

=1

µ2

(~B0T

)x,y

.

On the last two equations, the x, y subscripts mean thateach of the equations is really a pair of equations—onefor the x component and one for the y component. Also,

note that ~B0 = 1v k ×

~E0 in each case.

How we proceed from here depends on the directionof polarization. We will assume that the polarization isparallel to the plane of incidence. Since the plane of inci-dence is the xz plane, then the electric fields for all threewaves are in the xz-plane. This implies that the magneticfields for all three waves are in the ±y direction.

Notice the relations between the different electricfields as well as the wave vectors in the graphic below.For all three waves, the electric field vector, i.e. themagnitude of the electric field times the polarization di-rection is perpendicular to the wave vector. Also, asrequired by the third boundary condition, notice that

7.3. EM WAVES IN LINEAR MATERIALS 121

(E0I)x+(E0R)x = (E0T )x and the same for the y compo-nents. The magnetic field direction can be deduced usingthe right-hand rule and the fact that ~B0 = 1

vk × ~E0. Inall three cases, the magnetic field either points into thepage or out of the page.

z

xkR

kI

kT

~E0I

(E0I)x

(E0I)z

~E0R

~E0T

θRθI

θT

From the first boundary condition, we get

ε1

(−E0I sin θI + E0R sin θR

)= ε2

(−E0T sin θT

).

(7.4)Since the magnetic fields have no z components, the sec-ond boundary condition tells us nothing. The electricfields have no y components, so the third boundary con-dition gives us

E0I cos θI + E0R cos θR = E0T cos θT . (7.5)

From the fourth boundary condition, because the mag-netic fields have no x compoments, we get

1

µ1v1

(E0I − E0R

)=

1

µ2v2E0T . (7.6)

Using the laws of reflection and refraction, Eq. (7.4)and Eq. (7.6) both reduce to

E0I − E0R = βE0T ,

whereβ =

µ1v1

µ2v2=µ1n2

µ2n1.

Eq. (7.5) can be written as

E0I + E0R = αE0T ,

where

α =cos θTcos θI

.

Solving these two equations for the reflected and trans-mitted amplitudes gives us the Fresnel equations for

the case in which the polarization is parallel to the inci-dent plane

E0R =

(α− βα+ β

)E0I

E0T =

(2

α+ β

)E0I .

We can rewrite cos θT using a nice trick

cos θT =√

cos2 θT =

√1− sin2 θT .

From Snell’s law, sin θT = n1

n2sin θI . So

cos θT =

√1−

(n1

n2

)2

sin2 θI .

So we can write α as a function of the angle of incidence

α =cos θTcos θI

=

√1−

(n1

n2

)2

sin2 θI

cos θI.

If E0R = 0, then no reflection occurs. In this case,this occurs when α = β. The angle θB at which this oc-curs is called Brewster’s angle. In our case, it occurswhen

sin2 θB =1− β2(n1

n2

)2

− β2

.

For typical surfaces, µ1 ' µ2, then β ' n2

n1. Then

sin θB 'n2√n2

1 + n22

.

If you construct a right triangle with angle θB , oppositeside n2 and hypotenuse

√n2

1 + n22, then we see that

tan θB 'n2

n1.

The wave intensities are

II =1

2ε1v1E

20I cos θI

IR =1

2ε1v1E

20R cos θR

IT =1

2ε2v2E

20T cos θT .

The cosines are there because the wave fronts are at anangle to the surface. The reflection coefficient is then

R =IRII

=

(E0R

E0I

)2

=

(α− βα+ β

)2

.


since θI = θR. The transmission coefficient is

T =ITII

=ε2v2

ε1v1

(E0T

E0I

)2cos θTcos θI

=ε2v2

ε1v1

(2

α+ β

)2cos θTcos θI

= αβ

(2

α+ β

)2

.

In the last line, we used the fact that v = 1√µε , which

implies that ε2ε1

= µ1

µ2

(v1v2

)2

,

ε2v2

ε1v1=v2

v1

µ1

µ2

(v1

v2

)2

=v1

v2

µ1

µ2= β.

Take the square root of the complex number z =x+ iy.

One way to do it is to write√z = a+ bi. Then

z = (a+ bi)2 = (a2 − b2) + (2ab)i.

This gives us the system of equations

x = a2 − b2

y = 2ab,

which we need to solve for a and b. Plugging a = y2b

into x = a2 − b2, and then solving the quadratic-likeequation for b2 gives us

b2 = −1

2x± 1

2

√x2 + y2.

Plugging this back into x = a2 − b2 gives us

a2 =1

2x± 1

2

√x2 + y2.

Taking the square root of these, then since√z = a+ bi,

we have that

√z =

1√2

[±√x±

√x2 + y2 ± i

√−x±

√x2 + y2

].

This has a lot of plus/minus symbols, so there are anumber of possible solutions. The plus-minus symbolsinside the square roots are correlated—meaning if oneis positive then the other is positive. However, theseare not necessarily correlated with the outer plus-minussymbols. Testing all the possibilities to see if squar-ing them gives z = x + iy, we find that there are twosolutions:

√z = ± 1√

2

[√√x2 + y2 + x+ i

√√x2 + y2 − x

].

Example:

Take the square root of the complex number z =x+ iy, by converting it to polar form.

We can write z in polar form as

z = reiφ,

where r is the magnitude and φ is the phase angle

r =√x2 + y2

φ = tan−1(yx

).

Then taking the square root and converting the re-sult back to Cartesian form gives us√z =

√rei

φ2

=(x2 + y2

) 14

[cos

(φ

2

)+ i sin

(φ

2

)]=

(x2 + y2

) 14

[±√

1 + cosφ

2± i√

1− cosφ

2

]

We can simplify further since

cosφ = cos(

tan−1(yx

))=

x√x2 + y2

.

Plugging this in for cosφ and simplifying a little bitgives us

√z =

1√2

[±√√

x2 + y2 + x± i√√

x2 + y2 − x].

This gives us four possible solutions. Testing each oneby squaring them to see if they simplify to x + iy, wefind that the two final solutions are

√z = ± 1√

2

[√√x2 + y2 + x+ i

√√x2 + y2 − x

].

Example:

7.4. ABSORPTION AND DISPERSION 123

7.4 Absorption and Dispersion

EM Waves in Conductors

In a conductor, the free current and free charge densities~Jf and ρf are not zero like they are in linear materials.Any free charge density in a conductor dissipates veryquickly. In the following, we assume that any initial freecharge density has already dissipated. Maxwell’s equa-tions for a conductor are then

~∇ · ~E = 0

~∇ · ~B = 0

~∇× ~E = −∂~B

∂t

~∇× ~B = µσ ~E + µε∂ ~E

∂t.

These differ from Maxwell’s equations in nonconductingmaterials only by the µσ ~E in the last equation, whichcomes from the free current density ~Jf = σ ~E.

Taking the curl of the last two equations gives us themodified wave equations

∇2 ~E = µε∂2 ~E

∂t2+ µσ

∂ ~E

∂t

∇2 ~B = µε∂2 ~B

∂t2+ µσ

∂ ~B

∂t.

These allow plane wave solutions

~E(z, t) = ~E0ei(kz−ωt)

~B(z, t) = ~B0ei(kz−ωt),

where the wave number

k2 = µεω2 + iµσω,

is now a complex number.

Since the wave number is complex, it has real andimaginary components. We can write it as

k = k + iκ,

where

k = ω

√εµ

2

√√1 +

( σεω

)2

+ 1

κ = ω

√εµ

2

√√1 +

( σεω

)2

− 1.

Now the wave solutions are

~E(z, t) = ~E0e−κzei(kz−ωt)

~B(z, t) = ~B0e−κzei(kz−ωt).

The complex exponential, ei(kz−ωt) is sinusoidal, and theimaginary part of k results in a non-sinusoidal exponen-tial decay factor e−κz out front. This means the wave

is attenuated—its amplitude decreases with increasing z.As a measure of how strongly the wave is attenuated, welook at what depth the wave’s amplitude has decreasedby a factor of 1

e . This is called the skin depth

d =1

κ.

As before, Maxwell’s equations add further con-straints to our solutions. If we let the polarization bein the x direction, we find that

~E(z, t) = E0e−κzei(kz−ωt)x

~B(z, t) =k

ωE0e

−κzei(kz−ωt)y.

To find the real amplitudes, we start by writing k inpolar form

k = Keiφ,

where

K = |k| =√k2 + κ2 = ω

√εµ

√1 +

( σεω

)2

φ = tan−1(κk

).

The amplitudes are related by

B0 =k

ωE0

B0eiδB =

Keiφ

ωE0e

iδE

B0eiδB =

K

ωE0e

i(δE+φ)

Looking at the phases, we see that the magnetic field andelectric field are no longer in phase since

δB = δE + φ.

Looking at the real amplitudes, we see that

B0 =K

ωE0 =

√εµ

√1 +

( σεω

)2

E0.

The real fields can now be written as

~E(z, t) = E0e−κz cos (kz − ωt+ δE) x

~B(z, t) = B0e−κz cos (kz − ωt+ δE + φ) y.


Reflection from Conductors

For reflection from conducting surfaces, we need the moregeneral boundary conditions

ε1E⊥1 − ε2E⊥2 = σf

B⊥1 −B⊥2 = 0

~E‖1 − ~E

‖2 = 0

1

µ1

~B‖1 −

1

µ2

~B‖2 = ~Kf × n,

where σf is the free surface charge, ~Kf is the free sur-face current, and n points from medium 2 into medium 1.For our setup, we’ll assume that medium 2 is conductingand medium 1 is nonconducting and that the boundarybetween the two is at the xy-plane.

If we have an incident wave from the left,


~BI(z, t) =1

v1E0Ie

i(k1z−ωt)y,

we get the reflected wave


~BR(z, t) = − 1

v1E0Re

i(−k1z−ωt)y,

and the transmitted wave


~BT (z, t) =k2

ωE0T e

i(k2z−ωt)y.

Note the negative sign for ~BR and that the wave numberfor the wave transmitted into the conductor is complex.

The first two boundary conditions give us nothing,but the second two give us

E0I + E0R = E0T

E0I − E0R = βE0T ,

where

β =µ1v1

µ2ωk2.

We used here the fact that ~Kf = 0 for ohmic conductors.Adding/subtracting the two results gives us

E0R =1− β1 + β

E0I

E0T =2

1 + βE0I .

These are the same as the results for nonconducting ma-terials, except that now have a complex β.

For a perfect conductor, we have σ = ∞, which im-plies that k2 =∞, and therefore, β =∞. Then,

E0R = −E0I

E0T = 0.

So for a perfect conductor, the wave is 100% reflected.

Dispersion

If the speed of a wave in a medium depends on the fre-quency of the wave, then the medium is dispersive. Thisis the phenomenon of dispersion. Conductors are disper-sive, but here we focus on dielectrics.

For a wave packet, each sinusoidal component travelsat the ordinary wave velocity or phase velocity

v =ω

k.

But the wave packet as a whole travels at the group ve-locity

vg =dω

dk.

The energy in a dispersive medium always travels at thegroup velocity.

An electron bound to a molecule in a dielectric can bemodeled as a forced, damped, harmonic oscillator. New-ton’s law gives us

md2x

dt2= Fbinding + Fdamp + Fdrive,

where the binding force can be approximated as a restor-ing (spring) force

Fbinding = −mω20x, ω0 =

√kspringm

.

Here, m is the mass of the electron, and x is its displace-ment from equilibrium. We may also have some kind ofvelocity dependent damping force

Fdamp = −mγdxdt,

where γ is just a proportionality constant. Finally, an in-cident electromagnetic wave of frequency ω and polarizedin the x direction provides a sinudoidal driving force

Fdrive = qE = qE0 cos(ωt).

Putting them all together, we have

d2x

dt2+ γ

dx

dt+ ω2

0x =q

mE0 cos(ωt).

In complex form,

d2x

dt2+ γ

dx

dt+ ω2

0 x =q

mE0e

−iωt.

Trying the steady-state solution

x(t) = x0e−iωt,

7.4. ABSORPTION AND DISPERSION 125

gives us

x0 =q/m

ω20 − ω2 − iγω

E0.

The real part of q x(t)

p(t) =q2/m

ω20 − ω2 − iγω

E0e−iωt.

is the dipole moment as a function of time. Because ofthe complex denominator, the dipole moment is out ofphase with ~E and lags behind it by

tan−1

(γω

ω20 − ω2

).

When ω << ω0, this phase angle is very small, but whenω >> ω0, it approaches π.

This was all for one electron in a dielectric. Sup-pose we consider now a large number, fj , of electrons ineach molecule, and a total of N molecules in the material.Each electron may experience a slightly different incidentEM frequency ωj , and each may have a different dampingparameter γj . Then the complex polarization is

~P =Nq2

m

∑j

(fj

ω2j − ω2 − iγjω

)~E.

Then the real polarization is just the real part of this.Keep in mind that this polarization of the material iscompletely unrelated to the polarization of the incidentEM wave. The complex polarization is related to thecomplex field by a complex susceptibility

~P = ε0χe~E.

We define the complex permittivity as

ε = ε0 (1 + χe) ,

then the complex dielectric constant is

εr = 1 +Nq2

mε0

∑j

fjω2j − ω2 − iγjω

.

When ω is near one of the resonant frequencies, ωj , theimaginary term can become large.

In a dispersive material,

∇2 ~E = εµ0∂2 ~E

∂t2.

This is for a given frequency. The plane wave solutionsare

~E(z, t) = ~E0ei(kz−ωt),

wherek = k + iκ =

√εµ0 ω.

Plugging this in gives us

~E(z, t) = ~E0e−κzei(kz−ωt),

which shows that the wave is attenuated.

Since the intensity is proportional to E2 which isproportional to e−2κz, we call

α ≡ 2κ,

the absorption coefficient.

The wave velocity and the index of refraction are

v =ω

k

n =ck

ω.

For a gas, the second term in εr is small, so we canapproximate the square root in k = ω

c

√εr by the first

two terms in the binomial expansion. That is, we ap-proximate

√1 + x ' 1 + 1

2x, where x is a placeholder forthe second term in εr. Then we get

k ' ω

c

1 +Nq2

2mε0

∑j

fjω2j − ω2 − iγjω

n ' 1 +

Nq2

2mε0

∑j

fj(ω2j − ω2

)(ω2j − ω2

)2+ γ2

jω2

α ' Nq2ω2

mε0c

∑j

fjγj(ω2j − ω2

)2+ γ2

jω2.

Away from resonances we can ignore the damping,and the index of refraction simplifies to

n ' 1 +Nq2

2mε0

∑j

fjω2j − ω2

.

For most transparent materials, the resonances lie in theultraviolet, so for incident visible light, ω < ωj . Then wecan approximate

1

ω2j − ω2

=1

ω2j

(1− ω2

ω2j

)−1

' 1

ω2j

(1 +

ω2

ω2j

).

Then the index of refraction becomes

n ' 1 +Nq2

2mε0

∑j

fjω2j

+Nq2ω2

2mε0

∑j

fjω4j

.

Writing this in terms of the vacuum wavelength λ = 2πcω ,

n = 1 +A

(1 +

B

λ2

).

This is called Cauchy’s formula. The constants A andB can be deduced from the previous equation, and are re-spectively called the coefficient of refraction and thecoefficient of dispersion.


7.5 Guided Waves

Now we look at electromagnetic waves in hollow pipes orhollow wave guides. For now, we will assume the waveguide has an arbitrary but constant cross-section. Wewill assume the walls of the wave guide are perfect con-ductors. This means ~E = ~B = 0 within the walls them-selves, which implies that the boundary conditions at theinner wall are

~E‖

= 0

B⊥ = 0.

We will consider monochromatic EM wave propagatingdown the inside of a wave guide that is oriented along thez-axis

~E(x, y, z, t) = ~E0(x, y)ei(kz−ωt)

~B(x, y, z, t) = ~B0(x, y)ei(kz−ωt).

Confined waves are not generally transverse waves, so wehave to include longitudinal components E0z and B0z. So

~E0(x, y) = E0x(x, y) x+ E0y(x, y) y + E0z(x, y) z

~B0(x, y) = B0x(x, y) x+ B0y(x, y) y + B0z(x, y) z.

Our wave has to satisfy Maxwell’s equations

~∇ · ~E = 0

~∇ · ~B = 0

~∇× ~E = −∂~B

∂t

~∇× ~B =1

c2∂ ~E

∂t.

To apply the third Maxwell equation, we take the

curl of ~E, which is simply the cross product in Cartesiancoordinates.

~∇× ~E =

∣∣∣∣∣∣∣∣x y z∂∂x

∂∂y

∂∂z

E0xei(kz−ωt) E0ye

i(kz−ωt) E0zei(kz−ωt)

∣∣∣∣∣∣∣∣=

(∂Ez∂y− ikEy

)ei(kz−ωt)x

+

(ikEx −

∂Ez∂x

)ei(kz−ωt)y

+

(∂Ey∂x− ∂Ex

∂y

)ei(kz−ωt)z.

We have dropped the subscript zero and the tilde for no-tational convenience. Next, we take the partial derivative

of ~B with respect to t

∂ ~B

∂t= iω ~B0e

i(kz−ωt)

= iω(Bx x+By y + Bz) z

)ei(kz−ωt)

I have again dropped the subscript zeros and tildes fromthe components for notational convenience. Next, we re-

late our results for ~∇× ~E and ∂ ~B∂t using the third Maxwell

equation. After simplifying and equating vector compo-nents, we get the three equations

∂Ez∂y− ikEy = iωBx (7.7)

ikEx −∂Ez∂x

= iωBy (7.8)

∂Ey∂x− ∂Ex

∂y= iωBz. (7.9)

Next we repeat the process, computing ~∇ × ~B and∂ ~E∂t and then relating the two using the fourth Maxwell

equation. When we do so, we get another three equations

∂Bz∂y− ikBy = − iω

c2Ex (7.10)

ikBx −∂Bz∂x

= − iωc2Ey (7.11)

∂By∂x− ∂Bx

∂y= − iω

c2Ez. (7.12)

Multiplying Eq. (7.8) by k and Eq. (7.10) by ω, andthen subtracting the second result from the first, we get

Ex =i(

ωc

)2 − k2

(k∂Ez∂x

+ ω∂Bz∂y

). (7.13)

Similarly, we find

Ey =i(

ωc

)2 − k2

(k∂Ez∂y− ω∂Bz

∂x

)(7.14)

Bx =i(

ωc

)2 − k2

(k∂Bz∂x− ω

c2∂Ez∂y

)(7.15)

By =i(

ωc

)2 − k2

(k∂Bz∂y

+ω

c2∂Ez∂x

). (7.16)

So we have written Ex, Ey, Bx, and By, all in terms ofderivatives of Ez and Bz. So we only need to find Ez andBz, and then we can easily compute the others from thesefour equations.

