PHYS685 quiz2 2019 answer key

PHYS 685 Classical Electrodynamics Exam 2 (Electrostatics) Answer Key Due within 24 hours

1) You are expected to work on the problems independently.

2) You may refer to the Jackson textbook as well as lecture notes distributed/taken in class, from which any conclusions, formulas, theorems and math identities can be directly used.

3) You are not supposed to search the internet for answers to the problems. However, you can use online integration tables for your work.

4) You are only required to complete any three of the following five problems. If you complete a forth problem in addition, that one will be considered as an extra credit problem. Indicate which problems you pick to solve. Each problem is worth 10 points.

5) Your exam score counts for 15% towards your final grade.

Q1. In a spherical coordinate system, the axis of a charged ring has the direction θ' ̂θ+ϕ' ̂ϕ , and the center of the ring O' is located at (d ,θ' ,ϕ') . The radius of the ring is a , and charge q is uniformly distributed on the ring. Write the scalar potential Φ(r ,θ,ϕ) using spherical harmonics expansion.

Solution:

The addition theorem of spherical harmonics has:

Pl(cosγ)= 4π 2l+1m

∑

l

Ylm* (θ' ,ϕ')Ylm(θ,ϕ) where the angle γ is the inclusive angle between two

vectors (r ,θ,ϕ) and (r ' ,θ' ,ϕ') .

Based on the solution of the electrical potential of a charged ring on Jackson page 104, when the axis of the ring is along the z-axis:

Φ(r ,θ)= q 4πϵ0

∑

∞ _r

<

l

r>

l+1Pl(cosα)Pl(cosθ)

The definitions of r<

l _, r>

l _and

α can be found on Jackson pg. 104).

a

Φ(r ,θ,ϕ)

θ' ̂θ+ϕ' ̂ϕ

If the axis of the ring is tilted to θ' ̂θ+ϕ' ̂ϕ , we can write:

Φ(r ,β)= q 4πϵ0

∑

∞ _r

<

l

r>l+1

P_l(cosα)P_l(cosβ)

where β is the angle between (r ,θ,ϕ) and (r ' ,θ' ,ϕ') .

The potential at (r ,θ,ϕ) , Φ(r ,θ,ϕ)≡Φ (r ,β)

Therefore,

Φ(r ,θ,ϕ)= q 4π ϵ0

∑

∞

∑

l 4π 2l+1Ylm

*

(θ' ,ϕ')Y_lm(θ,ϕ) r<

l

r>

l+1Pl(cosα) or

Φ(r ,θ,ϕ)=_ϵq₀

∑

∞

∑

1 2l+1Ylm

*

(θ' ,ϕ')Y_lm(θ,ϕ) r<

l

r>l+1

P_l(cosα)

where r< ( r> ) is the smaller (larger) of r and

√

+a2

Q2. An infinitely long wire carrying a line charge density λ is placed between two parallel, infinite, grounded conducting planes with separation a . The charged wire is parallel to the planes, and is at a distance b from one of the conducting planes. Express the potential anywhere in terms of a , b , and λ (Hint: use the method of separation of variables).

Solution:

Set up a Cartesian coordinate system as follows:

Based on the symmetry, the potential is independent of z . We thus simply this into a 2D problem. Except at (0,b) , the Laplace equation is satisfied within the two parallel planes.

∇2

Φ(x , y)=∂

2_{Φ(x , y)}

∂x2 +

∂2_{Φ(x , y)} ∂y2 =0

Using the method of separation of variables, let us write

Φ(x , y)=X(x)Y(y)

Pluging this into the Laplace equation, we have

1 X

d2 X d x2 +

1 Y

d2Y d y2=0

Considering the boundary condition, it is necessary to write:

λ

a

b

λ y

1 X

d2X d x2 =k

2 _and 1 Y

d2Y d y2=−k

2

The solutions to the two ODEs are: X=A1ekx+A2e−kx

Y=B1sin(ky)+B2cos(ky)

A1, A2, B1, B2 are determined by the boundary conditions.

