1

Electric PotentialElectric Potential Energy

Electric Potential

Equipotential Surfaces

(≠ Electric Potential Energy)

The Computation of Electric Potential

Potential Gradient – Connection to Electric Field

Energy methods applied

to Electricity

2

Overview: A square of relationships

The electric field as a force per unit charge: Also recall work:0q 0

0

limF

Eq

W cosF ds F ds In this chapter we introduce the notions of electrical potential energy and electric potential. Their relationship to

force and electric field and each other are symbolically represented by this square:

F

E

†q × q

U

V

†q × q[F] = Newton

[U] = N·m = Joule

[E] = N/C [V] = J/C = Volt (V)

dx

/ x

dx

/ x

3

Electric Potential Energy (EPE)

In particle mechanics (EnPh131) you learned the power and utility of work-energy methods: they often replace

complicated integrations with simple algebra. We now turn to applying these methods to electrical problems.

W cosb b

a ba a

F dl F dl Basic defn: Work done by a force in moving from „a‟ to „b‟:

To evaluate this line integral in general, an explicit parametrization of the path is required: difficult in general.

Conservative force (Equivalent characterizations): The work done by the force is:

1) path independent

2) expressible in terms of a potential energy function that depends only on position

3) zero around any closed path

We will show that electrostatic forces are conservative, in the same way that gravitational forces are.

( ) ( )consW U U a U b

4

Electric Potential Energy (EPE)

Why is the notion of potential energy useful? Recall the work-energy theorem: totalW K

2 2

2 2

b

a

b b vb a

total neta a v

mv mvdv dsW F ds m ds m dv mv dv K

dt dt

One-line super-proof of W.E.T.:

If only conservative forces act (no friction, air resistance, etc.), then total consW W

But conservative forces are those that can be characterized by a potential energy: consW U

Thus, if only conservative forces act:

a a b b

a b

K U

K U K U

E E

Conservation of Energy

It remains to show that electrostatic forces (i.e. Coulomb‟s law) is conservative...

5

Electric Potential Energy (EPE): Uniform Electric FieldsWe saw in the last chapters that although the basic electrostatics law is inverse-square (i.e. a variable force), that

UNIFORM (i.e. constant) E-field configurations exist: the large parallel charged plates configuration in particular.

Keep the signs straight! (Remember there are two kinds of charge, versus one kind of mass.)

Why is the line integral defining the work path independent?

1) Because the electrostatic force is constant in magnitude, it‟s magnitude

comes out of the integral.

2) Because the electrostatic force is constant in direction (perpendicular

to the plates), only the y-component of the path contributes, and so in

conjunction with reason 1, the line integral reduces to the sum of the changes

in y-coordinate.

Mathematically:

0 0 0

b b b

es esa a a

W F dl q E dl q E dy q E y

E-field directed down.

0

0

es

es

W q E y

U q Ey

6

(Work done

= Area under graph)gravF mg const esF qE const

EPE of a Uniform Electric Field: Comparision with Gravity

Gravitational (uniform field)

UA = mghA

UB = mghB

+++++++++++++++++++++

-------------------------------------

Electric (uniform field)

UA = qEhA

UB = qEhB

d

F

7

YF23.53 (partial and modified) A particle with charge +7.60 nC, and mass 3.00×10-5 kg is in a uniform electric

field directed to the left. Another force, in addition to the electric force, acts on the particle so that when it is

released from rest, it moves to the right. After it has moved 8.00 cm, the additional force has done 6.50×10-5 J of

work and the particle is moving at 1.70 m/s.

(a) What work was done by the electric force?

(c) What is the magnitude of the electric field?

8

0

0, 2

0

1ˆ

4es q onq

qqF r

r

0 0, 2

0 0

1 1

4 4

b

aa b

r

es a br

qq qqdxW

rx r

ˆl xi

Electrical Potential Energy: Two point-charges

More generally now, since the electrostatic force is a variable force, to compute the work

done by it we must integrate the inverse square law:

ra Start with Coulomb‟s law:

qx+

q0

rb

Along the straight path , the work done by the es force is:

Note for qq0>0 this is repulsive, for qq0<0 this is attractive.

Note the 1/r dependence. You

must not confuse force (1/r2) with

energy (1/r).

9

Electrical Potential Energy (General)

Our computation considered a straight radial path. But since the es

force between the two point charges is itself radial, any nonradial

component in a path element does not contribute to the work.

0 0, 2 2

0 0

1cos cos ...

