Lecture 2 Carl Bromberg - Prof. of Physics

PHY481: Electrostatics

Introductory E&M review (2)

Lecture 2 Carl Bromberg - Prof. of Physics 1

Electric field from point charges The electric field Ep generated by point charges at

point P is the vector sum of Ei from each charge:

Find electric field at the origin due to the three charges q1-3

on corners of a square with side a.

E p =

1

4πε0

qi

ri2

i=1

n

∑ ri

E p =1

4πε0

−q1

a2k −

q2

2a2

j+ k

2

⎛⎝⎜

⎞⎠⎟−

q3

a2j

⎡⎣⎢

⎤⎦⎥

=−1

4πε0a2q1 + 2

4q2( )k + q3 + 2

4q2( ) j⎡⎣ ⎤⎦

Lecture 2 Carl Bromberg - Prof. of Physics 2

Dipole field on the bisector

p = qL = −qLi

– On the bisector, the vertical componentscancel, horizontal components add.

E =2

4πε0

q

r2cosθ i =

1

4πε0

qL

r3i

=−p

4πε0r3

– Far from the dipole

Note minus sign

E =

−p

4πε0 y3 r ≈ y

Field line’s direction is out of +q and into –q– Definition of dipole moment vector

r

–

y

xθ

L

+ p

r

2Ecosθ

cosθ =

L / 2

r

Lecture 2 Carl Bromberg - Prof. of Physics 3

Uniformly charged infinite plane For an infinite horizontal plane the only reasonable direction for

the electric field E is vertical. Electric field can be determined by integrating over the charge

distribution (try it yourself). It is not too surprising that thefield is the same at all distances above the plane.

z

xy

σ

E E =

σ2ε0

k (above)

ΔE =

σε0

The change in the electric field going from below to above E = −

σ2ε0

k (below)

Lecture 2 Carl Bromberg - Prof. of Physics 4

Parallel charge sheets Two infinite sheets of charge are separated by a

constant distance d. One sheet has a charge density +σ and the other a charge density –σ.– Outside, the electric fields point in opposite directions– Between the sheets the electric fields point in the same

direction.

E

−σσ

Eoutside =

σ2ε0

i +σ

2ε0

(− i) = 0

Field between the plates

Uniform electric field E, applies aconstant force on a small particle withcharge q and mass m.

F = qE and a =F

m=

q

mE

Outside plates field is zero

Einside =

σε0

i

Lecture 2 Carl Bromberg - Prof. of Physics 5

Torque on a small electric dipole

Ν = p × E = pE sinθ (−k)

An electric dipole p in a uniform electric fieldE experiences a net torque Ν and no net force.– Choose coordinates where p and E lie in the x/y

plane. p and E have an angle θ between them.

In addition to a torque, an electric field E witha divergence will generate, a net force F on anelectric dipole, p :

F = pi

∂E j

∂xi

e jGeneral expression needsoperators to be covered later

Cartesiancoordinates

+E p θ

qE

–qEx

−

y

Lecture 2 Carl Bromberg - Prof. of Physics 6

Energy of dipole in electric field

U = −p ⋅E = − pE cosθ

Potential energy U of the electric dipole p in uniform electricfield E:

+E p θ

qE

–qEx

−

y

Lecture 2 Carl Bromberg - Prof. of Physics 7

Gauss’s Law Electric field passing through a closed (mathematical) surface

– A surface enclosing NO net charge has a zero net field leaving orentering the surface.

– A surface enclosing a positive (negative) charge has a net fieldleaving (entering) the surface proportional to the enclosed charge.

E ⋅

S∫ dA =qencl

ε0

– For symmetric charge distributions, pick an enclosing surface whereE and dA are everywhere parallel to each other.

General expressionfor Gauss’s Law

dA

n

closed surface

Lecture 2 Carl Bromberg - Prof. of Physics 8

Coulomb’s Law <---> Gauss’s Law For symmetric charge distributions, pick enclosing surfaces, so

that E and dA are are parallel to each other.– For a point charge at the origin, use a spherical surface, radius R,

centered on the charge (makes direction of normal = radial)

E ⋅S∫ dA =

q

4πε0R2R2 sinθdθ

0

π∫ dφ

0

2π∫ =

q

ε0

This is a “proof” that Gauss’s law follows directly from the CoulombForce Law for point charges, and their derived electric fields.

dA = R2 sinθdθdφ n = r

q n = r

E =

q

4πε0R2r

Electric field at surface

Evaluate Gauss’sIntegral

Lecture 2 Carl Bromberg - Prof. of Physics 9

Field of a line of charge - use Gauss’s Law Consider an infinitely long line of charge with linear

charge density λ , and a cylindrical gaussian surface.– The electric field is parallel to the surface at the top

and bottom of the cylinder, E•dA is zero.– The electric field is perpendicular to the surface and

therefore parallel to the surface normal.

E ⋅S∫ dA = Er dφ

0

2π∫ dz

0

L

∫

E2πrL =qencl

ε0

=λL

ε0

E =λ

2πε0r; E =

λ2πε0r

r

qencl = λL

Lecture 2 Carl Bromberg - Prof. of Physics 10

Field of a charged spherical shell - Gauss’s Law Consider a radius R spherical shell with surface charge density σ.

– A spherical Gaussian surface with radius r < R, is inside the chargesurface, and encloses no charge -> Einside = 0.

– A spherical Gaussian surface with radius r > R, has the electric fieldnormal to its surface.

E ⋅S∫ dA = E r2 sinθdθ

0

π∫ dφ

0

2π∫

E4πr2 =q

ε0

E =q

4πε0r2 ; E =

q

4πε0r2r

Same electric field ascharge q at the origin

q = σ4πR2

Lecture 2 Carl Bromberg - Prof. of Physics 11

Uniform charge density sphere - Gauss’s Law Find the electric field inside of a sphere, radius R, with a

uniform charge density ρ throughout the volume.– Pick a spherical Gaussian surface, radius r,

inside the charged sphere

E ⋅S∫ dA = Er2 sinθdθ

0

π∫ dφ

0

2π∫

E4πr2 =qencl

ε0

=Qr3

ε0R3

E =Qr

4πε0R3 ; E =

Qr

4πε0R3r

ρ = Q / 43πR3( )

qencl = ρ 43πr3( ) = Q

r3

R3

– Electric field outside of a sphere

E =

Q

4πε0r2r Same electric field as

charge Q at the origin

Lecture 2 Carl Bromberg - Prof. of Physics 12

Infinite sheet (again) - Gauss’s Law Infinite sheet of charge with surface density σ.

– Pick a cylindrical Gaussian surface, radius r, passing throughthe sheet.

– The dot product E•dA is non zero only on TWO the ends.

E ⋅S∫ dA = 2E rdr

0

R

∫ dφ0

2π∫

2(EπR2 ) =qencl

ε0

=σ πR2( )

ε0

E =σ

2ε0

; E = ±σ

2ε0

k

+ above– below

qencl = σ πR2( )