Next, we apply the first Maxwell equation, ~∇· ~E = 0.We start by differentiating Eq. (7.13) and Eq. (7.14) toget

∂Ex∂x

=i(

ωc

)2 − k2

(k∂2Ez∂x2

+ ω∂2Bz∂x∂y

)∂Ey∂y

=i(

ωc

)2 − k2

(k∂2Ez∂y2

− ω∂2Bz∂x∂y

).

7.5. GUIDED WAVES 127

Our third derivative is

∂Ez∂z

=∂

∂zE0z(x, y)ei(kz−ωt)

= ikE0z(x, y)ei(kz−ωt)

= ikEz.

Adding these three results and simplifying gives us thedivergence of ~E, which we know from Maxwell’s law tobe zero.

~∇ · ~E = 0∂Ex∂x

+∂Ey∂y

+∂Ez∂z

= 0[∂2

∂x2+

∂2

∂y2+(ωc

)2

− k2

]Ez = 0.

Similarly, from ~∇ · ~B = 0, we get[∂2

∂x2+

∂2

∂y2+(ωc

)2

− k2

]Bz = 0.

In free space, Ez = Bz = 0. In a wave guide, we canonly set one of these to zero. If we set Ez = 0, we callthe resulting wave a TE wave for “transverse electric”.If we set Bz = 0, we call it a TM wave for “transversemagnetic”. If both are zero, they are called TEM waves,however, this cannot occur in a hollow wave guide.

Rectangular Wave Guide

TE Waves

Let’s consider TE waves in a waveguide with rectangularcross-section of widths a and b. Suppose the waveguide islying alongside the z-axis with 0 ≤ x ≤ a and 0 ≤ y ≤ b.

For TE (transverse electric) waves, we have Ez = 0and Bz 6= 0, so we want to solve the PDE[

∂2

∂x2+

∂2

∂y2+(ωc

)2

− k2

]Bz = 0.

We will assume a separable solution

Bz(x, y) = X(x)Y (y).

Plugging this into the PDE and dividing through by XYgives us

1

X

d2X

dx2= − 1

Y

d2Y

dy2+ k2 −

(ωc

)2

.

So with constants kx and ky, we get the pair of ODEs

1

X

d2X

dx2= −k2

x

1

Y

d2Y

dy2= −k2

y = −k2x + k2 −

(ωc

)2

,

where

k2 −(ωc

)2

+ k2x + k2

y = 0. (7.17)

The general solution for the X equation is

X(x) = A sin(kxx) +B cos(kxx).

The boundary condition is that B⊥ = 0 at the inner wallsof the wave guide, so Bx = 0 at x = 0 and x = a. SinceBx = 0 there, we know that Eq. (7.15) is zero. Since wealready have Ez = 0 and hence ∂Ez

∂y = 0, we also have∂Bz∂x = 0. But since X(x) is the x-part of Bz, this implies

that dXdx = 0 or

dX

dx= −kxA cos(kxx)− kxB sin(kxx) = 0,

which implies that A = 0 and kxa = mπ with m =0, 1, 2, . . .. So this part of our solution is

X(x) = B cos(mπx

a

).

Similarly, for the Y equation, we get kyb = nπ withn = 0, 1, 2, . . ., so

Y (y) = B′ cos(nπy

b

).

Putting them together, our solution is

Bz = B0 cos(mπx

a

)cos(nπy

b

).

This is called the TEmn mode.

The wave number can be obtained by plugging kxand ky into Eq. (7.17) to get

k =1

c

√ω2 − ω2

mn,

where

ωmn = cπ

√(ma

)2

+(nb

)2

,

is called the cutoff frequency. It is called this becauseif ω < ωmn, then k becomes imaginary. So no waves withfrequency less than ωmn can propagate in the wave guide.By convention, the first index on TEmn is associated withthe larger width of the wave guide, so we assume here thata ≥ b. Furthermore, at least one of these indices must benonzero. So the lowest cutoff frequency, i.e. the smallestωmn, for this wave guide occurs at TE10 where we have

ω10 =cπ

a.

For a wave guide, no frequencies less than this are possi-ble.

The wave velocity is

v =ω

k=

c√1−

(ωmnω

)2 > c.

This is larger than the speed of light. However, the energytravels at the group velocity

vg =dω

dk=

1

dk/dω= c

√1−

(ωmnω

)2

< c.


TM Waves

For TM (transverse magnetic) waves, we have Bz = 0and Ez 6= 0, so we want to solve the PDE[

∂2

∂x2+

∂2

∂y2+(ωc

)2

− k2

]Ez = 0.

We again assume a separable solution

Ez(x, y) = X(x)Y (y).

All we’ve done is substitute Ez for Bz, so we again havethe general solutions

X(x) = A sin(kxx) +B cos(kxx)

Y (y) = A′ sin(kyy) +B′ cos(kyy).

But now our boundary condition is the ~E‖

= 0 at theinner walls, which means Ey = Ez = 0 at x = 0, a andEx = Ez = 0 at y = 0, b. Multiplying the right sides ofX(x) and Y (y) together to get Ez(x, y) and then apply-ing the boundary conditions Ez = 0 at x = 0 and y = 0tells us that B = B′ = 0, so

Ez(x, y) = E0 sin(kxx) sin(kyy).

From the boundary conditions E(a, y) = E(x, b) = 0, weget

kx =mπ

a, ky =

nπ

b, m, n = 0, 1, 2, . . . ,

so our solution is

Ez(x, y) = E0 sin(mπx

a

)sin(nπy

b

).

Since kx and ky are the same as for the TE case,we get the same cutoff frequency ωmn, and also the samewave and group velocities. However, the lowest cutofffrequency is now

ω11 =

√a2 + b2

b,

since if either m or n are zero then Ez(x, y) = 0, which isnot allowed for TM waves.

7.6. SUMMARY: ELECTROMAGNETIC WAVES 129

7.6 Summary: Electromagnetic Waves

1D Waves

In general, the one-dimensional wave equation is

∂2f

∂z2=

1

v2

∂2f

∂t2,

where v is the speed of the wave. Any function of theform

f(z, t) = g(z − vt),

represents a propagating wave form of shape g(z) propa-gating to the right with speed v. The most general solu-tion to the wave equation is any function of the form

f(z, t) = g(z − vt) + h(z + vt).

Every solution of the wave equation can be put in thisform.

Sinusoidal Waves

Any wave can be written as a linear combination of sinu-soidal waves. This is why the study of sinusoidal wavesis so important.

The function

f(z, t) = A cos (kz − ωt+ δ) ,

represents a sinusoidal wave traveling to the right. Fora wave traveling to the left, we just have to change thesign of the wave number k. The argument of the cosineis called the phase and δ the phase constant. The wavenumber is

k =2π

λ,

where λ is the wavelength. The frequency ν, the periodT , the wave number k, the speed v, and the angular fre-quency ω, are related as follows

ν =1

T=kv

2π=v

λ=

ω

2π.

For calculations, it is often easier to deal with wavesin complex form,

f(z, t) = Aei(kz−ωt), where A = Aeiδ.

Then the real wave is simply the real part of f(z, t).

Reflection and Transmission

If you have a wave passing from one medium to another,you can have reflection and transmission. Then you haveto consider three different waves—the incident, reflected,and transmitted waves. For all three, the angular fre-quency ω is the same, but the speeds and wave numberscan be different in the different media.

For a 1D wave, e.g. on a string, the incident, re-flected, and transmitted waves have the form:

fI(z, t) = AIei(k1z−ωt)

fR(z, t) = ARei(−k1z−ωt)

fT (z, t) = AT ei(k2z−ωt).

Assuming the strings are continuous and smooth wherethey are joined, the boundary conditions are

f(0−, t) = f(0+, t)

∂f

∂z

∣∣∣0−

=∂f

∂z

∣∣∣0+.

These allow us to write the reflection and transmissioncoefficients in terms of the incident coefficient as

AR =k1 − k2

k1 + k2AI

AT =2k1

k1 + k2AI .

Alternatively, using the relation k2k1

= v1v2

, we can writethem in terms of the wave speeds v1 and v2.

To find the amplitudes of the real waves, we writethe complex numbers in polar form. For example, AR =AeiδR , and then take the magnitude of both sides. Re-member, that

∣∣eiδ∣∣ = 1. In our case, we get

AR =

∣∣∣∣v2 − v1

v2 + v1

∣∣∣∣AIAT =

2v2

v2 + v1AI .

Plugging these back in for AR and AT allows us to findthe phases δR and δT in terms of δI . In this case, we willget different results for δA depending on whether v2 > v1

or v2 < v1.

If the oscillation of the wave is in the ±x direction,we say the wave is polarized in the x direction. A wavetraveling to the right that is polarized in the x direction,would be written as

fI(z, t) = AIei(k1z−ωt)x.

More generally, a wave can be polarized in the n direc-tion, and its polarization angle is the angle θ for which

n = cos θ x+ sin θ y.


EM Waves in Vacuum

Electromagnetic plane waves in a vacuum are 1D waveslike a wave on a string. However, now we have one wavefunction for the electric field wave and another for themagnetic field wave, and Maxwell’s equations show ushow they are related.

Away from sources, Maxwell’s equations are

~∇ · ~E = 0

~∇ · ~B = 0

~∇× ~E = −∂~B

∂t

~∇× ~B = µ0ε0∂ ~E

∂t.

Taking the curl of the third and fourth equations give usthe wave equations for electromagnetic waves:

∇2 ~E = µ0ε0∂2 ~E

∂t2

∇2 ~B = µ0ε0∂2 ~B

∂t2.

This implies that electromagnetic waves travel with speed

v =1

√ε0µ0

= c.

After applying the constraints of Maxwell’s equationsto plane waves, we get the general electromagnetic waves.For a wave propagating in an arbitrary direction withwave vector ~k = ω

c k, wave number |~k|, and polarizationn, the wave functions are

~E(~r, t) = E0ei(~k·~r−ωt)n

~B(~r, t) =1

ck × ~E.

The real ~E and ~B fields are

~E(~r, t) = E0 cos(~k · ~r − ωt+ δ

)n

~B(~r, t) =1

ck × ~E.

Since, ~E is transverse, we know that k · n = 0.

For an electromagnetic wave traveling in the z direc-tion and polarized in the x direction, we have

~E(z, t) = E0ei(kz−ωt)x

~B(z, t) =1

cE0e

i(kz−ωt)y.

Taking the real parts gives us

~E(z, t) = E0 cos (kz − ωt+ δ) x

~B(z, t) =1

cE0 cos (kz − ωt+ δ) y.

Notice that

B0 =k

ωE0 =

1

cE0.

For a monochromatic plane wave, B2 = 1c2E

2 =µ0ε0E

2, so the energy density, Poynting vector, and mo-mentum density are

u = ε0E2

~S = cu k

~p =1

cu k =

1

c2~S.

For an electromagnetic wave propagating in the z direc-tion, k = z and E2 = E2

0 cos2 (kz − ωt+ δ). The averagevalues over many cycles are then

〈u〉 =1

2ε0E

20

〈~S〉 =1

2cε0E

20 z

〈~p〉 =1

2cε0E

20 z.

In general, the direction of the Poynting vector is thesame as the direction of wave propagation.

The intensity, or average power per unit area trans-ported by an electromagnetic wave, is

I = 〈S〉 =1

2cε0E

20 .

The intensity is just the average energy density times thespeed of the wave.

An electromagnetic wave exerts a momentum

P =1

A

〈p〉Ac∆t∆t

=1

2ε0E

20 =

I

c,

on an ideal absorber. On an ideal reflector, it exerts amomentum

P =2I

c.

EM Waves in Linear Materials

EM waves in linear materials behave the same as in avacuum, but now all c, ε0, and µ0 become the material-specific v, ε, and µ. In a linear material with no freecharge or free current Maxwell’s equations are

~∇ · ~E = 0

~∇ · ~B = 0

~∇× ~E = −∂~B

∂t

~∇× ~B = µε∂ ~E

∂t.

This implies that waves in such materials travel withspeed

v =1√µε

=c

n,


where

n =

√εµ

ε0µ0,

is the index of refraction. For most materials, µ ≈ µ0, son ≈ √εr.

The energy density, Poynting vector, and intensityare now

u =1

2

(εE2 +

1

µB2

)~S =

1

µ

(~E × ~B

)I =

1

2εvE2

0 ,

For a wave that passes from one material to another,the boundary conditions are

ε1E⊥1 = ε2E

⊥2

~E‖1 = ~E

‖2

B⊥1 = B⊥21

µ1

~B‖1 =

1

µ2

~B‖2.

Reflection & Transmission (Normal Incidence)

Suppose an electromagnetic wave of frequency ω passesfrom a linear material with wave number k1 to anotherwith wave number k2, where the xy-plane forms theboundary between the two materials, and the incidentelectromagnetic wave is normal to the boundary. Thenwe have three electromagnetic waves to deal with—an in-cident wave, a reflected wave, and a transmitted wave.


~BI(z, t) =1

v1E0Ie

i(k1z−ωt)y


~BR(z, t) = − 1

v1E0Re

i(−k1z−ωt)y


~BT (z, t) =1

v2E0T e

i(k2z−ωt)y.

Note the negative sign in ~BR(z, t), and that we now havev1 and v2 instead of c.

Next, we apply the four boundary conditions, andthis is the tricky part. However, we note that the fieldsare parallel to the surface between the two media. Sincethere are no perpendicular components, we don’t careabout the first or third boundary conditions. Further-more, the electric fields are oscillating only in the x di-rection and the magnetic fields are oscillating only in the

y direction. We end up with

E0I + E0R = E0T

E0I − E0R = βE0T ,

whereβ =

µ1v1

µ2v2=µ1n2

µ2n1.

Adding/subtracting the two results and taking the realparts gives us

E0T =2

1 + βE0I

E0R =

∣∣∣∣1− β1 + β

∣∣∣∣E0I .

For most materials, µ ≈ µ0, then β ≈ v1v2

.

If we want to know the fraction of energy that is re-flected and the fraction that is transmitted, we computethe reflection and transmission coefficients

R =IRII

=

(E0R

E0I

)2

T =ITII

=ε2v2

ε1v1

(E0T

E0I

)2

.

Remember thatR+ T = 1.

Reflection & Transmission (Oblique Incidence)

For oblique incidence, our three electromagnetic waveshave the more general form:

~EI(~r, t) = ~E0Iei(~kI ·~r−ωt)

~BI(~r, t) =1

v1kI × ~EI

~ER(~r, t) = ~E0Rei(~kR·~r−ωt)

~BR(~r, t) =1

v1kR × ~ER

~ET (~r, t) = ~E0T ei(~kT ·~r−ωt)

~BT (~r, t) =1

v2kT × ~ET .

We now have three angles—the angle of incidence θI , theangle of reflection θR, and the angle of transmission θT .All are measured with respect to the normal.

In addition to the previous boundary conditions, ourgeometry gives us three laws of optics:

Law 1: The incident, reflected, and transmitted wavevectors are all in the same plane. This plane in-cludes the normal to the surface.

Law 2: The incident angle is equal to the reflected an-gle

θI = θR.

This is called the law of reflection.


Law 3: The third law is also called the law of refrac-tion or Snell’s law

sin θTsin θI

=n1

n2.

Without loss of generality, we can put the boundarybetween the two media at the xy-plane, and we can setthe angle of incidence, i.e. the plane containing all threewave vectors, to be the xz-plane. At z = 0, the boundaryconditions require that ~kI ·~r = ~kR ·~r = ~kT ·~r. Then thez-component of the fields is perpendicular to the surfaceand the x and y components are parallel. Our boundaryconditions are now

ε1

(~E0I + ~E0R

)z

= ε2

(~E0T

)z(

~B0I + ~B0R

)z

=(~B0T

)z(

~E0I + ~E0R

)x,y

=(~E0T

)x,y

1

µ1

(~B0I + ~B0R

)x,y

=1

µ2

(~B0T

)x,y

.

Note that ~B0 = 1v k ×

~E0 in each case. How we pro-ceed from here depends on the polarization of the incidentwave.

If the polarization of the incident wave is parallel tothe plane of incidence, then all three electric fields arein the xz-plane along with all three wave vectors and allthree magnetic fields point in the ±y direction. By care-fully diagramming the setup, noting all the angles, usingthe right-hand-rule to establish the directions of the mag-netic fields, applying the four boundary conditions, andthe three laws of optics, we obtain the pair of equations

E0I − E0R = βE0T

E0I + E0R = αE0T ,

where

β =µ1v1

µ2v2=µ1n2

µ2n1, α =

cos θTcos θI

.

This gives us the Fresnel equations

E0R =

(α− βα+ β

)E0I

E0T =

(2

α+ β

)E0I .

If the polarization of the incident wave is perpendic-ular to the plane of incidence, then all three magneticfields will be in the xz plane of incidence along with allthree wave vectors. All three electric fields will be in the±y direction. Now, the laws of optics and the boundaryconditions give us

E0I + E0R = E0T

E0I − E0R = αβE0T ,

where

β =µ1v1

µ2v2=µ1n2

µ2n1, α =

cos θTcos θI

.

This gives us the Fresnel equations

E0R =

(1− αβ1 + αβ

)E0I

E0T =

(2

1 + αβ

)E0I .

Note: By writing cos θT =√

1− sin2 θT and thenapplying Snell’s law, we can write α in terms of θI .

If E0R

E0I= 0, then no reflection occurs. The angle of

incidence at which this occurs is called Brewster’s angle.In the case of polarization parallel to the surface, Brew-ster’s angle occurs when α = β. In the case of polarizationperpendicular to the surface, this does not occur becausewe cannot have 1 = αβ.

The wave intensities are

II =1

2ε1v1E

20I cos θI

IR =1

2ε1v1E

20R cos θR

IT =1

2ε2v2E

20T cos θT .

The cosines are there because the wave fronts are at anangle to the surface. The reflection and transmission co-efficients are again

R =IRII

T =ITII.