4 4

b

a

b b r

es a ba a r

qq qq drW F dl dl

r r

cos dl dr

Note that for electrostatic forces:

Key geometrical realization:

Thus:

cf. Faraday’s law in

Chapter 29

i.e. same computation along any path from a to b.

for any path element.

0C es ldF

closed path

10

Electrical Potential Energy (General)

Thus it makes sense to speak of the electrostatic potential

energy between two point charges:

0

0

1 1

4es

b a

qqU W

r r

0

0

1( )

4es

qqU r

r

where this definition assumes the choice U(∞)=0.

Recall only DIFFERENCES in potential energy have meaning.

There are configurations (involving charge distributions extending to infinity)

for which this choice is NOT possible. See later examples.

U represents the work done BY the field to move the test charge q0 to infinity from a distance r away from q. As

always, U is a shared property of the system q and q0 (unlike the potential V which we‟re about to discuss.)

This result also holds for replacements of q by any spherically symmetric charge distribution (by Gauss‟s law!)...

11

Electrical Potential Energy (General) What about a system of point charges?

If a system of charges q1, q2, ... is already assembled, the total E-field acting on a

point charge q0 is the vector sum of the fields due to individual charges, so

that total work done on q0 through any displacement is the algebraic sum

of the contributions from qi. Thus:0 01 2

0 1 2 0

...4 4

i

i i

q q qq qU

r r r

However, work is also required to assemble the charges qi in the first place.

If they are all infinitely far apart initially, the potential energy associated with

the assembly of the charges is

0

1

4

i j

system

i j ij

q qU

r

where the sum extends over all pairs of charges; we don‟t count self-interactions (i≠j), and we don‟t double count.

Why? Hint: Apply (*) iteratively by bringing

in one charge at a time from infinity.

(*)

12

YF23.5 A small metal sphere carrying a net charge of q1= – 2.80 μC is held

stationary by insulating supports. A second small metal sphere, with a net

charge of q2= –7.80 μC and mass 1.50 g, is projected towards q1. When the

two spheres are 0.800 m apart, q2 is moving toward q1 with speed 22.0 m/s.

Treat the two spheres as point charges and ignore gravity. (a) What is the

speed of q2 when they are 0.400 m apart? (b) How close does q2 get to q1?

(Application question: Conservation of Energy)

YF23.10 Four electrons are located at the corners of a square of 10.0 nm on a side, with an alpha particle

(charge +2e) at its midpoint. How much work is need to move the alpha particle to the midpoint of one of

the sides of the square. (W2010, system example)

Solution:

We have to assume the electrons are held fixed. This is a 5-body system so there are 4(5)/2 = 10 terms in the

potential energy of the system. But since the work done by an external force is equal to the change in the potential

energy, and the electrons are in fixed positions, the 3(4)/2 = 6 terms involving all pairs of electrons cancels. Then:

10

04

14

r

eqUU syse

3020

)(

4

12

)(

4

12

r

eq

r

eqUU sysef

where 2/)4/()4/( 22

1 LLLr

where 2/54/,2/ 22

32 LLLrLr

Thus:J1008.6

844

4)( 21

1320

2

0..

rrr

eUUUWW fsebyext

13

YF23.71/SJ25.62 (Self-Energy of a Sphere of Charge)

A solid sphere of radius R contains a total charge Q distributed uniformly throughout its volume. Find

the energy needed to assemble this charge by bringing infinitesimal charges from far away. This energy is

called the self-energy of the charge distribution. (Hint: After you have assembled a charge q in a sphere of radius

r, how much energy would it take to add a spherical shell of thickness dr having charge dq? Then integrate

to get the total energy.)

This computation is used in nuclear physics to estimate the electrostatic energy contained an assembly of protons

confined to a nucleus, and shows how large nuclear forces must be to accomplish this confinement.

Notice this result implies that it takes an infinite amount of energy to assemble a point charge! There is

something seriously nontrivial occurring here: either „point charges‟ are „impossible‟, or classical electrostatics

must break down at small distance scales. The latter approach leads to the study of Quantum Electodynamics

(QED). The former is suggested by string theory.

– Continuous/Integral example

14

Electric Potential

We introduced the electric field as a force per unit charge at a point P.0 0

0

limq

FE

q

We can similarly define the ELECTRIC POTENTIAL as potential energy per unit charge:0 0

0

limq

UV

q

[V] = Joule/Coulomb = Volt

In terms of work:0 0

( ) ab bb

aba

a

W UV V V

q qV V

Again, only potential differences (aka voltages) have physical meaning.