For the y direction, because the potential is zero for y=0 and y=a , we thus derive

B2=0 and k=nπ a

For the x direction, since there is charge at x=0 , let's divide the potential field into two regions for

x>0 and x<0 , respectively.

For x>0 , since Φ=0 , we write X=A1e−

kx

and for x<0 , we write X=A2e

kx

The potential becomes:

ΦI(x , y)=

∑

∞

C_nsinnπy a e

−n_aπx

for x>0

ΦII(x , y)=

∑

∞

D_nsinnπy a e

nπ

a x _{for x<0}

We have two more conditions (1) the potential is everywhere continuous; and (2) the gradient of the potential around the charge at x=0,y=b , is specified by the charge density.

To satisfy potential continuity at x=0 Cn=Dn

The potential thus has the following form:

ΦI(x , y)=

∑

∞

C_nsinnπy a e

−n_aπx

for x>0

ΦII(x , y)=

∑

∞

C_nsinnπy a e

nπ

a x _{for x} <0

On the plane x=0 , let's suppose the charge density is σ , the boundary condition becomes: ∂ΦI

∂x −

∂ ΦII

∂x =−σϵ0=−δ (y−b) λϵ0

∂ΦI

∂x ∣x=0=

∑

n=1

∞

−nπ

a Cnsinnπ_ay ∂ΦII

∂x ∣x=0=

∑

∞ _n

π a Cnsin

nπy a

Hence,

∑

∞ _2nπ

a Cnsin nπy

a =δ(y−b) λϵ0

Multiply both side by sin nπy

a and integrate over y from 0to a :

∫

0

a 2nπ

a Cnsinnπ_aysinnπ_ayd y=

∫

0

a

δ(y−b) λ_ϵ 0sin

nπy a d y

LHS:

∫

0

a 2nπ

a Cnsin nπy

a sin nπy

a d y= 2nπ

a Cn a

2=nπCn

RHS:

∫

0

a

δ (y−b) λ_ϵ 0sin

nπy

a d y= λϵ0sin n bπ

a

Cn= λ_n πϵ0

sin nπb a

The potential is thus:

ΦI(x , y)= λπϵ₀

∑

∞ ₁

nsin n bπ

a sin nπy

a e

−n_aπx

for x>0

ΦII(x , y)= λπϵ₀

∑

∞ ₁

nsin n bπ

a sin nπy

a e nπ

a x _{for x<0}

Or:

Φ(x , y)= λ_{π ϵ} 0

∑

n=1

∞ ₁

nsin n bπ

a sin nπy

a e

−n_aπ∣x∣

Q3. An infinitely long cylindrical conductor with radius a is placed in a uniform electrical field E⃗ . The axis of the cylinder is perpendicular to E⃗ . Calculate the induced surface charge density σ of the cylinder.

Solution:

Due to symmetry, the electrical potential is independent of z. Using cylindrical coordinates, the Laplace equation can be written as:

∇2

Φ(r ,ϕ)= ∂2Φ ∂r2+

1 r ∂ Φ∂r+

1 r2

∂2 Φ

∂ ϕ2=0 (see eqn. 3.71, Jackson pg. 111)

Use separation of variables, rewrite Φ=R(r)Q(ϕ)

We derive:

d2Q dϕ2+m

2_Q

=0 and

d2R d r2+

1 r

d R d r −

m2

r2 R=0 (The solution to this equation can be obtained by making r=e

u ₎

The general solution to the Laplace equation is:

Φ(r ,ϕ)=

∑

∞

(Amsinmϕ+Bmcosmϕ)(cmr m

+Dmr− m

)

For r→0 , because the ∂Φ

∂(rcosθ)=−E

The potential must satisfy: Φ(r ,θ)∣r→∞=constant+E rcosθ , therefore we conclude:

(1) A_m=0

(2) Bm=0 for m≥1

(r ,ϕ) ⃗

The potential has the following reduced form:

Φ(r ,ϕ)=B₀+(C₁r+D1

r )cosϕ , where C1=−E

On the other hand, on the conducting surface of the cylinder ( r=a ), the potential is constant (i.e., the potential is independent of ϕ ), we derive

C₁a+D1

a =0 , that is D1=E a 2

The potential outside the cylinder is thus:

Φ(r ,ϕ)=B0−E rcosϕ+

E a2_cos ϕ r

The electrical field of the potential can be written as:

⃗

E=−∇ Φ(r ,ϕ)=−∂Φ ∂r r̂−

1 r ∂Φ∂ ϕ ̂ϕ

σ=ϵ0Er∣r=a=−ϵ0

(

∂r

)

r2)Ecosϕ∣r=a

Q4. A solid conducting hemisphere with radius R sits on the ground due to gravity. The density of the hemisphere is ρ . Now, in order to lift the hemisphere from the ground (i.e., the surface of the earth, which is conducting), let's apply a uniform electric field E⃗₀ whose direction is perpendicular to the ground. How big is E0 so that the hemisphere can be lifted up? (The acceleration of gravity is

g )

Solution (see also Homework #5, Q3):

The Laplace equation is satisfied outside the hemisphere.

Taking O as the origin of the coordinate system, and the z axis perpendicular to the plane. It can be seen that the electrical potential Φ is only a function of (r ,θ)

The solution to the Laplace equation is:

Φ(r ,θ)=

∑

∞

(a_nrn + bn

rn+1)Pn(cosθ)

when r→∞ , Φ corresponds to the uniform electronic field E , hence an=0 for n≥2 .

Φ(r ,θ)=a₀−E₀rcosθ+

∑

∞ _b

n

rnPn(cosθ)

On the surface of the hemisphere, r=R ,

ϕ(R ,θ)=a0−E0Rcosθ+

∑

n=0 bn

Rn+1 Pn(cosθ)=V

from which we derive for n≥2, bn=0 . Hence, ϕ(r ,θ)=a0+R

r (V−a0)−(1− R3

r3)E0rcosθ

In addition, because Φ(r ,π/2)=V for θ=π/2 and r>R, we obtain: a0=V . The potential becomes:

Φ(r ,θ)=V−(1−R3

r3)E0rcosθ

The charge density on the hemisphere is:

σ=−ϵ0∂ Φ

∂r ∣r=R=3ϵ0E0cosθ

F=

∫

∫s

2σErd S)cosθ=πR

2

∫

0

π /2

σE_rsinθcosθdθ

E_r=−( ∂ Φ

∂r )∣r=R= σϵ₀

F=π_ϵR2 0

∫

0

π/2

σ2_{sinθcosθdθ=}9πR 2

ϵ0 2_E

0 2 ϵ0

∫

0

π /2

sinθcos3_θdθ=9 4πR

2_ϵ 0E0

2

If F>2 3πR

3

ρg , the hemisphere can be lifted up. Hence we need:

Q5. A pair of opposite charges of magnitude q and distance d form a dipole. The dipole is placed at the center of a grounded conducting spherical shell with radius R. Solve for the potential anywhere within the sphere (using either Legendre polynomials or spherical harmonics) and calculate the force between the two charges.

Solution:

Use the method of images and the expansion of the Greens function on page 119 (eqn. 3.114), we can directly write the potential inside the sphere:

Φ(r ,θ)= q 4πϵ0

∑

∞

[

l

rl>+1 −1

R

(

)

l+1

]

(

₎

Since Pl(−x)=(−1)mPl(x) , the potential is simplified to:

Φ(r ,θ)= q 2πϵ0l=

∑

∞

[

l+1− 1 R

(

2R2 (r d)

)

l+1

]

Considering the two image changes outside the sphere, the net force between the two charges is:

F= q2 4πϵ0

[

4 d2−

R2 /(d/2)

((2R/d−d/2)−d/2)2−

R2 /(d/2)

((2R/d+d/2)−d/2)2

]

q R