The exact results for the intensities and the coefficientsdepends on the polarization of the incident wave.

EM Waves in Conductors

In a conductor, EM waves are attenuated. Maxwell’sequations for conductors are:

~∇ · ~E = 0

~∇ · ~B = 0

~∇× ~E = −∂~B

∂t

~∇× ~B = µσ ~E + µε∂ ~E

∂t.

These differ from Maxwell’s equations in nonconductingmaterials only by the µσ ~E in the last equation, whichcomes from the free current density ~Jf = σ ~E.

Taking the curl of the last two of Maxwell’s equationsgives us modified wave equations that have plane wavesolutions. After applying the constraints of Maxwell’s


equations and for polarization in the x direction, we getthe electromagnetic wave

~E(z, t) = E0e−κzei(kz−ωt)x

~B(z, t) =k

ωE0e

−κzei(kz−ωt)y.

where

k = k + iκ = Keiφ

k = ω

√εµ

2

√√1 +

( σεω

)2

+ 1

κ = ω

√εµ

2

√√1 +

( σεω

)2

− 1

K = |k| =√k2 + κ2 = ω

√εµ

√1 +

( σεω

)2

φ = tan−1(κk

).

Because of the e−κz in the wave solutions, the wavesare attenuated. The wave amplitude is decreased by afactor of 1/e at the material’s skin depth

d =1

κ.

The amplitudes are related by

B0 =k

ωE0 = B0e

iδB =K

ωE0e

i(δE+φ).

Since δB = δE + φ, the electric and magnetic fields arenot in phase for an EM wave in a conductor.

The real fields can be written as

~E(z, t) = E0e−κz cos (kz − ωt+ δE) x

~B(z, t) = B0e−κz cos (kz − ωt+ δE + φ) y.

For reflection from conducting surfaces, we need themore general boundary conditions

ε1E⊥1 − ε2E⊥2 = σf

B⊥1 −B⊥2 = 0

~E‖1 − ~E

‖2 = 0

1

µ1

~B‖1 −

1

µ2

~B‖2 = ~Kf × n,

where σf is the free surface charge, ~Kf is the free sur-face current, and n points from medium 2 into medium 1.For our setup, we’ll assume that medium 2 is conductingand medium 1 is nonconducting and that the boundarybetween the two is at the xy-plane.

Then we have the incident, reflected, and transmit-ted waves:


~BI(z, t) =1

v1E0Ie

i(k1z−ωt)y


~BR(z, t) = − 1

v1E0Re

i(−k1z−ωt)y


~BT (z, t) =k2

ωE0T e

i(k2z−ωt)y.

Note the negative sign for ~BR and that the wave numberfor the wave transmitted into the conductor is complex.

Using the fact that ~Kf = 0 for ohmic conductors,we get the results

E0R =1− β1 + β

E0I

E0T =2

1 + βE0I ,

where

β =µ1v1

µ2ωk2.

These are the same as the results for nonconducting ma-terials, except that now we have a complex β.

If the speed of a wave in a medium depends on thefrequency of the wave, then the medium is dispersive. Fora wave packet, each sinusoidal component travels at theordinary wave velocity or phase velocity

v =ω

k.

But the wave packet as a whole travels at the group ve-locity

vg =dω

dk.

The energy in a dispersive medium always travels at thegroup velocity.

Wave Guides

For an EM wave propagating down a wave guide withwalls that are perfect conductors, the boundary condi-tions on the wall are

~E‖

= 0

B⊥ = 0.

We will assume the EM wave is monochromatic (i.e.only one ω), and that our wave guide is along the z-axis.


Cartesian Coordinates

The EM wave inside the wave guide has the form

~E(x, y, z, t) = ~E0(x, y)ei(kz−ωt)

~B(x, y, z, t) = ~B0(x, y)ei(kz−ωt),

where

~E0(x, y) = E0x(x, y) x+ E0y(x, y) y + E0z(x, y) z

~B0(x, y) = B0x(x, y) x+ B0y(x, y) y + B0z(x, y) z.

Our wave has to satisfy Maxwell’s equations. Apply-

ing ~∇ × ~E = −∂ ~B∂t and ~∇ × ~B = 1

c2∂ ~E∂t , to the electric

and magnetic waves given above gives us

∂Ez∂y− ikEy = iωBx

ikEx −∂Ez∂x

= iωBy

∂Ey∂x− ∂Ex

∂y= iωBz

∂Bz∂y− ikBy = − iω

c2Ex

ikBx −∂Bz∂x

= − iωc2Ey

∂By∂x− ∂Bx

∂y= − iω

c2Ez.

Next, we write Ex, Ey, Bx, and By, in terms of derivativesof Ez and Bz. We do this, for example, by multiplyingthe second equation by k and the fourth equation by ω,and then adding the two to eliminate the By terms. Weget:

Ex =i(

ωc

)2 − k2

(k∂Ez∂x

+ ω∂Bz∂y

)Ey =

i(ωc

)2 − k2

(k∂Ez∂y− ω∂Bz

∂x

)Bx =

i(ωc

)2 − k2

(k∂Bz∂x− ω

c2∂Ez∂y

)By =

i(ωc

)2 − k2

(k∂Bz∂y

+ω

c2∂Ez∂x

).

Next, we take the spatial derivatives of all six compo-nents and plug them into Maxwell’s equations ~∇ · ~E = 0and ~∇ · ~B = 0. This gives us the two PDEs[

∂2

∂x2+

∂2

∂y2+(ωc

)2

− k2

]Ez = 0[

∂2

∂x2+

∂2

∂y2+(ωc

)2

− k2

]Bz = 0.

In free space, Ez = Bz = 0. In a wave guide, we canonly set one of these to zero.

For TE waves, we set Ez = 0, and solve the BzPDE above, assuming a separable solution Bz(x, y) =X(x)Y (y). If we have a rectangular wave guide along thez-axis with 0 ≤ x ≤ a and 0 ≤ y ≤ b, then for the sep-arated ODEs, we apply the boundary condition B⊥ = 0at x = 0, a and y = 0, b, to get the re-combined solution

Bz(x, y) = B0 cos(mπx

a

)cos(nπy

b

), m, n = 0, 1, 2, . . .

which is called the TEmn mode, and

k =1

c

√ω2 − c2π2

[(ma

)2

+(nb

)2].

For TM waves, we set Bz = 0, and solve the EzPDE above, assuming a separable solution Ez(x, y) =X(x)Y (y). If we have a rectangular wave guide along thez-axis with 0 ≤ x ≤ a and 0 ≤ y ≤ b, then for the sepa-

rated ODEs, we apply the boundary condition ~E‖

= 0 atx = 0, a and y = 0, b, to get the re-combined solution

Ez(x, y) = E0 sin(mπx

a

)sin(nπy

b

), m, n = 1, 2, 3, . . .

which is called the TMmn mode, and

k =1

c

√ω2 − c2π2

[(ma

)2

+(nb

)2].

The cutoff frequency ωmn is the frequency at whichk = 0. It is called this because for ω < ωmn, k becomesimaginary, and no waves propagate.

Chapter 8

Potentials and Fields

8.1 The Potential Formulation

In general, Maxwell’s equations are

~∇ · ~E =1

ε0ρ

~∇ · ~B = 0

~∇× ~E = −∂~B

∂t

~∇× ~B = µ0~J + µ0ε0

∂ ~E

∂t.

Recall that in general, if ~∇× ~F = 0, then we can write

~F = −~∇f,

where f is an arbitrary scalar potential. If ~∇ · ~F = 0,then we can write

~F = ~∇× ~G,where ~G is an arbitrary vector potential.

In electrostatics, ~∇ × ~E = 0, so we were able towrite ~E = −~∇V . But, now after adding time depen-dence, we can no longer do this. However, we still havethat ~∇ · ~B = 0, so we can write the magnetic field as thecurl of a vector potential

~B = ~∇× ~A. (8.1)

Plugging this into the third Maxwell equation givesus

~∇× ~E = − ∂

∂t

(~∇× ~A

)~∇×

(~E +

∂ ~A

∂t

)= 0.

This implies that we can write

~E +∂ ~A

∂t= −~∇V,

so

~E = −~∇V − ∂ ~A

∂t. (8.2)

In electrostatics, we had Poisson’s equation

∇2V = − ρ

ε0.

What do we get now? Plugging Eq. (8.2) into the firstMaxwell equation gives us

∇2V = − ρ

ε0− ∂

∂t

(~∇ · ~A

).

Plugging Eq. (8.1) and Eq. (8.2) into the fourthMaxwell equation and then applying the vector identity

~∇×(~∇× ~A

)= ~∇

(~∇ · ~A

)−∇2 ~A,

gives us

(∇2 ~A− µ0ε0

∂2 ~A

∂t2

)− ~∇

(~∇ · ~A+ µ0ε0

∂V

∂t

)= −µ0

~J .

135

136 CHAPTER 8. POTENTIALS AND FIELDS

Calculate the charge and current distributions ifthe potentials area

V = 0

~A =

−µ0k

4c (ct− |x|)2z for |x| < ct

0 for |x| > ct.

To calculate the electric field, we use Eq. (8.2) toget

~E =

−µ0k

2 (ct− |x|) z for |x| < ct

0 for |x| > ct.

To calculate the magnetic field, we use Eq. (8.1) to get

~B =

−µ0k

2c (ct− |x|) y for − ct < x < 0µ0k2c (ct− |x|) y for 0 < x < ct

0 for |x| > ct.

To get this result, recall that ∂∂x |x| = −1 for x < 0 and

∂∂x |x| = 1 for x > 0.

Taking derivatives, we find

~∇ · ~E = 0

~∇ · ~B = 0

~∇× ~E = ∓µ0k

2y

~∇× ~B =µ0k

2cz

∂ ~E

∂t= −µ0kc

2z

∂ ~B

∂t= ±µ0k

2y.

Thus, we see that Maxwell’s equations are satisfied ifthe charge and current are zero. That is,

ρ = 0

~J = 0.

However, there is a discontinuity in the magnetic fieldat x = 0. This implies the existence of a surface current~K in the yz-plane. From the boundary condition

1

µ1

~B‖1 −

1

µ2

~B‖2 = ~Kf × n,

this implies that kt y = ~K × x or a surface current

~K = kt z.


Example:

Gauge Transformations

The equations (8.1) and (8.2) do not uniquely specify thepotentials. This gives us the gauge freedom to addadditional conditions to the potentials V and ~A. Imag-ine that we have two pairs of potentials V, ~A and V ′, ~A ′.How much can these two sets differ from each other andstill yield the same electric and magnetic fields ~E and ~B?

We start by letting

~A ′ = ~A+ ~a

V ′ = V + b.

For ~A and ~A ′ to give the same magnetic field, sotheir curls must be the same. That is, ~∇ × ~A ′ =~∇× ~A+ ~∇×~a = ~∇× ~A. This implies that ~∇×~a = 0,which implies that ~a is the gradient of some scalar func-tion λ

~a = ~∇λ.

We must also get the same electric field. So from

Eq. (8.2), we get

−~∇V ′− ∂ ~A ′

∂t= −~∇V − ~∇b− ∂ ~A

∂t− ∂~a

∂t= −~∇V − ∂ ~A

∂t.

This implies

~∇b+∂~a

∂t= 0,

or

~∇(b+

∂λ

∂t

)= 0.

So the piece in parentheses does not depend on position,but it can depend on time. We can call it k(t), then

b = −∂λ∂t

+ k(t).

We can absorb k(t) into λ by redefining λ

λ→ λ+

ˆ t

0

k(t′) dt′.

This does not change ~∇λ, which is all we care about.

So, we get the gauge transformation

~A ′ = ~A+ ~∇λ

V ′ = V − ∂λ∂t .

8.1. THE POTENTIAL FORMULATION 137

This means we can always add the gradient of any scalarfunction to ~A, and this does not change the resulting ~Eand ~B, provided that we also subtract the time derivativeof that scalar function from V .

In electrostatics, to find ~E, it was sufficient to knowV . In electrodynamics, this is no longer the case. Wealso need to know ~A.

Tip:

Supposea

V = 0

~A = − 1

4πε0

qt

r2r.

Find the fields, and the charge and current distribu-tions.

We can use Eq. (8.2) to calculate the electric fieldand Eq. (8.1) to calculate the magnetic field. We findthat

~E =1

4πε0

q

r2r

~B = 0.

Notice that these are the fields of a static point chargeat the origin, despite the fact that our scalar potentialis V = 0 rather than V = 1

4πε0

qr .

Taking all the derivatives, we get

~∇ · ~E =q

ε0δ3(~r)

~∇ · ~B = 0

~∇× ~E = 0

~∇× ~B = 0

∂ ~E

∂t= 0

∂ ~B

∂t= 0.

For the first one, we used the fact that

~∇ ·(r

r2

)= 4πδ3(~r).

Plugging all of these derivatives into the Maxwellequations given at the beginning of this chapter, we findthat they are all satisfied for the charge and current dis-tributions

ρ = q δ3(~r)

~J = 0.

Now, suppose we use the scalar gauge function

λ = − 1

4πε0

qt

r.

Then our new potentials are

V ′ = V − ∂λ

∂t=

1

4πε0

q

r

~A ′ = ~A+ ~∇λ = 0.

If we go through the whole process again, we find thatwe get the same ~E and ~B, and so we get the samederivatives and Maxwell’s equations are satisfied in thesame way with the same ρ and ~J .

aThis is problems 10.3 and 10.5 in Griffiths.

Example:

The Coulomb Gauge

For the Coulomb gauge, we choose

~∇ · ~A = 0,

as in magnetostatics. We again get

∇2V = − ρ

ε0,

as in magnetostatics, but now the solution is time depen-dent

V (~r, t) =1

4πε0

ˆρ(~r ′, t)

udτ ′.

In the Coulomb gauge, the equation for the vector

potential becomes

∇2 ~A− µ0ε0∂2 ~A

∂t2= −µ0

~J + µ0ε0~∇(∂V

∂t

).

The Lorentz Gauge

In the Lorentz gauge, we choose

~∇ · ~A = −µ0ε0∂V

∂t,


and then, our equations for V and ~A become

∇2V − µ0ε0∂2V

∂t2= − ρ

ε0

∇2 ~A− µ0ε0∂2 ~A

∂t2= −µ0

~J .

If we define the d’Alembertian operator

2 ≡ ∇2 − µ0ε0∂2

∂t2,

then the two equations can be written compactly as

2V = − ρ

ε0

2 ~A = −µ0~J .

We will use the Lorentz gauge for the rest of thecourse.

8.2 Continuous Distributions

Retarded Potentials

For static distributions, the two equations above reduceto

∇2V = − ρ

ε0

∇2 ~A = −µ0~J ,

for which we know the solutions are

V (~r) =1

4πε0

ˆρ(~r ′)

udτ ′

~A(~r) =µ0

4π

ˆ ~J(~r ′)

udτ ′.

Remember that dτ ′ is a small volume element of thecharge distribution, and u is the distance from the lo-cation ~r ′ of this volume element to the field point ~r atwhich you are measuring things.

In the time-dependent case, the conditions at thefield point ~r depend on the conditions at ~r ′ at an earliertime. This is because electromagnetic influences are notinstantaneous. Rather, they travel at the speed of light.We now define this delay as the retarded time

tr ≡ t−u

c.

Our solutions V and ~A must be modified for time-dependent charge and current distributions. The re-tarded scalar potential is

V (~r, t) =1

4πε0

ˆρ(~r ′, tr)

udτ ′. (8.3)

Here, V (~r, t) is the potential at time t at the field point~r, and ρ(~r ′, tr) is the charge distribution at ~r ′ at theretarded time tr. The retarded vector potential is

~A(~r, t) =µ0

4π

ˆ ~J(~r ′, tr)

udτ ′. (8.4)

These retarded potentials seem plausible as the so-lutions, but to prove that, we have to verify that theysatisfy the equations

2V = − ρ

ε0

2 ~A = −µ0~J

as well as the Lorentz condition. We will verify thatV (~r, t) satisfies the first equation.

Taking the gradient of V (~r, t) and noting that bothnumerator and denominator depend on ~r (so we have touse the product rule), gives us

~∇V =1

4πε0

ˆ [(~∇ρ) 1

u+ ρ~∇

(1

u

)]dτ ′.

The gradient of ρ is

~∇ρ =∂ρ

∂xx+

∂ρ

∂yy +

∂ρ

∂zz

=∂ρ

∂tr

∂tr∂xx+

∂ρ

∂tr

∂tr∂yy +

∂ρ

∂tr

∂tr∂zz

= ρ

[∂tr∂xx+

∂tr∂yy +

∂tr∂zz

]= ρ~∇tr.

since ∂ρ∂tr

= ∂ρ∂t = ρ. But tr = t − u

c , so ~∇tr = − 1c~∇u.

By writing ~u = x x+ y y + z z, then u =√x2 + y2 + z2,

then it is easy to show that

~∇u =~u

u= u.

So, we have that

~∇ρ = ρ~∇tr = −1

cρ u.

It can also be shown that

~∇(

1

u

)= − u

u2.

Plugging these results into ~∇V gives us

~∇V =1

4πε0

ˆ [− ρc

u

u− ρ u

u2

]dτ ′.

Next, we take the divergence, again applying theproduct rule where appropriate,

∇2V = −1

c

1

4πε0

ˆ [u

u· ~∇ρ+ ρ~∇ ·

(u

u

)]dτ ′

− 1

4πε0

ˆ [u

u2· ~∇ρ+ ρ~∇ ·

(u

u2

)]dτ ′.

8.2. CONTINUOUS DISTRIBUTIONS 139

Similarly to what we did above, we find that

~∇ρ = −1

cρ~∇u = −1

cρu.

It can also be shown that

~∇ ·(u

u

)=

1

u2

~∇ ·(u

u2

)= 4πδ3(~u).

Plugging these in gives us

∇2V =1

c

1

4πε0

ˆ [ρ

uc− ρ

u2

]dτ ′

− 1

4πε0

ˆ [− ρ

u2c+ 4πρδ3(~u)

]dτ ′

=1

4πε0

ˆ [ρ

uc2− 4πρδ3(~u)

]dτ ′.

But1

4πε0

ˆρ

udτ ′ =

∂2V

∂t2,

so we have the result

∇2V =1

c2∂2V

∂t2− ρ

ε0,

confirming that

2V = − ρ

ε0.