=“potential of point a with respect to point b”

Two interpretations of Vab:

1) the work done BY the electric force when a UNIT charge moves from a to b

2) the work done AGAINST the electric force to move a UNIT charge slowly from b to a

Like energy, potential is scalar

15

Electric Potential (Point charges)

0 0

1

4pt

U qV

q r r = distance from the point charge q to

the point at which the potential is evaluated.

0 0

1

4

i

i i

qUV

q r

potential due to a collection of point

charges. Note: this is a scalar sum.

Positive charges accelerate spontaneously from regions of high

electric potentials to regions of low electric potential.

Negative charges accelerate spontaneously from regions of low

electric potentials to regions of high electric potential.

In either case, potential energy spontaneously decreaes, as it is converted

to kinetic energy. ALWAYS THINK CAREFULLY about your signs.

+ +

+

V

V–

U

U

16

Electric Potential (General)

If the charge distribution is continuous then as usual the sum is replaced by an integral over charge elements:

0

1

4

dqV

r r = distance from charge element dq to the field point where we compute V.

Notes:

1) this definition implies V=0 at points infinitely far away from all charges. If the charge distribution itself

extends to infinity, then this definition cannot be used, and we must use a different reference potential.

2) Remember the potential at a point is simply the potential energy a unit charge would have if placed there. Like

the electric field, the potential exists even when there are no charges placed at that point.

3) This scalar integral is much easier to compute than the vector integration we used to compute the E-field.

4) The electron volt = the energy acquired by an electron as it is accelerated through a potential difference of one

volt. The term is a misnomer: it‟s a unit of energy not a unit of potential difference (or voltage).

1 eV = 1.602×10-19 J

17

Computation of Electric Potential

If the collection of charges/charge distribution are given, then is the most straightforward

way to obtain V. Implicitly, this expression assumes V(∞) = 0, and can thus only be used for distributions that do

not extend to infinity (see line charge example).

0

1

4

dqV

r

However IF the electric field is already known, then it is often easier to compute V from a knowledge of the E-field:

0

b b

a ba a

W F dl q E dl (dividing by q0)

cosb b

ab a ba a

V V V E dl E dl along any path between

a and b.

Units warning: [E] = N/C = 1 V/m

This also gives an independent check of calculations. Be VERY careful with the signs, and the ordering

Va – Vb, not the other way around. The difference initial – final is arising because of the minus sign in Wcons= – ΔU

Point a is 1.5 V

higher in

potential than b

18

YF23.21 Two point charges q1 = +2.40 nC and q2 = –6.50 nC are held 0.100 m apart.

Point A is midway between them; point B is 0.08 m from q1 and 0.60 m from q2. Take

the electric potential to be zero at infinity. Find (a) the potential at A, (b) the potential at

B, (c) the work done by the electric field on a charge of 2.50 nC that travels from B to A.

(Note: A charge „released‟ at B would not travel spontaneously directly to A.

An external force would also have to be present to ensure it arrives at A.)

Discrete charge distribution

19

(YF23.79/SJ25.7eg) Continuous Distribution (no symmetry)

Electric charge is distributed uniformly along a thin rod of length a, with total charge Q.

Take the potential to be zero at infinity. Find the potential at the following points:

(a) point P a distance x to the right of the rod, and

(b) point R, a distance y above the right-hand end of the rod.

(c) In (a) & (b) what does your result reduce to as x or y becomes much larger than a?

Note: You could first compute the

E-field at P and R, and then

integrate along a path starting

at P or R to infinity, but here this

is much harder, since the former

is a vector calculation, and there is

no symmetry to assist you.

20

(Spherical Symmetry) YF Example 23.6 – Point Charge

Starting with the expression for the E-field about a point charge, determine the

potential (taking V=0 at infinity) at a distance r from a point charge q.

(Spherical Symmetry) YF Example 23.8 – Charged Conducting Sphere

Inside the sphere the E-field is zero (since it‟s a conductor) but that only means that

V = const, not necessarily that V=0. In fact since E is roughly speaking the „derivative‟

of V‟ (technically the gradient as we‟ll see shortly), and since E exists everywhere,

V must be continuous. Thus V inside must have the same value as on the surface.

Outside the sphere, we‟re guaranteed the same result as if the sphere was

replaced with a point charge.

21

Spherical Symmetry (YF23.78/22.42) Pitfall alert – “Potential Matching”

Consider a solid conducting sphere (radius a, charge +q) inside a hollow conducting sphere (inner radius b, outer

radius c, no net charge). Take V=0 as r→∞. (22.42) Calculate the E-field on each region r<a, a<r<b, b<r<c, r>c.

Use the E-field to calculate the potential V at the following values of r:

(a) r=c (b) r=b (c) r=a (d) r=0.

These are implicitly all potentials relative to infinity.

Spherical Symmetry (YF23.72) Consider the solid, uniformly charged insulating sphere of radius R we studied last

chapter From the E-field expression obtained,(a) determine V as a function of r both inside and outside the sphere,

assuming V=0 at infinity. (b) Graph V and E as fcts of r.

Problem-solving tip: Set these

problems up so that you work

inward/outward from your reference

potential, and demand continuity at

each interface.

Time-permitting

22

Cylindrical Symmetry (YF Example 23.10) – Pitfall Alert

1You might note that many of the problems with cylindrical symmetry are specified with

a wire thickness. This gives you a natural place to define a reference potential, and

cuts off the divergence associated with charge distributions running to infinity.

Find the potential at a distance r from a very long line of charge with

a uniform linear charge density λ, using the E-field expression from Ch 21/22:

Choose the reference potential V(R)=0 where R is the radius of the line charge1 .

Here we cannot take the reference potential to be V=0 at infinity, because the charge

distribution itself runs to infinity, and the integral is logarithmically divergent.dqV

r

i.e.2 2

( )dx

V rr x

is ill-defined.

0

1ˆ( )

2E r r

r

23

YF23.61 – partial (Coaxial Cylinders) A long metal cylinder with radius a is supported on an insulating stand on

the axis of a long, hollow metal tube with radius b. The positive charge per unit length on the inner cylinder is λ,

and there is an equal negative charge per unit length on the other cylinder.

(a) Calculate the potential V(R) for (i) r < a, (ii) a < r < b, (iii) r > b. Take V=0 at r = b.

(b) Show that the potential of the inner cylinder with respect to the outer is

0

ln2

ab

bV

a

24

Planar symmetry – Parallel plate configuration (YF Example 23.9)

Find the potential (difference) at any height y between two oppositely charged plates.

Ans: ( ) b abV y V Ey V Ed

(easy way to measure σ.)

0 abV

d

Uniform Ring of Charge (YF Example 23.11 & YF23.69)

Find a potential at a point P on the ring axis at a distance x from the center. (Take V(∞)=0)

Warning: Only for this geometry is there a linear relation between E and V.

Method 1:

Method 2:

(direct integration)

(use E-field from ch21)2 2 3/2 2 2

0 0

1( )

4 ( ) 4x x

Q x QV x E dl dx

x a x a

2

2 2 2 200 0 0

1 1 1( )

4 4 4

dq a QV x d

r x a x a

/ (2 )Q a

2 2u x a

25

Phys 230 Fall 2009 Midterm (Short Answer 6 & 7) – Application of ring result (E-cons) + Superposition principle (Motivated by YF23.34)

A charge of –450 μC is uniformly distributed on an insulating ring of 2.40 m radius. A point charge of +360 μC is placed at

the center of the ring. Points A and B are located on the axis of the ring (taken to be the x-axis) as indicated, and are

respectively 1.80 m and 3.20 m from the center of the ring. Q ring= – 450 μC

Qpoint = + 360 μC

A B1.8 m 1.4 m+

6. (3 marks) If the potential is taken to be zero at infinity, what point(s) on the finite positive x-axis lie on the V=0

equipotential surface for this charge distribution?

7. (4 marks) If a point charge of mass 8.99×10-4 kg and charge +40.0 μC is released from rest at point A (while the ring and

point charge at the origin are held fixed), how fast is it going when it reaches point B?

(Right now this wording, i.e. “equipotential surface”, is a fancy way of asking

where is V=0 on the x-axis. See next section…)

(Must use energy conservation.)

26

Equipotential Surfaces

Topographical maps such as this one

near Jasper, Alberta indicate „contours‟

of constant elevation. Near the surface

of the earth, these are therefore also

„contours‟ of constant gravitational

potential energy, i.e. equipotentials.

Where the contours are close together,

(such as near the summit of Whistlers

the rate of change of elevation is large

(i.e. steep), in directions perpendicular

to the contours. Where they‟re far apart,

such as near the Jasper townsite itself,

the ground is more level.

more contours/horizontal displacement = „steep‟

less contours/horizontal displacement = „level‟

In exact analogy we can talk about

electrical equipotential surfaces, usually

represented in two dimensions as the

cross-sections of those surfaces.