For the ~A equation, the proof is similar.

The advanced potentials

V (~r, t) =1

4πε0

ˆρ(~r ′, ta)

udτ ′

~A(~r, t) =µ0

4π

ˆ ~J(~r ′, ta)

udτ ′,

whereta ≡ t+

u

c,

is the advanced time, are also solutions. However, theyviolate causality.

If the current distribution is simply a current I(t) ina wire, then the retarded vector potential takes the form

~A(~r, t) =µ0

4π

Î(t− u

c

)u

d~l,

where the integral is taken along the wire, and u is thedistance from the field point ~r to the current element d~l.


Consider an infinite straight wire. At t = 0, a con-stant current I0 is turned on. Calculate the resultingelectric and magnetic fields. a

The electric and magnetic fields are given by

~B = ~∇× ~A

~E = −~∇V − ∂ ~A

∂t.

Since the wire is not electrically charged, the scalar po-tential is zero. For the vector potential, we use theretarded vector potential. We will use cylindrical coor-dinates and let the wire be along the z-axis. Then theretarded vector potential is

~A(s, t) =µ0

4π

ˆ ~J(~r ′, tr)

udτ ′ =

µ0

4π

ˆ ∞−∞

I(tr)

udz z.

At a field point P , a distance s from the wire, ittakes a time t = s

c for the existence of the current atthe closest point in the wire to register at P . For t < s

c ,

the signal has not yet reached P , so ~A = 0 there atthat time. For any time t > s

c , only the current in closeparts of the wire contribute to the vector potential at~A. For portions of the wire more distant, the signal (ofthe existence of current in that portion of the wire) hasnot yet propagated to P due to the finite speed of light.

The distance u is the distance from the field pointP to the current element at dz. We can make a righttriangle with base s, height z, and hypotenuse u. Thesignal from the current element dz travels a distanceu to reach P . The element at dz does not begin con-tributing to ~A at P until t > u

c . But u =√s2 + z2,

which gives us (tc)2 > z2 + s2. In other words, only thesegment of wire

|z| ≤√

(ct)2 − s2,

contributes to ~A at P . So

~A(s, t) =µ0I04π

ˆ √(ct)2−s2

−√

(ct)2−s2

1√s2 + z2

dz z.

The integrand is even, so we multiply by two and inte-grate over half of it

~A(s, t) =µ0I02π

ˆ √(ct)2−s2

0

1√s2 + z2

dz z.

This integral has the solution

~A(s, t) =µ0I02π

ln

[ct+

√(ct)2 − s2

s

]z.

The electric field is then

~E(s, t) = −∂~A

∂t=

µ0I0c

2π√

(ct)2 − s2z.

The magnetic field is

~B(s, t) = ~∇× ~A =µ0I02πs

ct√(ct)2 − s2

φ.

In the limit t → ∞, we see that we recover thestatic case

~E = 0, ~B =µ0I02πs

φ.


Example:

8.2. CONTINUOUS DISTRIBUTIONS 141

Calculate the vector potential outside an infinitewire carrying a current I(t) = kt.a

The current in the wire at a point z = ±z′ at timet is kt. The current in the wire at ±z′ at time t as seenfrom the point P a distance s from z = 0 is the currentat ±z′ at the earlier time t − u

c , where u =√s2 + z′ 2

is the distance from ±z′ to P . In other words, as seenfrom P , the current in the wire depends not only on tbut also on z

IP (z, t) = k(t− u

c

)= k

(t− 1

c

√s2 + z2

).

This is in contrast to the previous example, where thecurrent as seen from P depended only on t.

Below is a plot of what the current looks like at Pat some specific time t:

IP (z)

z√(ct)2 − s2−

√(ct)2 − s2

k(t− s

c

)

In contrast, in the previous example, it looked like:

IP (z)

z√(ct)2 − s2−

√(ct)2 − s2

I0

Now, our vector potential is

~A(s, t) =µ0

4π

ˆ √(ct)2−s2

−√

(ct)2−s2

k(t− 1

c

√s2 + z2

)√s2 + z2

dz z

=µ0kt

2πz

ˆ √(ct)2−s2

0

1√s2 + z2

dz

−1

c

µ0k

4πz

ˆ √(ct)2−s2

−√

(ct)2−s2dz

=µ0k

2πt ln

(ct+

√(ct)2 − s2

s

)z

−1

c

µ0k

2π

√(ct)2 − s2 z

aThis is problem 10.9 (a) in Griffiths.

Example:

Jefimenko’s Equations

Given the retarded potentials Eq. (8.3) and Eq. (8.4), onecan in principle calculate the fields by differentiating viaEq. (8.1) and Eq. (8.2). However, the integrands dependon ~r both through u and tr, so this is not always trivial.

In the previous section, we calculated ~∇V and cal-

culating ∂ ~A∂t is easy. Putting these two together gives us

the time-dependent generalization of Coulomb’s law, andthe formula to calculate ~E without first going throughEq. (8.2)

~E(~r, t) =1

4πε0

ˆ [ρ(~r ′, tr)

u2u+

ρ(~r ′, tr)

cuu−

~J(~r ′, tr)

c2uu

]dτ ′.

In the static case, the last two terms are zero and werecover Coulomb’s law (in integral form).

Next, we want to find a similar equation for ~B. Thecurl of ~A is

~∇× ~A =µ0

4π

ˆ [1

u

(~∇× ~J

)− ~J × ~∇

(1

u

)]dτ ′.

The x component of the curl of ~J is(~∇× ~J

)x

=∂Jz∂y− ∂Jy

∂z.

But we can write

∂Jz∂y

=∂Jz∂tr

∂tr∂y

= Jz∂tr∂y

= −1

cJz∂u

∂y,


and similarly for∂Jy∂z . So(

~∇× ~J)x

= −1

c

(Jz∂u

∂y− Jy

∂u

∂z

)=

1

c

(~J × (~∇u)

)x,

and since we noted earlier that (~∇u) = u, we have that

~∇× ~J =1

c~J × u.

So our curl of ~A becomes the time-dependent generaliz-tion of the Biot-Savart law

~B(~r, t) =µ0

4π

ˆ [ ~J(~r ′, tr)

u2+~J(~r ′, tr)

cu

]× u dτ ′.

8.3 Point Charges

Lienard-Wiechert Potentials

In the previous sections, we calculated the retarded po-tentials of a continuous distribution. Now, we want todo the same for a single charged particle q moving alongsome path

~w(t).

The retarded time can be found implicitly from

|~r − ~w(tr)| = c(t− tr),

where ~w(tr) is the retarded position of the particle,and

~u ≡ ~r − ~w(tr)

is the vector from the retarded position of the particle tothe field point ~r.

At time t, an observer at the field point ~r sees thecharge q at its retarded position ~w(tr). The observer seesthe particle only at a single point at any given moment intime. That is, only one point on the particle’s trajectorycan communicate with the field point at a given time.

From Eq. (8.3), the potential for a point charge canbe written as

V (~r, t) =1

4πε0u

ˆρ(~r ′, tr) dτ

′.

For a point particle, we are able to pull u outside of theintegral. However, contrary to intuition, it is no longertrue that

´ρ(~r ′, tr) dτ

′ equals q—the total charge of theparticle. This is because the retardation requires us toevaluate ρ at different times at different points in the dis-tribution. Rather, we find that for a finite distribution,

ˆρ(~r ′, tr) dτ

′ =q

1− u·~vc

,

where ~v is the velocity of the charged particle at the re-tarded time.

To prove this, we start with the potential in the form

V (~r, t) =1

4πε0

ˆρ(~r ′, t′)

uδ(t′ − t′r) dt′ dτ ′.

More explicitly,

V (~r, t) =1

4πε0

ˆρ(~r ′, t′)

|~r − ~r ′|δ(t′ − t′r) dt′ dτ ′.

For a point particle traveling on a trajectory ~w(t), we canwrite the charge distribution as

ρ(~r, t) = qδ3 (~r − ~w(t)) .

Plugging this in gives us

V (~r, t) =q

4πε0

ˆδ3 (~r ′ − ~w(t′))

|~r − ~r ′|δ(t′ − t′r) dt′ dτ ′.

Doing the τ ′ integration gives us

V (~r, t) =q

4πε0

ˆδ(t′ − t′r)|~r − ~w(t′r)|

dt′.

Note that the retarded time t′r is a function of the fieldpoint ~r(t) and the source point ~w(t′).

From |~r − ~w(tr)| = c(t− tr), we have that

tr = t− 1

c|~r − ~w(tr)| .

So

t′ − t′r = t′ − t+1

c|~r − ~w(t′)| .

To evaluate the remaining integral, we use the iden-tity

δ (g(x)) =∑i

δ(x− xi)g′(xi)

,

where the xi are the zeros of g(x). In our case,

g(x) = t′ − t+1

c|~r − ~w(t′)|

= t′ − t+1

c

√(~r − ~w(t′)) · (~r − ~w(t′)).

Differentiating gives us

g′(x) = 1− 1

c

(~r − ~w(t′))

|~r − ~w(t′)|· d~wdt

= 1− 1

c

(~r − ~w(t′)) · ~v|~r − ~w(t′)|

= 1− ~u · ~vuc

=uc− ~u · ~v

uc.

So we end up with

δ(t′ − t′r) =δ(t′ − tr)uc−~u·~vuc

,

8.3. POINT CHARGES 143

and the integral gives us

V (~r, t) =1

4πε0

qc

uc− ~u · ~v.

Similarly, for the vector potential, we start with

~A(~r, t) =1

4πε0

ˆ ~J(~r ′, t′)

uδ(t′ − t′r) dt′ dτ ′.

For a point charge,

~J(~r, t) = q~v δ3 (~r − ~w(t)) .

We end up getting

~A(~r, t) =µ0

4π

qc~v

uc− ~u · ~v=~v

c2V (~r, t).

Constant Velocity

We will now consider the case of a point charge moving atconstant velocity.1 Let t = 0 correspond to the momentwhen the charge is passing through the origin, then itspath is given by

~w(t) = ~vt.

The implicit relation for the retarded time is then

|~r − ~vtr| = c(t− tr).

Squaring both sides gives us

r2 − 2~r · ~vtr + v2t2r = c2(t2 − 2ttr + t2r).

We can now solve for tr using the quadratic formula

tr =(c2t− ~r · ~v)−

√(c2t− ~r · ~v)2 + (c2 − v2)(r2 − c2t2)

c2 − v2.

An examination of the limit v = 0 implies that we shouldtake the minus sign in the quadratic formula as shownabove.

Recall that |~r − ~w(tr)| = c(t − tr) = u where~u = ~r − ~w(tr). This implies that

u =~u

u=~r − ~w(tr)

|~r − ~w(tr)|.

In our case,

u =~r − ~vtrc(t− tr)

.

So we can write the denominator of the potential as

uc− ~u · ~v = uc

(1− u · ~v

c

)= c2(t− tr)

(1− ~v

c· [~r − ~vtr]c(t− tr)

)= c2(t− tr)− ~v · [~r − ~vtr]= c2t− ~v · ~r − (c2 − v2)tr

Plugging in the value we found for tr gives us

uc− ~u · ~v =√

(c2t− ~r · ~v)2 + (c2 − v2)(r2 − c2t2).

So for a point charge moving at constant velocity ~v, thescalar potential is

V (~r, t) =1

4πε0

qc√(c2t− ~r · ~v)2 + (c2 − v2)(r2 − c2t2)

,

and the vector potential is

~A(~r, t) =µ0

4π

qc~v√(c2t− ~r · ~v)2 + (c2 − v2)(r2 − c2t2)

.

We can expand the quantity under the radical to get√(c2t− ~r · ~v)2 + (c2 − v2)(r2 − c2t2) =

√−2c2t~r · ~v + (~r · ~v)2 + c2r2 − v2r2 + c2v2t2.

Let~R ≡ ~r − ~vt,

be the vector from the non-retarded position of the particle to the field point ~r. Then we can write

R2 = (~r − ~vt) (~r − ~vt) = r2 − 2t~r · ~v + v2t2.

So

~r · ~v =R2 − r2 − v2t2

−2t,

which implies that

−2c2t~r · ~v = c2R2 − c2r2 − c2v2t2.

1This is example 10.3 in Griffiths.


Substituting this into the equation above gives us

√(c2t− ~r · ~v)2 + (c2 − v2)(r2 − c2t2) =

√c2R2 + (~r · ~v)2 − v2r2.

We can also write ~R ·~v = (~r − ~vt) ·~v = ~r ·~v− v2t = Rv cos θ, where θ is the angle between ~R and ~v. Then squaringthe last two sides,

R2v2 cos2 θ = (~r · ~v)2 − 2v2t~r · ~v + v4t2.

Rearranging and simplifying gives us

(~r · ~v)2

= R2v2 cos2 θ + 2v2t~r · ~v − v4t2

= R2v2 cos2 θ + 2v2t

(R2 − r2 − v2t2

−2t

)− v4t2

= R2v2 cos2 θ − v2R2 + v2r2

= −R2v2 sin2 θ + v2r2.

Substituting this result into the radical gives us

√(c2t− ~r · ~v)2 + (c2 − v2)(r2 − c2t2) = cR

√1− v2 sin2 θ

c2.

Plugging this in, tells us that for a point charge moving at constant velocity ~v, the potential can be written as2

V (~r, t) =1

4πε0

q

R√

1− v2 sin2 θ/c2.

2This was problem 10.14 in Griffiths.

8.3. POINT CHARGES 145

Calculate the Lienard-Wiechert potentials for apoint charge in hyperbolic motion along the x-axisa

~w(t) =√b2 + (ct)2 x.

Assume that ~r is on the x-axis and to the right of themoving charge.

We start with the Lienard-Wiechert potential

V (~r, t) =1

4πε0

qc

uc− ~u · ~v.

In our case,

u = x−√b2 + c2t2r = c(t− tr).

Solving for tr gives us

tr =b2 − (x− ct)2

2c(x− ct),

then the velocity is d~wdt evaluated at tr or

~v =c2tr√b2 + c2t2r

x.

But we know that√b2 + c2t2r = x− c(t− tr),

so

~v =c2tr

x− c(t− tr)x.

Then

uc− ~u · ~v = uc− uv

= c2(t− tr)− c(t− tr)c2tr

x− c(t− tr)

=c2(t− tr)(x− ct)x− c(t− tr)

=c2(t− tr)(x− ct)ctr + (x− ct)

.

Plugging in tr from above gives us

uc− ~u · ~v =c(x2 − c2t2 − b2)(x− ct)

b2 + (x− ct)2.

Plugging this result into the Lienard-Wiechert po-tential gives us

V (x, t) =1

4πε0

q[b2 + (x− ct)2

]c(x2 − c2t2 − b2)(x− ct)

.

Then the vector potential is

~A(x, t) =~v

c2V (x, t)

=1

4πε0

q

c

b2 − (x− ct)2

(x2 − c2t2 − b2)2(x− ct).

aThis is problem 10.16 in Griffiths.

Example:

Fields

Given that the electric and magnetic fields are computedfrom the scalar and vector potentials as

~E = −~∇V − ∂ ~A

∂t~B = ~∇× ~A,

and we now know the potentials for a moving point charge

V (~r, t) =1

4πε0

qc

uc− ~u · ~v

~A(~r, t) =~v

c2V (~r, t),

we are can now compute the electric and magnetic fieldsof a moving point charge.

However, the differentiation is not straightforward,because both ~u and ~v depend on the retarded time tr,which is defined implicitly via |~r − ~w(tr)| = c(t − tr).

Calculating the derivatives ~∇V , ~∇ × ~A, and ∂ ~A∂t is te-

dious and messy, so I will only quote the results fromGriffiths. Refer to Griffiths for the details.


~∇V =1

4πε0

qc

(uc− ~u · ~v)3

[(uc− ~u · ~v)~v − (c2 − v2 + ~u · ~a)~u

]∂ ~A

∂t=

1

4πε0

qc

(uc− ~u · ~v)3

[(uc− ~u · ~v)

(−~v +

u~a

c

)+u

c(c2 − v2 + ~u · ~a)~v

]~∇× ~A = − 1

4πε0c

q

(~u · ~N)3~u×

[(c2 − v2)~v + (~u · ~a)~v + (~u · ~N)~a

],

where

~a =d~v

dtr,

is the acceleration of the point charge at the retardedtime, and

~N = cu− ~v.

With these derivatives, we find that the electric fieldof a moving point charge is

~E(~r, t) =q

4πε0

u

(~u · ~N)3

[(c2 − v2) ~N + ~u× ( ~N × ~a)

].

The magnetic field of a moving point charge is

~B(~r, t) =1

cu× ~E(~r, t).

Plugging these fields into the Lorentz force law

~F = Q~E +Q~V × ~B,

gives us the force exerted by the moving charge q on atest charge Q moving with velocity ~V .

Notice that for a stationary particle with ~v = ~a = 0,our electric field reduces to the static case

~E =1

4πε0

q

u2u.

as it should.

Constant Velocity

To find the fields of a point charge moving at a constantvelocity3, we set ~a = 0 in the above equations to get

~E(~r, t) =q

4πε0

u

(~u · ~N)3

[(c2 − v2) ~N

].

We can write

u ~N = u(cu− ~v) = c~u− u~v.

Then, we can write the path of the particle is ~w = ~vt,which allows us to substitute ~u = ~r−~vtr and u = c(t−tr)using the implicit relations defined earlier.

u ~N = c(~r − ~vtr)− c(t− tr)~v = c(~r − ~vt).

Note that

~u · ~N = uc− ~u · ~v.

In a previous example, we found that for a point chargemoving at constant velocity,

uc− ~u · ~v = Rc

√1− v2 sin2 θ

c2,

where~R = ~r − ~vt,

and θ is the angle between ~R and ~v.

So we can write the electric field as

~E(~r, t) =q

4πε0

1− v2

c2(1− v2 sin2 θ

c2

) 32

R

R2.

Compared to a stationary particle, a fast moving particlehas its field reduced by a factor of 1−v2/c2 in the forwardand backward directions (θ = 0, 180), and increased bya factor of 1/

√1− v2/c2 in the perpendicular directions

(θ = 90, 270). So where a stationary particle has anelectric field with surfaces of constant E being spheres, afast-moving charge has an electric field with surfaces ofconstant E which look more like pancakes—flattened inthe direction of motion.