On an equipotential surface the voltage

is constant. This does not imply the E-

field on the surface is constant.

Claim: E-field lines and equipotential surfaces are mutually perpendicular.

Reasoning: By defn. the potential energy is constant on the surface. Thus the E-field can do no work on a charge

moving along the surface. Thus the E-field cannot have a component parallel to the surface.

Note: Field lines are lines, equipotentials are surfaces. If the lines are „curved‟ so are the surfaces. In the special

case of a uniform field (whose field lines are straight), the equipotential surfaces are parallel planes.

27

Equipotential Surfaces – 3D Visualization

We generally only

draw 2D

cross-sections

of the equipotential

surfaces....

28

Equipotentials and Conductors Under electrostatic conditions:

1) The E-field just outside a conductor must be perpendicular to the surface everywhere.

2) The surface of a conductor is always an equipotential surface.

Don‟t confuse Gaussian surfaces (arbitrarily chosen) with

equipotential surfaces (fixed by the charge distribution).

If the E-field had a parallel component, then violating conservative nature

of electroSTATIC fields.

3) If a conductor contains a cavity, and no charge is present inside the cavity, then

there is no net charge anywhere on the surface of the cavity.1 (Ch22 claim.)

1This is the basis of electostatic shielding: you can safely touch the

inside walls of a charged conducting box without being shocked.

Key: First prove by contradiction that every point in the cavity is at the same

potential as the inside surface of the conductor.

0ldE

29

Potential Gradient: Calculation of Electric field from Potential

We‟ve hinted several times that essentially the „derivative‟ of the potential is the E-field. It‟s more complicated than

that because the potential is a scalar function, but the E-field is a vector function (i.e. 3 scalar functions)1 .

1You should be asking yourself how can the potential contain so much

information? Ans: In electrostatics, not all components of the E-field are truly

independent! The components are constrained to satisfy Gauss‟s law which

in differential form is

On one hand:b

aba

V E dl On the other:a b

ab a bb a

V V V dV dV

Since a and b are arbitrary, we must have: x zydV E dl d Ex dy dzEE

From multivariable calculus:V

dV dxVV

xdy

zdz

y

(total/exact „differential‟)

Thus: , ,x y z

V V VE E E

x y z

(in Cartesian coordinates)

E V

Warning: If your expression for V was derived by

holding y and z fixed (i.e. it wasn‟t completely general),

for example, then you cannot calculate Ey or Ez from it.

Often in problems with spherical or cylindrical

symmetry, we can calculate the radial component

of the electric field from:r

VE

r

[E] = 1 N/C or V/m

0/ E

30

30

Conceptual Example

(The negative sign in means the electric field points in the direction in which V decreases most rapidly.)E V

31

SJ25.67 When an uncharged conducting sphere of radius a is placed at the origin of of an xyz coord system that

lies in an initially uniform electric field , the resulting electric potential is V(x,y,z) = V0 for points inside the

sphere and 3

00 0 2 2 2 3/2

( , , )( )

E a zV x y z V E z

x y z

for points outside the sphere, where V0 is the constant electric potential on the conductor. Use this equation to

determine the x,y, and z components of the resulting electric field.

0ˆE E k

Supplementary This is an exercise in computing partial derivatives.

32

Phys 230 Fall 2009 Midterm (Long Answer #1, parts a & d) – continued from Chapter 21

b) A non-uniformly charged insulated disk of radius R lying in the yz plane with its center at the origin at x=0, has a

surface charge density given by:( )

Cr

r

c) Use the E-field due to a ring and b) to determine the E-field due to this charged disk at any point on the

positive x-axis. Use the substitution r = x tan θ. (Symmetry dictates the E-field will still only have an x-component

on the x-axis.)

where r is the distance from the center of the disk to a point in the yz plane.

220

1

4 Rxx

QEx

Result from c):

d) Find the potential V(x) relative to infinity due to this charged disk at any point on the positive x-axis.

Use either c) and the substitution x = R tan θ, or the result stated in a) and use the substitution r = x tan θ. [4]

8. a) The potential along the x-axis of a uniformly charged ring of radius r lying in yz plane with its center at the origin

x=0 and whose axis is the x-axis (as in question 6/7), relative to infinity, is given by

2204

1)(

xr

qxV

(see slide 24)

Use this to determine the x-component of the E-field due to the ring on the positive x-axis. [3]