For the magnetic field,

u =~u

u=~r − ~vtr

u=~r − ~vt+ (t− tr)~v

u=~R

u+~v

c.

Since ~E points along ~R, the cross product ~R× ~E is zero.So the magnetic field is

~B =1

cu× ~E =

1

c2~v × ~E.

3This is example 10.4 in Griffiths.

8.4. SUMMARY: POTENTIALS AND FIELDS 147

8.4 Summary: Potentials and Fields

Gauges and Fields

In general, we can find the electric and magnetic fieldsfrom the scalar and vector potentials using the equations

~E = −~∇V − ∂ ~A

∂t~B = ~∇× ~A.

Given V and ~A, we can use these equations to calculate~E and ~B. To verify that Maxwell’s equations are sat-isfied, take all six derivatives of ~E and ~B. To get thecharge and current distributions, figure out what ρ and~J are needed to satisfy Maxwell’s equations.

If we have the gauge transformation

~A ′ = ~A+ ~∇λ

V ′ = V − ∂λ∂t .

where λ is an arbitrary scalar function, then ~A ′, V givethe same electric and magnetic fields as ~A, V . TheCoulomb gauge has the condition

~∇ · ~A = 0.

The Lorentz gauge has the condition

~∇ · ~A = −µ0ε0∂V

∂t.

To identify whether a given combination of potentials~A, V are a Coulomb and/or Lorentz gauge, just compute~∇ · ~A and compare with the above equations.

In the Lorentz gauge, our ODEs for ~A and V can bewritten compactly as

2V = − ρ

ε0

2 ~A = −µ0~J ,

where

2 ≡ ∇2 − µ0ε0∂2

∂t2,

is the d’Alembertian operator.

Continuous Distributions

In the time-dependent case, the conditions at the fieldpoint ~r depend on the conditions at ~r ′ at an earlier time.This is because electromagnetic influences are not instan-taneous. Rather, they travel at the speed of light.

For continuous distributions, the retarded potentialsare

V (~r, t) =1

4πε0

ˆρ(~r ′, tr)

udτ ′

~A(~r, t) =µ0

4π

ˆ ~J(~r ′, tr)

udτ ′.

wheretr ≡ t−

u

c,

is the retarded time.

If the current distribution is simply a current I(t) ina wire, then the retarded vector potential takes the form

~A(~r, t) =µ0

4π

Î(t− u

c

)u

d~l,

where the integral is taken along the wire, and u is thedistance from the field point ~r to the current element d~l.

To calculate the exact scalar and vector potentials ofa time dependent current,

• Calculate the retarded potentials using the defini-tions given above.

• If nothing is electrically charged, then V = ~∇V =0, and we only have to calculate ~A.

• For line current, I(t)→ I(t− u

c

).

• Use d = rt (or in this case, u = ct) to determine the

integration limits when computing ~A.

The fields are then

~E(~r, t) =1

4πε0

ˆ [ρ(~r ′, tr)

u2u+

ρ(~r ′, tr)

cuu−

~J(~r ′, tr)

c2uu

]dτ ′

~B(~r, t) =µ0

4π

ˆ [ ~J(~r ′, tr)

u2+~J(~r ′, tr)

cu

]× u dτ ′.

These are known as Jefimenko’s equations.

To calculate the exact electric and magnetic fields ofa time-dependent current distribution, we would typicallycalculate the retarded potentials and then calculate thefields from those rather than using Jefimenko’s equations.After calculating the fields, check limits (like t → 0) toensure that your result is sensible.

Point Charges

In the previous sections, we calculated the retarded po-tentials of a continuous distribution. Now, we want todo the same for a single charged particle q moving alongsome path

~w(t).

The retarded time can be found implicitly from

|~r − ~w(tr)| = c(t− tr),

where ~w(tr) is the retarded position of the particle,and

~u ≡ ~r − ~w(tr)


is the vector from the retarded position of the particle tothe field point ~r.

The potentials, known as the Lienard-Wiechert po-tentials, of the particle are

V (~r, t) =1

4πε0

qc

uc− ~u · ~v

~A(~r, t) =µ0

4π

qc~v

uc− ~u · ~v=~v

c2V (~r, t).

To calculate the Lienard-Wiechert potentials for amoving point particle, follow this procedure:

1. Find the retarded time by solving

|~r − ~w(tr)| = c(t− tr),

for tr.2. Calculate

~u = ~r − ~w(tr)

~v(tr) =d~w

dtr.

Plug these into uc − ~u · ~v and replace tr by whatwas found in step 1.

3. Plug this into the equations for V and ~A.

To calculate the exact fields of a moving point charge,we calculate the Lienard-Wiechert potentials, and calcu-late the fields from those.

Miscellaneous

Useful formulae:

~∇u =~u

u= u

~∇(

1

u

)= − u

u2

~∇ ·(u

u

)=

1

u2

~∇ ·(u

u2

)= 4πδ3(~u)

θ = − sin θ x+ cos θ y.

Chapter 9

Radiation

In a previous section, we looked at electromagneticwaves, but we didn’t consider how those waves were gen-erated in the first place. We will now focus on radia-tion–the generation of electromagnetic waves.

By definition, radiation is an irreversible flow of elec-tromagnetic energy from the source to infinity. The totalpower passing through the surface of a sphere of radiusr centered on the source is the integral of the Poyntingvector

P (r) =

˛~S · d~a =

1

µ0

˛ (~E × ~B

)· d~a.

Then the power radiated is the limit of this quantity asthe size of the enclosing sphere is taken to infinity

Prad = limr→∞

P (r).

The area of a sphere goes as A ∼ r2, so for Prad to benonzero, the Poynting vector ~S must decrease no fasterthan 1

r2 in the limit r →∞. For example, if S ∼ 1r3 , then

P (r) ∼ 1r and Prad = 0. In electrostatics, E ∼ 1

r2 . Inmagnetostatics, B ∼ 1

r2 . So in the static case, S ∼ 1r4 ,

then P (r) ∼ 1r2 , and Prad = 0. In other words, static con-

figurations do not radiate. In the time-dependent case,looking at the Jefimenko equations, we see that ~E and ~Bhave 1

r (or 1u ) terms. If there is radiation, it comes from

these terms. So to calculate radiation, we construct the1r2 term in ~S from the 1

r terms in ~E and ~B, and then weintegrate this term over a spherical surface and take thelimit of the spherical surface to infinity.

9.1 Electric Dipole Radiation

Consider an electric dipole consisting of two chargedmetal spheres—one with charge +q(t), and the other withcharge −q(t). The two spheres are connected by a wireof length d.

−q(t)

+q(t)

d

u+

u−

~r

We will take the wire to be along the z axis and thetwo spheres to be equidistant from the origin. Then ~ris the vector from the origin to the field point, u+ is thedistance from the top sphere to the field point, and u− isthe distance from the bottom sphere to the field point.

Now suppose the charges are being moved back andforth between the two spheres at an angular frequency ωsuch that the charge on the top sphere is

q(t) = q0 cos(ωt).

Thus, we have an oscillating electric dipole

~p(t) = p0 cos(ωt) z,

wherep0 = q0d.

The Scalar Potential

The retarded scalar potential is

V (~r, t) =1

4πε0

q+(tr+)

u++q−(tr−)

u−

=

q0

4πε0

cos[ω(t− u+

c

)]u+

−cos[ω(t− u−

c

)]u−

.

The law of cosines implies that

u± =

√r2 ∓ rd cos θ +

(d

2

)2

.

We will look at a hierarchy of distances

r >>c

ω>> d,

149

150 CHAPTER 9. RADIATION

where r is your distance from the oscillating dipole, andd is the “size” of the distribution.

Approximation # 1: (d << r)

In the first approximation we look at, the separationdistance in the dipole is much smaller than the distancefrom the dipole. In this approximation,

u± ' r(

1∓ d

2rcos θ

).

Then1

u±' 1

r

(1± d

2rcos θ

),

and

cos[ω(t− u±

c

)]' cos

[ω(t− r

c

)± ωd

2ccos θ

].

We can expand this using a trig identity as

cos[ω(t− u±

c

)]' cos

[ω(t− r

c

)]cos

[ωd cos θ

2c

]∓ sin

[ω(t− r

c

)]sin

[ωd cos θ

2c

].

Approximation # 2:(d << c

ω

)In this limit,

cos

[ωd cos θ

2c

]' 1

sin

[ωd cos θ

2c

]' ωd cos θ

2c,

so

cos[ω(t− u±

c

)]' cos

[ω(t− r

c

)]∓ωd cos θ

2csin[ω(t− r

c

)].

Plugging our results into V (~r, t) (given earlier) and dis-carding O(d2), gives us

V (r, θ, t) =p0 cos θ

4πε0r

−ωc

sin[ω(t− r

c

)]+

1

rcos[ω(t− r

c

)].

Approximation # 3:(r >> c

ω

)In this approximation, the second term in V (r, θ, t)

becomes negligible due to the factor of 1r , and the above

result becomes

V (r, θ, t) = −p0ω cos θ

4πε0crsin[ω(t− r

c

)].

Noting that ~p0 · r = p0 cos θ, we can write this in thecoordinate-free form

V (~r, t) = − ω

4πε0crsin[ω(t− r

c

)]~p0 · r.

The Vector Potential

For the vector potential, recall that the retarded vectorpotential for a current in a wire is

~A(~r, t) =µ0

4π

Î(t− u

c

)u

d~l.

In our case, we have current

~I(t) =dq

dtz = −q0ω sin(ωt)z,

so

~A(~r, t) = −µ0q0ω

4π

ˆ d2

− d2

sin[ω(t− u

c

)]u

dz z.

Applying Approximation #1 gives us

u ' r

(1− d

2rcos θ

)1

u' 1

r

(1 +

d

2rcos θ

).

Then following the same procedure as we did forsin[ω(t− u

c

)], we find that

sin[ω(t− u

c

)]' sin

[ω(t− r

c

)]+ωd cos θ

2ccos[ω(t− r

c

)].

Plugging this in and integrating gives us

~A(r, θ, t) = −µ0p0ω

4πrsin[ω(t− r

c

)]z,

where we have discarded terms of order O(d2).

Noting that ~p0 = p0 z, we can write this in thecoordinate-free form

~A(r, θ, t) = −µ0ω

4πrsin[ω(t− r

c

)]~p0.

The Fields

The electric field is calculated as

~E = −~∇V − ∂ ~A

∂t.

Taking the gradient of V gives us

~∇V = − p0ω

4πε0c

−cos θ

r2sin[ω(t− r

c

)]r

−ω cos θ

rccos[ω(t− r

c

)]r

− sin θ

r2sin[ω(t− r

c

)]θ

9.2. MAGNETIC DIPOLE RADIATION 151

Applying Approximation #3 gives us

~∇V ' p0ω2 cos θ

4πε0c2rcos[ω(t− r

c

)]r.

Similarly,

∂ ~A

∂t= −µ0p0ω

2

4πrcos[ω(t− r

c

)]z,

where z = cos θ r − sin θ θ.

Plugging ~∇V and ∂ ~A∂t into the equation for the elec-

tric field gives us

~E = −µ0p0ω2 sin θ

4πrcos[ω(t− r

c

)]θ.

Noting that ~p0 × r = p0 sin θ φ and φ × r = θ, weknow that (~p0 × r) × r = p0 sin θ θ, so we can write thisresult in the coordinate-free form

~E = −µ0ω2

4πrcos[ω(t− r

c

)](~p0 × r)× r.

To find ~B, we use ~B = ~∇× ~A. Following the sameprocedure to calculate the curl of ~A as we did to find ~∇V ,we find that

~B = −µ0p0ω2 sin θ

4πrccos[ω(t− r

c

)]φ.

Using ~p0 × r = p0 sin θ φ, we can write this in thecoordinate-free form

~B = −µ0ω2

4πrccos[ω(t− r

c

)](~p0 × r) .

Notice that ~E and ~B are orthogonal to each otherand to their direction of travel, and E0

B0= c. So these

equations describe an electromagnetic plane wave. Sincethey are emanating from a point, these are really spheri-cal waves.

The Poynting vector is

~S =~E × ~Bµ0

=µ0p

20ω

4 sin2 θ

16π2r2ccos2

[ω(t− r

c

)]r.

So the energy flow is indeed outward from the oscillatingdipole.

The intensity is the time average (over one completecycle) of the Poynting vector, so

〈~S〉 =µ0p

20ω

4 sin2 θ

32π2r2cr.

The intensity as a function of θ is a “donut”. It is highestin the equatorial plane, and there is zero radiation in thez direction.

The total power that is radiated by the oscillatingdipole is found by integrating the average intensity overa sphere of radius r

〈P 〉 =

ˆ〈~S〉 · d~a

=

ˆµ0p

20ω

4 sin2 θ

32π2r2cr · r2 sin θ dθ dφ r

=µ0p

20ω

4

12πc.

Notice that the power is independent of the radius, andwe can take the radius of the sphere to be infinity.

9.2 Magnetic Dipole Radiation

Suppose we have a circular wire loop of radius b centeredon the origin and lying in the xy-plane. If we drive analternating current

I(t) = I0 cos(ωt),

around this loop, then we have an oscillating magneticdipole

~m = πb2I(t)z = m0 cos(ωt),

wherem0 = πb2I0.

The wire loop is uncharged so the scalar potential iszero. The retarded vector potential is

~A(~r, t) =µ0

4π

Î0 cos

[ω(t− u

c

)]u

d~l ′.

The vector to the field point (where we are measuring~A) is

~r = r sin θ cosφ i+ r sin θ sinφ j + r cos θ k.

The location of the current differential on the loop in thexy-plane can be written as

~r ′ = b cosφ′ i+ r sinφ′ j,

where φ′ is the angle between the positive x-axis and thepoint on the loop.

Then the differential can be written as

d~l ′ = d~r ′ = b(− sinφ′ i+ cosφ′ j

)dφ′,

and

~u = ~r − ~r ′

= (r sin θ cosφ− b cosφ′)i

+r sin θ sinφ− b sinφ′)j

+r cos θ k.


The magnitude, after simplifying, is

u =√r2 + b2 − 2rb sin θ cos(φ− φ′)

= r

√1 +

b2

r2− 2b

rsin θ cos(φ− φ′).

In the limit b << r, this becomes

u ' r[1− b

rsin θ cos(φ− φ′)

].

Then1

u' 1

r

[1 +

b


].

We can also approximate

cos[ω(t− u

c

)]= cos

[ω(t− r

c

)+ωb

csin θ cos(φ− φ′)

]' cos

[ω(t− r

c

)]cos

[ωb


]− sin

[ω(t− r

c

)]sin

[ωb


]' cos

[ω(t− r

c

)]− sin

[ω(t− r

c

)] ωbc

sin θ cos(φ− φ′).

Plugging these in gives us

~A(~r, t) =µ0I0b

4πr

ˆ cos[ω(t− r

c

)]− sin

[ω(t− r

c

)] ωbc

sin θ cos(φ− φ′)[

1 +b


](− sinφ′ i+ cosφ′ j

)dφ′

=µ0I0b

2

4πr

πr

sin θ cos[ω(t− r

c

)]− πω

csin θ sin

[ω(t− r

c

)](− sinφ i+ cosφ j

)=

µ0I0b2

4πr

πr

sin θ cos[ω(t− r

c

)]− πω

csin θ sin

[ω(t− r

c

)]φ

=µ0m0

4πr

1

rsin θ cos

[ω(t− r

c

)]− ω

csin θ sin

[ω(t− r

c

)]φ.

In the static limit (ω = 0), we get the expected result

~A(~r, θ) =µ0

4π

m0 sin θ

r2φ.

In the radiation zone (r >> ω/c), the first term in~A is zero

~A(r, θ, t) = −µ0m0ω

4πrcsin θ sin

[ω(t− r

c

)]φ.

Then the electric field is ~E = −∂ ~A∂t or

~E =µ0m0ω

2

4πrcsin θ cos

[ω(t− r

c

)]φ,

and the magnetic field is ~B = ~∇× ~A or

~B = −µ0m0ω2

4πrc2sin θ cos

[ω(t− r

c

)]θ.

Notice that these are an electromagnetic wave with E0

B0=

c.

The energy flux is

~S =1

µ0

~E × ~B =µ0

c

m0ω

2

4πrcsin θ cos

[ω(t− r

c

)]2

r.

Taking the time average gives us the intensity

〈~S〉 =µ0m

20ω

4

32π2c3sin2 θ

r2r.

Integrating this over a sphere gives us the radiated power

〈P 〉 =µ0m

20ω

4

12πc3.

Note: For an electric dipole and a magnetic dipoleof similar dimensions, the power radiated by the electricdipole is much, much greater than the power radiated bya magnetic dipole. In fact,

PmagPelec

∼(ωb

c

)2

,

where b gives the length scale. Because of the c in thedenominator, this is a very small quantity, implying thatthe power radiated by a magnetic dipole is much weakerthan that of an electric dipole of similar size.

9.3. ARBITRARY RADIATION SOURCES 153

9.3 Arbitrary Radiation Sources

Now consider an arbitrary, time-dependent charge/cur-rent distribution near the origin. As before, ~r is the fieldpoint, ~r ′ specifies an element in the configuration, and~u = ~r − ~r ′. We will assume that r >> r′. The retardedpotentials are

V (~r, t) =1

4πε0

ˆρ(~r ′, t− u

c

)u

dτ ′

~A(~r, t) =µ0

4π

ˆ ~J(~r ′, t− u

c

)u

dτ ′.

We now define the retarded time at the origin as

t0 = t− r

c.

Next, we expand ρ and ~J in Taylor series in time, to firstorder in r′, about t0 to get

V (~r, t) ' 1

4πε0

[Q

r+r · ~p(t0)

r2+r · ~p(t0)

rc

],

where Q is the total charge, and

~p(t0) =

ˆ~r ′ρ(~r ′, t0) dτ ′,

is the electric dipole moment at time t0. Similarly,

~A(~r, t) ' µ0

4π

~p(t0)

r.

The full derivation is given in Griffiths.

This leads us to an electric field

~E ' µ0

4πr

(r ×

(r × ~p

)),

and a magnetic field

~B ' − µ0

4πrc

(r × ~p

),

where ~p is evaluated at t0. Again, the full derivation isgiven in Griffiths.

What we’ve done is a multipole expansion of the re-tarded potentials to first order in r′. If it turns out that~p = 0, then these formulae don’t work, and we have toobtain high order formulae by repeating the process andexpanding to second (or higher) order in r′.

For example, for an oscillating electric dipole, wehave

p(t) = p0 cos(ωt),

then

p(t0) = −ω2p0 cos(ωt).

Plugging this in gives us the same results that we got inthe section on electric dipole radiation.

If we define the z-axis (in spherical coordinates) to

be in the direction of ~p(t0), then we can write the fieldsin the form

~E(r, θ, t) ' µ0p(t0)

4πrsin θ θ

~B(r, θ, t) ' µ0p(t0)

4πrcsin θ φ.

Then the Poynting vector is

~S =1

µ0

~E × ~B ' µ0

16π2r2c[p(t0)]

2sin2 θ r.

The power radiated is obtained by integrating the Poynt-ing vector over a spherical surface with large r and thentaking r to infinity

P =

ˆ~S · d~a ' µ0p

2

6πc.

9.4 Point Charges

Recall that the electric and magnetic fields of a pointcharge in arbitrary motion are

~E(~r, t) =q

4πε0

u

(~u · ~N)3

[(c2 − v2) ~N + ~u× ( ~N × ~a)

]~B(~r, t) =

1

c~u× ~E,

where ~N = cu − ~v. The first term in the electric fieldis called the velocity field. The second term is calledthe acceleration field. The Poynting vector can thenbe written as

~S =1

µ0c

[E2u− (u · ~E)~E

].

But not all of this moving energy is radiated energy. Someof it is energy that is being carried along in the fields. Ra-diated energy is energy that leaves (to infinity) and nevercomes back.

Only the 1r2 terms in ~S contribute to radiation. In

the electric field then, only the acceleration field remainsin the radiation field

~Erad =q

4πε0

u

(~u · ~N)3~u× ( ~N × ~a).

The velocity field also carries energy along, but this en-ergy is not radiated away. Then

~Srad =1

µ0c(Erad)

2u.

We want to calculate the total power radiated by theparticle at time tr. Suppose we center a large sphere of


radius u on the moving particle at time tr. At a latertime

t− tr =u

c,

the particle is at the surface of the sphere. If the chargeis instantaneously at rest at the time tr, then ~N = cu,and we get

~Erad =q

4πε0c2uu× (u× ~a)

=µ0q

4πu[(u · ~a)u− ~a] .

So we get

~Srad =1

µ0c

( µ0q

4πu

)2 [a2 − (u · ~a)2

]u

=µ0q

2a2

16π2c

sin2 θ

u2u.

Here, θ is the angle between u and ~a. From this equa-tion, we see that no power is radiated in the forward orbackward directions. Rather, the power is radiated outin sort of a donut shape—more to the sides than forwardor backward.

Then the total radiated power is

P =

˛~Srad · d~a

=µ0q

2a2

16π2c

ˆsin2 θ

u2u2 sin θ dθ dφ.

This simplifies to

P =µ0q

2a2

6πc.

which is called the Larmor formula. It is only when aparticle is moving near the speed of light that this is nolonger a good approximation for the radiated power.

9.5 Radiation Reaction and Self-Force

We now know that an accelerating particle radiates. Sincethe particle is losing energy by this radiation, the parti-cle’s kinetic energy must be decreasing. In other words,an accelerating charged particle feels a resistive force thatan uncharged particle does not.

For a nonrelativistic (i.e. v << c) charged particle,the power lost to radiation is given by the Larmor formula

P =µ0q

2a2

6πc.

This suggests a resistive radiation force ~F rad, that ac-counts for the energy loss

~F rad · ~v = −µ0q2a2

6πc.

However, this reasoning is not completely correct. Anaccelerating particle, in addition to losing kinetic energydue to radiation (which is carried away to infinity) mayalso temporarily “lose” kinetic energy to the velocity fieldwhich it carries with it. The relation above accounts onlyfor the energy lost to radiation, but to understand the re-sistive radiation force, we need to account for both. Thisequation is, however, valid on average. We can say that

ˆ t2

t1

~F rad · ~v dt = −µ0q2

6πc

ˆ t2

t1

a2 dt, (9.1)

provided that the system is the same at times t1 and t2.This can be accomplished, for example, if our system isa periodic (oscillating) system and (t2 − t1) is an integernumber of periods. We can integrate the right side byparts as

ˆ t2

t1

a2 dt =

ˆ t2

t1

d~v

dt· d~vdtdt

= ~v · d~vdt

∣∣∣t2t1−ˆ t2

t1

d2~v

dt2· ~v dt

= −ˆ t2

t1

~a · ~v dt.

The boundary term drops out by the assumption that thesystem is the same at times t1 and t2. We can now writeEq. (9.1) as

ˆ t2

t1

(~F rad −

µ0q2

6πc~a

)· ~v dt = 0.

Clearly, one solution is

~F rad =µ0q

2

6πc~a.

This is in fact the right solution, and it is called theAbraham-Lorentz formula for the radiation reactionforce.

This derivation tells us what the recoil force is, butit doesn’t tell us why it occurs. To do that, Griffiths ex-plores a dumbbell shaped charge distribution and in theend takes the distance between the two charges to zero.Suffice it to say that the radiation reaction self force is“the net force exerted by the fields generated by differentparts of the charge distribution acting on one another”(Griffiths).

The radiation reaction force leads to an interestingproblem. Suppose there is no external force acting on acharged particle. Then from Newton’s second law,

Frad =µ0q

2

6πca = ma.

However, this ODE implies that

a(t) = a0et/τ ,

9.5. RADIATION REACTION AND SELF-FORCE 155

where

τ =µ0q

2

6πmc.

This implies that a charged particle will spontaneously ac-celerate, even without the presence of a force. We couldresolve this by asserting that a0 must be zero. However,that leads to even worse problems. Then, if an externalforce is applied, the particle begins accelerating beforethe force is applied. That is, future forces reach back intime to accelerate particles in the present. Apparently,these problems persist even in the relativistic versions ofthe theory.

Often in physics, the (self) forces exerted on a par-ticle by the particle’s own fields are neglected, and theresults are accurate enough. Including these self-forcesleads to problems, which in many cases remain unre-solved. The Abraham-Lorentz force is one such example.In QED, this problem is swept under the rug by a pro-cess called renormalization. However, in other casessuch as for the gravitational force, even renormalizationcannot resolve the self-force problem.

A charged particle attached to the end of aspring of natural frequency ω0 and driving frequencyω should feel a damping force due to the radiationreaction. Calculate that damping force. a

The equation of motion for a driven damped har-monic oscillator is typically written as

x+ 2βx+ ω20x =

1

mFdriving,

where β is the damping parameter.

In our case, the equation of motion of the springis

mx = Fspring + Frad + Fdriving

= −mω20x+mτ

...x + Fdriving.

The system is being driven at frequency ω, so we canwrite the solution as

x(t) = x0 cos(ωt+ δ).

This implies that

x = −ω2x,

and so we can write the equation of motion as

x+ τω2x+ ω20x =

1

mFdriving.

By comparison, we see that the radiation dampingparameter is

β =1

2τω2.


Example:


9.6 Summary: Radiation

Radiation is the generation of electromagnetic waves.Radiation is an irreversible flow of EM energy from asource to infinity. The total power passing through thesurface of a sphere of radius r centered on the source isthe integral of the Poynting vector

P (r) =

˛~S · d~a =

1

µ0

˛ (~E × ~B

)· d~a.

Then the power radiated is the limit of this quantity asthe size of the enclosing sphere is taken to infinity

Prad = limr→∞

P (r).

To calculate radiation, we construct the 1r2 term in ~S

from the 1r terms in ~E and ~B, and then we integrate this

term over a spherical surface and take the limit of thespherical surface to infinity. It is only the 1

r terms in ~E

and ~B that contribute to radiation.

Oscillating Dipoles

Electric dipole radiation is produced by, e.g. an oscillat-ing electric dipole. If we have a time-dependent chargedistribution q(t), then we have a time dependent electricdipole p(t) = dq(t). Magnetic dipole radiation is pro-duced by, e.g. an oscillating magnetic dipole. If we havea time-dependent current distribution I(t), then we havea time-dependent magnetic dipole m = AI(t). Here weuse r for the distance from the field point to the centerof the distribution, and d is the width of the distribution.To find the power radiated:

1. Calculate the retarded scalar potential. If there isno charge, then V = 0, and you can go immediatelyto calculating the vector potential.

• Make the approximation d << r to approxi-mate u and 1

u• Make the approximation d << c

ω to approxi-mate e.g. the sine and cosine pieces as smallangles.• Make the approximation r >> c

ω to discardany terms containing 1

r .

2. Calculate the retarded vector potential

• Make the same approximations as for thescalar potential

3. Calculate the fields from the retarded potentials

~E = −~∇V − ∂ ~A

∂t~B = ~∇× ~A.

Again, apply the three approximations as appropri-ate.

4. Calculate the energy flux as the Poynting vector

~S =~E × ~Bµ0

.

5. Calculate intensity as the time average of the Poynt-ing vector

〈~S〉.6. Calculate the total radiated power by integrating

the average intensity over a sphere of radius r

〈P 〉 =

ˆ〈~S〉 · d~a.

7. If the radiated power still depends on r, take thelimit r →∞.

Arbitrary Sources

For an arbitrary time-dependent charge/current distribu-tion near the origin, we can do a multipole expansion ofρ and ~J from the retarded potentials via a Taylor seriesin time to first order in r′. We get

V (~r, t) ' 1

4πε0

[Q

r+r · ~p(t0)

r2+r · ~p(t0)

rc

]

~A(~r, t) ' µ0

4π

~p(t0)

r.

where Q is the total charge, and

~p(t0) =

ˆ~r ′ρ(~r ′, t0) dτ ′,

is the electric dipole moment at time t0. Then the fieldsbecome

~E ' µ0

4πr

(r ×

(r × ~p

))~B ' − µ0

4πrc

(r × ~p

).

If we define the z-axis (in spherical coordinates) to

be in the direction of ~p(t0), then we can write the fieldsin the form

~E(r, θ, t) ' µ0p(t0)

4πrsin θ θ

~B(r, θ, t) ' µ0p(t0)

4πrcsin θ φ.

Then the Poynting vector is

~S =1

µ0

~E × ~B ' µ0

16π2r2c[p(t0)]

2sin2 θ r.

The power radiated is obtained by integrating the Poynt-ing vector over a spherical surface with large r and thentaking r to infinity

P =

ˆ~S · d~a ' µ0p

2

6πc.

9.6. SUMMARY: RADIATION 157

Point Charges

For a point charge in arbitrary motion,

~E(~r, t) =q

4πε0

u

(~u · ~N)3

[(c2 − v2) ~N + ~u× ( ~N × ~a)

]~B(~r, t) =

1

c~u× ~E,

where ~N = cu− ~v.

The Poynting vector can then be written as

~S =1

µ0c

[E2u− (u · ~E)~E

].

But not all of this moving energy is radiated energy. Onlythe 1

r2 terms in ~S contribute to radiation. Simplifying to

get ~Erad with only 1r2 terms, calculating the simplified

~Srad and then integrating over a sphere to get the totalradiated power gives us the Larmor formula

P =µ0q

2a2

6πc

for the power radiated by an accelerating point charge.

Since an accelerating charge is losing energy dueto radiation, it experiences a resistive radiation reactionforce

~F rad =µ0q

2

6πc~a.

This is called the Abraham-Lorentz formula for theradiation reaction force.

Chapter 10

Relativistic Electrodynamics

10.1 Special Theory of Relativity

There is a remarkable coincidence in electrodynamics thatis explained by the special theory of relativity. Supposeyou have a loop of wire and a magnet in motion relativeto each other. If the magnet is stationary and the loopmoves through the magnet, then the flux rule tells us thata motional emf

E = −dΦ

dt

is established in the loop. If the loop is stationary and themagnet is moved past the loop, then according to Fara-day’s law, the changing magnetic field induces an electricfield, and the resulting electric force would generate anemf in the loop of also

E = −dΦ

dt.

Before relativity, this was viewed as a remarkable coinci-dence.

Galilean relativity—the idea that classical mechanicsobeys the same laws in all inertial frames—was acceptedlong before the special theory of relativity. However, itwas believed that electrodynamics did not obey relativ-ity. For example, if a charged particle attached to a carmoving at constant velocity flies past an observer besidethe road, the observer beside the road would see a mag-netic field produced by the moving charge. However, anobserver inside the car at rest with respect to the parti-cle, would see only an electric field and would insist thatthe particle is not producing a magnetic field. Here wehave two observers who disagree about the result of anelectrodynamics experiment.

Einstein’s two postulates of special relativity are

1. The laws of physics (including electrodynamics) ap-ply in all inertial frames

2. The speed of light in a vacuum is the same in all in-ertial frames regardless of the motion of the source

Suppose there is a frame A moving with speed vABwith respect to a frame B which is moving with speed vBC

with respect to a frame C. What is the speed of frame Awith respect to C? For example, suppose you (frame A)are walking down the aisle of a train (frame B) that ismoving with respect to the ground (frame C). What isyour speed with respect to the ground? According to theGalilean velocity addition rule, your speed with respectto the ground is

vAC = vAB + vBC .

However, according to Einstein’s velocity additionrule, speeds actually add as

vAC =vAB + vBC1 + vABvBC

c2.

There are several important consequences of Ein-stein’s special relativity:

• A pair of events which are simultaneous in on in-ertial frame are not necessarily simultaneous in an-other inertial frame.

• Moving clocks run slow. This is the phenomenon oftime dilation. For two frames in relative motion,time measured in the two frames is related by

t = γτ,

where τ is the proper time measured in the restframe of the moving clock, and t is the time mea-sured by the observer who sees the moving clock.The factor

γ =1√

1− v2

c2

,

is called the Lorentz factor.• Moving objects are shortened. This is the phe-

nomenon of Lorentz contraction. The lenth Lof a moving object is related to its length ` mea-sured in its rest frame by

L = γ`.

Note, a moving object is Lorentz contracted only inthe dimension that is parallel to its direction of mo-tion. Perpendicular dimensions are not contracted.

158

10.1. SPECIAL THEORY OF RELATIVITY 159

To remember the forms of the time dilation andLorentz contraction equations, just remember the twosayings “Moving clocks run slower” and “Movinglengths are shorter,” and use the fact that the Lorentzfactor is always greater than one, γ > 1.

Tip:

Suppose you are in a frame S and an event occurs atthe spacetime location (x, y, z, t) in your frame. There isa second frame S′ moving with speed v in the x directionrelative to your frame S. What are the coordinates of thesame event in the moving frame?

The Galilean transformation equations tell us that

x′ = x− vty′ = y

z′ = z

t′ = t,

however, these are not correct. Special relativity gives usthe true transformations called the Lorentz transfor-mations

x′ = γ(x− vt)y′ = y

z′ = z

t′ = γ(t− vx

c2

).

The inverse Lorentz transformations (i.e. given the space-time event (x′, y′, z′, t′) in S′ we can obtain the coor-dinates in S) can be obtained simply by switching theprimes and changing the sign of v.

x = γ(x′ + vt′)

y = y′

z = z′

t = γ

(t′ +

vx′

c2

).

We get more convenient equations if we make thesubstitutions

x0 = ct

x1 = x

x2 = y

x3 = z

β =v

c.

Then we can write the Lorentz transformation equations

x0 ′ = γ(x0 − βx1)

x1 ′ = γ(x1 − βx0)

x2 ′ = x2

x3 ′ = x3.

or in the form of a matrix multiplication asx0 ′

x1 ′

x2 ′

x3 ′

=

γ −γβ 0 0

−γβ γ 0 0

0 0 1 0

0 0 0 1

x0

x1

x2

x3

.

We can also write them in the compact form

xµ ′ =

3∑ν=0

Λµνxν .

Here, Λ is the 4×4 Lorentz transformation matrix shownabove. The superscript µ refers to the row, and ν refersto the column.

We can derive the Lorentz equations by consideringa pulse of light emitted at the origin at t = 0. In frameS, the light front is the expanding spherical shell

x2 + y2 + z2 = c2t2.

In frame S′ whose origin coincides with that of frame Sat t = 0, we expect the light front to have the form

(x′)2 + (y′)2 + (z′)2 = c2(t′)2.

As always, we assume that S′ is moving in the x direc-tion with speed v relative to S. From time dilation andLorentz contraction, we know that the x and t compo-nents mix together, so we try a solution of the form

x′ = ax+ bt

y′ = y

z′ = z

t′ = gx+ ft.

At x′ = 0, the first equation gives us x = − ba t, which

implies that v = − ba . At x = 0, the first and third equa-

tions give us x′ = bt and t′ = ft. This implies that t = t′

f

and so x′ = bf t′. But this is distance equals speed times

time again, implying that −v = bf . Comparing this with

v = − ba implies that a = f . So we can write

0 = (x′)2 + (y′)2 + (z′)2 − c2(t′)2

= (ax+ bt)2 + y2 + z2 − c2(gx+ at)2.

Fully expanding this and comparing it with

0 = x2 + y2 + x2 − c2t2,

implies that

a2 − c2g2 = 1

c2a2 − b2 = c2

2ab− 2c2ag = 0.

160 CHAPTER 10. RELATIVISTIC ELECTRODYNAMICS

Assuming a 6= 0, the third equation implies b = c2g.From v = − b

a , we have that b = −av, and plugging thisin, −av = c2g or g = −av/c2. Substituting these resultsfor b and g into the first equation gives us

a =1√

1− v2

c2

= γ.

This implies that

b = −γvg = −γv

c2.

Plugging these back into our original trial solutions givesus the Lorentz transformations.

More generally, we define a 4-vector as any set offour components that transforms under Lorentz trans-formations in the same way as the position 4-vector(x0, x1, x2, x3)

aµ ′ =

3∑ν=0

Λµνaν .

For transformation along the x-axis, then

a0 ′ = γ(a0 − βa1)

a1 ′ = γ(a1 − βa0)

a2 ′ = a2

a3 ′ = a3.

This is written in terms of a contravariant 4-vector withthe index upstairs

aµ = (a0, a1, a2, a3).

We can also write covariant 4-vectors with the indexdownstairs

aµ = (a0, a1, a2, a3).

The two are related by a minus sign on the zeroth com-ponent

aµ = (a0, a1, a2, a3) = (−a0, a1, a2, a3).

Whenever you raise or lower the temporal index, you at-tach a minus sign

a0 = −a0.

Whenever you raise or lower a spatial index, there is nosign change

a1 = a1

a2 = a2

a3 = a3.

The dot product of two 4-vectors aµ and bµ is writtenas

aµbµ = −a0b0 + a1b2 + a2b2 + a3b3.

Note the minus sign for the temporal term. Also, notethat we are using Einstein summation notation. Finally,it doesn’t matter which one 4-vector is covariant. Thatis, aµb

µ = aµbµ.

The dot product of any two 4-vectors is invariantunder Lorentz transformations. That is, it is the same inany inertial frame.

If an event A occurs at aµ = (a0, a1, a2, a3) and eventB occurs at bµ = (b0, b1, b2, b3), then the difference be-tween the two

∆xµ = aµ − bµ,

is the displacement 4-vector. The dot product of thedisplacement 4-vector with itself is called the intervalbetween the two events, and is written as

I = −(∆x0)2 + (∆x1)2 + (∆x2)2 + (∆x3)2 = −c2t2 + d2,

where t is the time difference between the two events andd is the spatial separation between the two events. Likeall 4-vector dot products, the interval between two spec-ified events is the same in all inertial frames.

The spacetime interval between two events alwaysfalls into one of three categories:

Timelike: If I < 0, then the interval is called timelikesince this is the sign we get when d = 0. For atimelike separation, there always exists an inertialframe (traveling between the two locations) whereinthe two events occur at the same location.

Spacelike: If I > 0, then the interval is called space-like since this is the sign we get when t = 0. Fora spacelike separation, there always exists an iner-tial frame wherein the two events occur at the sametime.

Lightlike: If I = 0, then the interval is called lightlike.

10.2 Relativistic Mechanics

Suppose you have an object (attached to frame S′) mov-ing with speed u relative to your frame S. We define theordinary velocity as

~u =d~l

dt,

where both the distance d~l and the time dt are measuredin your frame S. Note that this is an ordinary 3-vector.We define the proper velocity as

~η =d~l

dτ,

where the distance d~l is still measured in your frame S,but now the time dτ is the proper time which is mea-sured in the rest frame S′ of the moving object. The two

10.2. RELATIVISTIC MECHANICS 161

velocities are related by

~η =1√

1− u2

c2

~u.

The proper velocity ~η is also an ordinary 3-vector, how-ever, it is also the spatial part of a 4-vector called the4-velocity

ηµ =dxµ

dτ,

which has a zeroth component

η0 =dx0

dτ= c

dt

dτ=

c√1− u2

c2

.

To transform from S to S′ moving at speed v in thex-direction, we use

η0 ′ = γ(η0 − βη1)

η1 ′ = γ(η1 − βη0)

η2 ′ = η2

η3 ′ = η3.

This is much nicer than the transformation rules for theordinary velocity.

In order for conservation of momentum to work inrelativity, we define the relativistic momentum as

~p = m~η =m~u√1− u2

c2

.

For a closed system, this relativistic momentum is al-ways conserved. This is the spatial part of the energy-momentum 4-vector

pµ = mηµ = (E/c, ~p) ,

whose temporal component is E/c, where

E =mc2√1− u2

c2

,

is the relativistic energy. For a closed system, thisenergy is always conserved.

For a closed system, the relativistic momentum andrelativistic energy are both conserved.

Tip:

When an object is stationary, its total energy iscalled its rest energy

Erest = mc2.

When in motion, the remainder

Ekin = E −mc2 = mc2

1√1− u2

c2

− 1

,

is the kinetic energy.

Taking the dot product of the energy-momentum 4-vector with itself leads us to the important relativisticenergy-momentum relation

E2 = p2c2 +m2c4,

which allows us to calculate E or p given the other, with-out having to know the velocity.

If two masses m that are both moving at 35c,

collide head-on and stick together after the collision,what is the mass M of the composite object after thecollision? a

Conservation of momentum doesn’t give us any-thing because the total momentum is zero before andafter the collision. Before the collision, both objectsare moving at 3

5c, so they each have energy

mc2√1−

(35

)2 =5

4mc2.

Afterwards, they are at rest, so the total rest energyis

Mc2.

So

Mc2 =5

4mc2 +

5

4mc2.

So the final mass is

M =5

2m.

The final mass is greater than the sum of the initialmasses, implying that kinetic energy has been con-verted into mass.


Example:

In the example above, kinetic energy was convertedinto mass. Classically, when two objects collide, we saythat the kinetic energy has been converted to thermal en-ergy. In special relativity, we can still say that. Sinceenergy and mass are equivalent, an object has slightlymore mass when it is heated than when it is cold. Thisis simply because the hotter object holds more internalthermal energy, and internal energy U is related to massby U = mc2.


For a massless particle such as a photon, the rela-tivistic energy is

E = pc.

In special relativity, momentum and total energy arealways conserved. However, kinetic energy is not alwaysconserved. As in classical mechanics, a collision is elas-tic if kinetic energy is conserved. In an elastic collision,kinetic energy is conserved, and since the total energyis always conserved, the rest energy must be conserved.This implies that mass is conserved. For an elastic col-lision between particles, this means the same particlescome out of the collision as went into the collision.

If kinetic energy is conserved then mass is conserved.

Tip:

Newton’s second law

~F =d~p

dt,

holds in relativity provided that the momentum ~p we useis the relativistic momentum.

Consider the case of a particle of mass m subject toa constant force F . If the particle starts from rest at theorigin at t = 0, then integrating dp

dt = F gives us

p = Ft.

In the classical case, p = mdxdt , and integrating mdx

dt = Fttells us that the particle’s trajectory x(t) is a quadratic/-parabolic curve. What about the relativistic case? Rela-tivistic momentum implies that

p =mu√1− u2

c2

= Ft.

Inverting to get u as a function of t gives us

u =(F/m)t√

1 + (Ft/mc)2.

Replacing u = dxdt and integrating, gives us

x(t) =mc2

F

√

1 +

(Ft

mc

)2

− 1

.

The trajectory is now a hyperbola. Hence, in special rel-ativity, motion due to a constant force is called hyper-bolic motion. This is what occurs when a charged par-ticle is accelerated in a particle accelerator. In the limitt << mc

F , we get x ' F2m t

2, so we recover the classicalresult.

Newton’s third law does not, in general, hold in rel-ativity.

Work is still the line integral of the force

W =

ˆ~F · d~l.

We can write

W =

ˆd~p

dt· d~l =

ˆd~p

dt· d~l

dtdt =

ˆd~p

dt· ~u dt.

But

d~p

dt· ~u =

d

dt

m~u√1− u2

c2

· ~u=

m~u(1− u2

c2

) 32

· d~udt

=d

dt

mc2√1− u2

c2

=

dE

dt.

Here we used the fact that

d

dtu2 =

d

dt~u · ~u = 2~u · d~u

dt,

and applied the product rule when differentiating. Ourresult implies that

W =

ˆdE

dtdt = Ef − Ei.

This tells us that the work-energy theorem is valid inrelativity. Since the rest energy is constant, for E, we canuse either the total energy or just the kinetic energy.

Lorentz transforming force from a frame S to S′ be-comes messy because we have to Lorentz transform bothdp′ and dt′. For example, for the y component, we get

F ′y =dp′ydt′

=dpy

γ(dt− β

c dx)

=dpydt

γ(

1− βcdxdt

)=

Fy

γ(

1− βuxc

) .For the other two components, we get

F ′z =Fz

γ(

1− βuxc

)F ′x =

Fx − β(~u·~F )c

1− βuxc

.

10.3. RELATIVISTIC ELECTRODYNAMICS 163

We can avoid the messiness by defining a proper forcecalled the Minkowski force. It is simply the derivativeof the relativistic 4-momentum with respect to propertime

Kµ =dpµ

dτ.

Its spatial part is

~K =dt

dτ

d~p

dt=

1√1− u2

c2

~F ,

and its zeroth component is

K0 =dp0

dτ=

1

c

dE

dτ.

While nicer, the ordinary force is typically a more usefulquantity than the Minkowski force.

A charged particle in a uniform magnetic fieldundergoes uniform circular motion.a The magneticforce

F = QuB,

provides the centripetal force. However, we can no

longer use the classical formula mu2

R for the cen-tripetal force. For a particle in circular motion, wecan write dp = p dθ. Then the centripetal force is

F =dp

dt= p

dθ

dt= p

u

R.

This is the form of the centripetal force that is validin special relativity. Classically, the two are the same,but relativistically, they are not.

Equating the two gives us

p = QBR.

This looks the same as the classical result, however,now p is the relativistic momentum.


Example:

10.3 Relativistic Electrodynamics

Classical electrodynamics is already compatible with spe-cial relativity. Special relativity implies that magnetismis a relativistic phenomenon. To see this, let us pretendthat we are unaware of the existence of magnetism andapply special relativity to a specific system.

Suppose we have a wire in which both positive andnegative charges are moving. In our frame S, the pos-itive charges are moving to the right with speed v, andthe negative charges are moving to the left with speed v.Assuming the charges are evenly and closely spaced, wesay the wire has a positive line charge of λ and a negative

line charge of −λ. Since the positive charges are movingto the right and the negative charges are moving to theleft, we have a net current of

I = 2λv.

Now suppose a charge q is moving to the right with speedu < v through our frame S. It is moving parallel to andat a distance of s from the wire.

Now consider the frame S′ attached to the movingcharge. So frame S′ is moving to the right at speed u withrespect to frame S. In S′, the charge q is at rest. In thisframe, the Einstein velocity addition rule tells us that thevelocities of the positive and negative charges are

v± =v ± u1∓ vu

c2.

Now, the magnitudes of the velocities of the negative andpositive charges are no longer the same. Now, the velocityof the negative charges is greater than the velocity of thepositive charges. This implies that the Lorentz factors forthe positive and negative line densities are different

γ± =1√

1− v2±c2

,

and thus, the Lorentz contraction of the space betweenthe negative charges is now greater than the Lorentz con-traction of the space between the positive charges. Thatis, in S′ it appears that the negative charges are packedmore closely than the positive charges, and so the wireappears to carry a net negative charge. In frame S′, theline densities are now

λ± = ± (γ±)λ0,

where λ0 is the charge density of the positive line chargein its own rest frame (which, by the way, is neither Snor S′). In frame S, the charges are already moving withspeed v, so

λ = γλ0,

where

γ =1√

1− v2

c2

.

We can rewrite λ± as

γ± = γ1∓ vu

c2√1− u2

c2

.

So in frame S′, the net line density is

λ′ = λ+ + λ− = λ0 (γ+ − γ−) =−2λuv

c2√

1− u2

c2

.

This is all to say that in frame S′, the wire has a netcharge, and so there is an electric field

E′ =λ′

2πε0s.


So the electric force on the charge q in frame S′ is

F ′ = qE = − λv

πε0c2s

qu√1− u2

c2

.

Since there is a force on charge q in frame S′, theremust be a force on charge q in frame S. Using the trans-formation equations for force, which were found in a pre-vious section, we find that the force in frame S is relatedto the force in frame S′ by

F =

√1− u2

c2F ′ = − λv

πε0c2qu

s= −qu

(µ0I

2πs

).

Since the wire is uncharged in S, this force F is clearlynot an electric force. It is a completely different force,and we recognize it as the magnetic force. Hence, mag-netism can be thought of as a relativistic effect of movingelectric charges.

Field Transformations

For a frame S′ moving with speed v in the x directionrelative to frame S, it can be derived via moving parallelplate capacitors and a solenoid, that the Lorentz trans-formations for the electric and magnetic fields are

E′x = Ex

E′y = γ (Ey − vBz)E′z = γ (Ez + vBy)

B′x = Bx

B′y = γ(By +

v

c2Ez

)B′z = γ

(Bz −

v

c2Ey

).

Instead of ~E and ~B acting as the spatial parts of two4-vectors, instead, it happens that the components of ~Eand ~B are mixed together.

In the special case that ~E = 0 in frame S, we canwrite

~E ′ = ~v × ~B ′.In the special case that ~B = 0 in frame S, we can write

~B ′ = − 1

c2

(~v × ~E ′

).

Electromagnetic Waves

Consider an electromagnetic plane wave with angular fre-quency ω polarized in the y direction and traveling in thex direction through the vacuum.

~E(x, y, z, t) = E0 cos (kx− ωt) y

~B(x, y, z, t) =1

cE0 cos (kx− ωt) z,

wherek =

ω

c.

Only Ey and Bz are nonzero, so when we apply thetransformation equations, we get E′x = E′z = B′x = B′y =0. For E′y, we get

E′y = γ (Ey − vBz)

= γE0 cos (kx− ωt)− γv

cE0 cos (kx− ωt)

= γ(

1− v

c

)E0 cos (kx− ωt)

=

√1− v/c1 + v/c

E0 cos (kx− ωt) .

Similarly, for B′z, we get

B′z =

√1− v/c1 + v/c

E0

ccos (kx− ωt) .

So the fields in S′ are

~E ′ =

√1− v/c1 + v/c

E0 cos (kx− ωt) y′

~B ′ =

√1− v/c1 + v/c

E0

ccos (kx− ωt) z′.

Applying the inverse Lorentz transformations for x andt, we find that

kx− ωt = k (γx′ + γvt′)− ω(γt′ +

γv

c2x′)

= γk(

1− v

c

)x′ − γω

(1− v

c

)t′

= k

√1− v/c1 + v/c

x′ − ω

√1− v/c1 + v/c

t′.

So we can write the fields in S′ in terms of the coordinatesof S′ as

~E ′(x′, y′, z′, t′) = E′0 cos (k′x′ − ω′t′) y′

~B ′(x′, y′, z′, t′) =E′0c

cos (k′x′ − ω′t′) z′.

where

E′0 = E0

√1− v/c1 + v/c

k′ = k

√1− v/c1 + v/c

ω′ = ω

√1− v/c1 + v/c

.

In S′, the wavelength is

λ′ =2π

k′=

2π

k

√1 + v/c

1− v/c= λ

√1 + v/c

1− v/c.


In S′, the speed of the electromagnetic wave is

ω′

k′=ω√

1−v/c1+v/c

k√

1−v/c1+v/c

=ω

k= c.

So if you did as Einstein imagined and rode alonga lightwave at the speed of light what would you see?As v → c, we see that λ′ → ∞, E′0 → 0, and ω′ → 0.So traveling alongside the light wave, you see it havingspeed c, zero amplitude, zero angular frequency, and in-finite wavelength.

The Field Tensor

~E and ~B are not the spatial components of 4-vectors asone might have expected. Rather, as we’ve seen from thefield transformation equations, components from both aremixed together. The proper relativistic representation ofthe fields is as a rank-4 tensor

Fµν =

F 00 F 01 F 02 F 03

F 10 F 11 F 12 F 13

F 20 F 21 F 22 F 23

F 30 F 31 F 32 F 33

.

Recall that the transformation from S to S′ for a4-vector (aµ → aµ′) is accomplished as

aµ = Λµνaν ,

where Λµν is the Lorentz transformation matrix. To trans-form a second-rank tensor from S to S′, we need twofactors of Λ

Fµν ′ = ΛµλΛνσFλσ.

A rank-2 tensor in 4 dimensions has 16 components.However, as we see from our field transformation equa-tions, we only need six components. This is accomplishedby using an antisymmetric tensor, for which

Fµν = −F νµ,

or in full form:

Fµν =

0 F 01 F 02 F 03

−F 01 0 F 12 F 13

−F 02 −F 12 0 F 23

−F 03 −F 13 −F 23 0

.

We can obtain these components, by going throughall six independent cases of Fµν ′ = ΛµλΛνσF

λσ, and thencomparing the results with the six field transformationequations we derived earlier. For example, to find com-ponent F 01, we would start with

F 01′ = Λ0λΛ1

σFλσ.

The right side is in Einstein summation notation, so weexpand in λ and σ. We start by expanding λ = 0, 1, 2, 3.

F 01′ = Λ00Λ1

σF0σ + Λ0

1Λ1σF

1σ + Λ02Λ1

σF2σ + Λ0

3Λ1σF

3σ.

But, we know that Λ00 = γ, Λ0

1 = −γβ, and Λ02 = Λ0

3 = 0,so

F 01′ = γΛ1σF

0σ − γβΛ1σF

1σ.

Next, we expand σ = 0, 1, 2, 3 to get

F 01′ = γΛ10F

00 + γΛ11F

01 + γΛ12F

02 + γΛ13F

03

−γβΛ10F

10 − γβΛ11F

11 − γβΛ12F

12 − γβΛ13F

13.

Again, we enter the known values from the Lorentz trans-formation matrix Λ, and this simplifies to

F 01′ = −γ2βF 00 + γ2F 01 + γ2β2F 10 − γ2βF 11.

Since we are dealing with an antisymmetric tensor, weknow that F 00 = F 11 = 0 and F 01 = −F 10, so thissimplifies to

F 01′ = γ2F 01 − γ2β2F 01 = γ2(1− β2

)F 01 = F 01.

We now have the relationship between F 01′ and F 01, sowe are done with the first of the six independent compo-nents. We repeat this for the other five components, andthen we will have a set of six transformation equations.We then compare these with the six field transformationequations that we found earlier to relate the componentsof Fµν to the components of ~E and ~B.

The full field tensor is

Fµν =

0 Ex/c Ey/c Ez/c

−Ex/c 0 Bz −By−Ey/c −Bz 0 Bx

−Ez/c By −Bx 0

.

There’s a second version of the field tensor which has dif-ferent components but yields the same transformations.This is called the dual tensor, and it is

Gµν =

0 Bx By Bz

−Bx 0 −Ez/c Ey/c

−By Ez/c 0 −Ex/c−Bz −Ey/c Ex/c 0

.

To go from Fµν to the dual tensor Gµν , just replace Ei/cwith Bi and replace Bi with −Ei/c.


Electrodynamics in Tensor Notation

Suppose we have a small volume of charge Q. In its restframe, it has volume V0, so the proper charge densityis

ρ0 =Q

V0.

Note that charge is an invariant quantity, so Q is the samein all frames.

Now, suppose this charge is moving past you withspeed v. Then in your frame, the volume is Lorentz con-tracted in the direction of motion, and you see a volumeof V = V0/γ. This implies that you measure a chargedensity of

ρ =Q

V= γρ0,

and a current density of

~J = ρ~v = γρ0~v = ρ0~η.

These are the components of the current density 4-vector

Jµ = ρ0ηµ =

(cρ, ~J

).

In nonrelativistic 3D space, the continuity equationis written as

~∇ · ~J = −∂ρ∂t.

We can write the left side more compactly as

~∇ · ~J =∂Jx∂x

+∂Jy∂y

+∂Jz∂z

=

3∑i=0

∂J i

∂xi.

The right side we can write as

∂ρ

∂t=

1

c

∂J0

∂t=∂J0

∂x0.

So we can combine these two to write the continuity equa-tion (in Einstein summation form) as

∂µJµ = 0.

where

∂µ ≡∂

∂xµ.

Note that

∂F 0ν

∂xν=

∂F 00

∂x0+∂F 01

∂x1+∂F 02

∂x2+∂F 03

∂x3

=1

c

[∂Ex∂x

+∂Ey∂y

+∂Ez∂z

]=

1

c~∇ · ~E = µ0J

0 = µ0cρ.

In other words, we get Gauss’s law

~∇ · ~E =ρ

c.

Similarly, if we expand ∂F 1ν

∂xν , we obtain the x-component

of Ampere’s law with Maxwell’s correction. We use ∂F 2ν

∂xν

and ∂F 3ν

∂xν to obtain the y and z components. From ∂G0ν

∂xν ,we obtain the third of Maxwell’s equations, and from∂G1ν

∂xν , ∂G2ν

∂xν , and ∂G3ν

∂xν , we obtain the x, y, and z com-ponents of Faraday’s law. Thus, we can express all fourof Maxwell’s equations in the compact pair of equations

∂νFµν = µ0J

µ, ∂νGµν = 0.

In terms of the field tensor and the proper velocity,the Minkowski force on a charge q is

Kµ = qηνFµν .

This is the Lorentz force law in relativistic notation for ifwe work out the µ = 1, 2, 3 cases, we see that the spatialcomponent of Kµ is

~K = γq(~E + ~v × ~B

)= γ ~F .

which implies that

~F = q(~E + ~v × ~B

).

Relativistic Potentials

We can combine the scalar potential V and the vectorpotential ~A into the 4-vector potential

Aµ =

(V

c, ~A

).

This allows us to write the electromagnetic field tensor interms of the 4-potential as

Fµν = ∂µAν − ∂νAµ,

where

∂µ =∂

∂xµ.

Beware! These derivatives are with respect to the co-variant vectors xµ. Most of the earlier derivatives weencountered were with respect to the contravariant vec-tors xµ. The only difference is that now we attach a mi-nus sign whenever we deal with the temporal componentx0 = −x0.

Remember that Fµν has only six independent com-ponents. If we expand F 01, F 02 and F 03 using this defi-nition, we obtain

~E = −~∇V − ∂ ~A

∂t.

If we expand F 12, F 13 and F 23 using this definition, weobtain

~B = ~∇× ~A.


In the potential formulation, we can write all ofMaxwell’s equations in the single equation

∂ν∂νAµ = −µ0J

µ,

where

∂ν∂ν = 2 =∂

∂xν

∂

∂xν= ∇2 − 1

c2∂2

∂t2,



10.4 Summary: Relativistic Electrodynamics

Suppose there is a frame A moving with speed vABwith respect to a frame B which is moving with speed vBCwith respect to a frame C. What is the speed of frame Awith respect to C? For example, suppose you (frame A)are walking down the aisle of a train (frame B) that ismoving with respect to the ground (frame C). What isyour speed with respect to the ground? Einstein’s veloc-ity addition rule tells us that

vAC =vAB + vBC1 + vABvBC

c2.

Two important implications of special relativity are:

• Moving clocks run slow. This is the phenomenon oftime dilation. For two frames in relative motion,time measured in the two frames is related by

t = γτ,

where τ is the proper time measured in the restframe of the moving clock, and t is the time mea-sured by the observer who sees the moving clock.

• Moving objects are shortened. This is the phe-nomenon of Lorentz contraction. The lenth Lof a moving object is related to its length ` mea-sured in its rest frame by

L = γ`.

Note, a moving object is Lorentz contracted onlyin the dimension that is parallel to its direction ofmotion.

The Lorentz factor is

γ =1√

1− v2

c2

> 1.

Suppose an event occurs at (x, y, z, t) in frame S. If asecond frame S′ is moving with speed v in the x directionrelative S, then the coordinates of that same event in S′

are given by the Lorentz transformations:

x′ = γ(x− vt)y′ = y

z′ = z

t′ = γ(t− vx

c2

).

The inverse Lorentz transformations (i.e. S′ → S) can beobtained simply by switching the primes and changingthe sign of v. We can write the Lorentz transformations

in matrix form asx0 ′

x1 ′

x2 ′

x3 ′

=

γ −γβ 0 0

−γβ γ 0 0

0 0 1 0

0 0 0 1

x0

x1

x2

x3

.

Here, x0 = ct is the time component, and β = vc .

A 4-vector as any set of four components that trans-forms under Lorentz transformations in the same way asthe position 4-vector. A contravariant 4-vector has theindex upstairs

aµ = (a0, a1, a2, a3).

A covariant 4-vector has the index downstairs.

aµ = (a0, a1, a2, a3).

The two are related by a minus sign on the zeroth com-ponent

aµ = (a0, a1, a2, a3) = (−a0, a1, a2, a3).

Whenever you raise or lower the temporal index, youattach a minus sign (a0 = −a0). Whenever you raiseor lower a spatial index, there is no sign change (a1 =a1, a2 = a2, a3 = a3). For a Lorentz transformation alongthe x-axis,

a0 ′ = γ(a0 − βa1)

a1 ′ = γ(a1 − βa0)

a2 ′ = a2

a3 ′ = a3.

The dot product of two 4-vectors aµ and bµ is written as

aµbµ = −a0b0 + a1b2 + a2b2 + a3b3.

The dot product of any two 4-vectors is invariant underLorentz transformations. That is, it is the same in anyinertial frame.

If an event A occurs at aµ = (a0, a1, a2, a3) and eventB occurs at bµ = (b0, b1, b2, b3), then the difference be-tween the two

∆xµ = aµ − bµ,

is the displacement 4-vector. The dot product of the dis-placement 4-vector with itself is called the “interval” be-tween the two events, and is written as

I = −(∆x0)2 + (∆x1)2 + (∆x2)2 + (∆x3)2 = −c2t2 + d2,

where t is the time difference between the two events andd is the spatial separation between the two events.

10.4. SUMMARY: RELATIVISTIC ELECTRODYNAMICS 169

Mechanics

If an object (attached to frame S′) moves with speed vrelative to your frame S, its ordinary velocity is

~v =d~l

dt,

where both the distance d~l and the time dt are measuredin your frame S. Its proper velocity is

~η =d~l

dτ= γ~v,

where the distance d~l is still measured in your frame S,but now the time dτ is the proper time which is mea-sured in the rest frame S′ of the moving object. Theproper velocity ~η is the spatial part of a 4-vector calledthe 4-velocity

ηµ =dxµ

dτ,

which has a zeroth component

η0 =dx0

dτ= c

dt

dτ= γc.

Relativistic momentum is defined as

~p = m~η = γm~v.

This is the spatial part of the energy-momentum 4-vector

pµ = mηµ = (E/c, ~p) ,

whose temporal component is E/c, where

E = γmc2,

is the relativistic energy. For a closed system, the rela-tivistic momentum and relativistic energy are both con-served. For a massless particle like the photon,

E = pc.

When an object is stationary, its total energy is itsrest energy

Erest = mc2.

When in motion, the remainder

Ekin = E −mc2 = mc2(γ − 1),

is the relativistic kinetic energy. Taking the dot productof the energy-momentum 4-vector with itself leads us tothe important relativistic energy-momentum relation

E2 = p2c2 +m2c4,

which allows us to calculate E or p given the other, with-out having to know the velocity.

Newton’s second law

~F =d~p

dt,

holds in relativity provided that we use the relativisticmomentum. Lorentz transforming a force from a frame Sto S′ is messy because, for example, to get F ′x = dpx/dt

′,we have to Lorentz transform both the top dpx and thebottom dt′ into dpx and dt. And we potentially have todo this for all three components of the force.

Work is still the line integral of the force

W =

ˆ~F · d~l,

and the work-energy theorem is still valid in relativity

W =

ˆdE

dtdt = Ef − Ei.

Here, we can use either the total energy or the kineticenergy since the rest energy is constant.

We can define a proper force (also called the“Minkowski force”). It is simply the derivative of therelativistic 4-momentum with respect to proper time

Kµ =dpµ

dτ.

Its spatial part is

~K =dt

dτ

d~p

dt= γ ~F ,

and its zeroth component is

K0 =dp0

dτ=

1

c

dE

dτ.

Electrodynamics

For a frame S′ moving with speed v in the x directionrelative to frame S, it can be derived via moving parallelplate capacitors and a solenoid, that the Lorentz trans-formations for the electric and magnetic fields are

E′x = Ex

E′y = γ (Ey − vBz)E′z = γ (Ez + vBy)

B′x = Bx

B′y = γ(By +

v

c2Ez

)B′z = γ

(Bz −

v

c2Ey

).

In the special case that ~E = 0 in frame S, we can write

~E ′ = ~v × ~B ′.

In the special case that ~B = 0 in frame S, we can write

~B ′ = − 1

c2

(~v × ~E ′

).


Consider an electromagnetic plane wave with angu-lar frequency ω polarized in the y direction and travelingin the x direction through the vacuum.

~E(x, y, z, t) = E0 cos (kx− ωt) y

~B(x, y, z, t) =1

cE0 cos (kx− ωt) z,

where k = ωc . Only Ey and Bz are nonzero, so we only

have to apply those two field transformations. After alsoapplying the inverse Lorentz transformations to obtainkx − ωt in terms of x′ and t′, we can write the fields inS′ in terms of the coordinates of S′ as

~E ′(x′, y′, z′, t′) = E′0 cos (k′x′ − ω′t′) y′

~B ′(x′, y′, z′, t′) =E′0c

cos (k′x′ − ω′t′) z′.

E′0 = E0

√1− v/c1 + v/c

k′ = k

√1− v/c1 + v/c

ω′ = ω

√1− v/c1 + v/c

λ′ =2π

k′=

2π

k

√1 + v/c

1− v/c= λ

√1 + v/c

1− v/c

ω′

k′=

ω√

1−v/c1+v/c

k√

1−v/c1+v/c

=ω

k= c.

So if we did as Einstein imagined and rode with speedv parallel to a lightwave, then as v → c, we would seethat λ′ → ∞, E′0 → 0, and ω′ → 0. If you’re travelinganti-parallel to the lightwave, then all the v would getnegative signs.

To transform from S to S′, use

Fµν ′ = ΛµλΛνσFλσ.

Remember, we are using Einstein summation no notation,so this should be expanded in terms of λ = 0, 1, 2, 3 andσ = 0, 1, 2, 3.

The field tensor is

Fµν =

0 Ex/c Ey/c Ez/c

−Ex/c 0 Bz −By−Ey/c −Bz 0 Bx

−Ez/c By −Bx 0

.To go from Fµν to the dual tensor Gµν , just replace Ei/cwith Bi and replace Bi with −Ei/c.

The current density 4-vector is

Jµ = ρ0ηµ =

(cρ, ~J

),

where the relativistic charge density is

ρ =Q

V= γρ0,

and the relativistic current density is

~J = ρ~v = γρ0~v = ρ0~η.

Here, subscript zero indicates the value in the frame ofthe moving charge distribution.

In relativistic notation, the continuity equation (~∇ ·~J = −∂ρ∂t ) is written as

∂µJµ = 0.

In relativistic notation, Maxwell’s equations can bewritten as

∂νFµν = µ0J

µ, ∂νGµν = 0.

In terms of the field tensor and the proper velocity,the Minkowski force on a charge q is

Kµ = qηνFµν .

This is the Lorentz force law in relativistic notation andhas spatial component

~K = γ ~F , ~F = q(~E + ~v × ~B

).

We can combine the scalar potential V and the vec-tor potential ~A into the 4-vector potential

Aµ =

(V

c, ~A

).

This allows us to write the electromagnetic field tensor interms of the 4-potential as

Fµν = ∂µAν − ∂νAµ.

In the potential formulation, we can write all of Maxwell’sequations in the single equation

∂ν∂νAµ = −µ0J

µ,

where

∂ν∂ν = 2 =∂

∂xν

∂

∂xν= ∇2 − 1

c2∂2

∂t2,


Remember:

∂µ ≡ ∂

∂xµ

∂µ ≡ ∂

∂xµ.

For the second one, the derivatives are with respect to thecovariant vectors xµ instead of the contravariant vectorsxµ. The only difference is the minus sign whenever wedeal with the temporal component x0 = −x0.

Chapter 11

Check Your Understanding

Why is the sky blue? Why are sunsets red?

171

Index

4-vector potential, 166, 1704-velocity, 161

Abraham-Lorentz formula, 154Absorption, 123Absorption coefficient, 125Acceleration field, 153Advanced potentials, 139Advanced time, 139Ampere model, 81Ampere’s law, 74, 77Angle of incidence, 120Angle of reflection, 120Angle of refraction, 120Angular frequency, 113Angular momentum, 108Angular momentum density, 108Atomic polarizability, 55Auxiliary Field, 83Auxiliary field, 83

Biot-Savart law, 70, 71Bound surface charge, 57Bound volume charge, 57Boundary conditions, 20, 79

Capacitance, 27Capacitor, 27Cauchy’s formula, 125Charge conservation, 101Coefficient of dispersion, 125Coefficient of refraction, 125Conductivity, 89Conductor, 25Conductors, 123, 124Conservation Laws, 101Conservation of charge, 101Continuity equation, 70, 101Contravariant, 160Coulomb gauge, 137Coulomb’s law, 5, 7Covariant, 160Curl, 14Curl-less field, 2Current density 4-vector, 166Current loop, 81Cylindrical coordinates, 45

D’Alembertian operator, 167d’Alembertian operator, 138Damping parameter, 155Diamagnetism, 81, 82Dielectric constant, 60Dielectrics, 55Dipole, 7Dipole moment, 48

Dirac delta function, 1Dispersion, 123, 124Divergence theorem, 1, 10, 22Divergence-less field, 2

Elastic collision, 162Electric dipole radiation, 149Electric displacement, 59Electric field, 5Electric potential, 15Electric susceptibility, 59Electrodynamics, 89Electromagnetic induction, 92Electromagnetic Waves, 112Electromotive force, 89, 90Electrostatic pressure, 27Electrostatics, 1Emf, 90Energy, 21, 101Energy density, 117

Faraday cages, 26Faraday’s law, 92, 93Ferromagnetism, 81, 85First uniqueness theorem, 35Flux, 9Free charge, 59Frequency, 113Fundamental theorem for gradi-

ents, 16

Gauge freedom, 136Gauge transformation, 136Gauss’s law, 9, 10Gilbert model, 81Green’s theorem, 1Group velocity, 124Guided waves, 126

Harmonic function, 34Harmonic functions, 34Helmholtz theorem, 2Hidden momentum, 108Hyperbolic motion, 162

Ideal dipole, 48Index of refraction, 118Induced dipole, 55Induced electric field, 93Inductance, 94Induction, 92Intensity, 117Ionization, 55Irrotational field, 2

Jefimenko’s equations, 141

Kinetic energy, 161

Laplace’s equation, 18, 34Larmor formula, 154Law of reflection, 120, 131Law of refraction, 120, 132Legendre polynomial, 43Legendre polynomials, 47Length contraction, 158, 168Lienard-Wiechert potentials, 142Line integral, 2Linear dielectrics, 59Linear Material, 118Linear materials, 84Linear Media, 118Lorentz contraction, 158, 168Lorentz factor, 158Lorentz force law, 67Lorentz gauge, 137

Magnetic dipole, 81Magnetic dipole moment, 81Magnetic dipole radiation, 151Magnetic field, 80Magnetic susceptibility, 84Magnetic vector potential, 77Magnetization, 80–82Magnetostatics, 67Maxwell’s equations, 97Method of Images, 36Minkowski force, 163, 169Momentum, 103Momentum density, 117Momentum flux density, 107motional emf, 90Multipole expansion, 79

Neumann formula, 94Nonlinear material, 85Nonlinear materials, 84

Ohm’s law, 89

Paramagnetism, 81, 82Period, 113Permeability, 84Permeability of free space, 71Permittivity, 60Permittivity of free space, 5Phase, 113Phase constant, 113Phase velocity, 124Physical dipole, 48Poisson’s equation, 18, 34, 77Polarization, 55, 57, 115Polarization angle, 116

172

INDEX 173

Polarization vector, 116Potential difference, 15Poynting vector, 117Principle of superposition, 5Proper charge density, 166Proper velocity, 160

Quadrupole, 48

Radiation, 149Radiation reaction, 154Reflection, 114, 119, 120, 124Reflection coefficient, 119Relative permeability, 84Relative permittivity, 60Relativistic energy, 161Relativistic momentum, 161Relativity, 158Renormalization, 155Resistivity, 89Rest energy, 161Retarded position, 142, 147

Retarded potentials, 138Retarded scalar potential, 138Retarded time, 138, 142, 147Retarded vector potential, 138Rodrigues formula, 43Rotator, 14

Scalar field, 2Self-force, 154Separation of Variables, 38Sinusoidal waves, 113Skin depth, 123Snell’s law, 120, 132Solenoidal field, 2Special relativity, 158Stokes’ theorem, 2Superposition principle, 19Surface current density, 70Susceptibility tensor, 60

TE wave, 127TEM wave, 127

Time constant, 95Time dilation, 158, 168TM wave, 127Transmission, 114, 119, 120Transmission coefficient, 119Transverse wave, 115

Vacuum permeability, 71Vector field, 2Vector fields, 1Vector potential, 77, 79Velocity, 160Velocity field, 153Volume current density, 70

Wave equation, 112Wave number, 113, 117Wave vector, 117Wave velocity, 124Wavelength, 113Work, 21, 162

This handbook grew out of the extensive notes that I took as a physicsundergraduate student.

Documents

A Hostetler Handbook...vergence theorem. Using the divergence theorem on a sphere of radius Rabout the origin gives us V r ~V d˝ = r^ r2 d~a = 2ˇ 0 d˚ ˇ 0 r^ r2 R2 sin r^d = 4ˇ: