PHYS 3327 PRELIM 1

Prof. Itai Cohen, Fall 2016

Wednesday, 10/12/16 Name:

Read all of the following information before starting the exam:

• Put your name on the exam now.

• Show all work, clearly and in order, if you want to get full credit.

• Box or otherwise indicate your final answers.

• Question 5(g) is a extra credit question worth 5 points. The total exam score cannot

exceed 100, but the extra credit can help you make up points lost elsewhere.

Problem # Score

1 /12

2 /08

3 /20

4 /25

5 /35

5(g) Extra credit /05

Total /100

• It is your responsibility to make sure that you have all of the pages!

• Good luck!

Question 1: Concentric charged spheres [12 points]

A uniformly charged sphere of radius a with total charge −Q is surrounded by a concentric

spherical shell of radius b, which has a total charge +Q distributed uniformly on its surface.

(a) [6 points] Find the net electric field produced by the charge configuration, ~E(~r), in

each of the following regions:

(i) outside the shell, r > b.

Ans. By symmetry, the electric field must point in the radial direction, and is a function

of r alone. Taking a spherical Gaussian surface at radius r, we get

E(4πr2) = 4πQencl = 0 . =⇒ E = 0.

(ii) between the sphere and the shell, a < r < b.

Ans. Here Qencl = −Q. Hence, E(4πr2) = −4πQ , =⇒ ~E(~r) = −Qr2r .

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(iii) inside the sphere, r < a.

Ans. The charge density inside the inner sphere is ρ = −Q/(43πa3). Hence the total charge

inside radius r is Qencl = ρ(43πr3) = −Q(r3/a3). Therefore,

E(4πr2) = −4πQ(r3/a3) , =⇒ ~E(~r) = −Qra3

r .

(b) [6 points] Sketch graphs of the radial component of the electric field, E(r), and the

electric potential Φ(r) as a function of r, assuming Φ(r →∞) = 0.

You don’t need to find an expression for Φ(r). However, pay attention to qualitative

features such as slope, curvature, and relative magnitudes in your plots, particularly

at r = 0, r = a, and r = b.

Ans. The plots are shown below. The electric field is given by the negative gradient of the

potential, E(r) = −dΦdr

. Since the electric field is zero for r > b, Φ(r > b) = Φ(r →∞) = 0. Note that Φ(r) is continuous everywhere, and smooth at r = a. Also, it has

zero slope at r = 0, since the electric field there vanishes.

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Question 2: Current loop in a magnetic field [8 points]

A current-carrying loop of arbitrary shape is partially immersed in a uniform magnetic field~B. The magnetic field is pointing into the page, perpendicular to the plane of the loop.

(a) [4 points] Write an expression for the magnetic force, d~F , on a small segment d~l of the

loop inside the field region. Sketch d~l and d~F on the figure.

Ans. From the Lorentz force law, we get d~F = Id~l× ~B. The vectors are shown in the figure.

(b) [4 points] Integrate your expression to find the net magnetic force on the loop, in terms

of I, a, B, x and y (see figure). Hint: I and ~B are constants!

Ans. To find the net force, we integrate d~F over the part of the loop inside the field region.

Both I and ~B can be taken out of the integral since they are both constants. Thus we

get

~F = I

(∫d~l

)× ~B .

However,∫d~l is just summing over all the little vectors from the left end to the right

end of the loop. Hence,∫d~l = ax. Therefore,

~F = Ia x× (−B z) = IaB y .

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Question 3: Polarized dielectric cylinder [20 points]

Consider an infinitely long dielectric cylinder of radius R. It has a frozen-in polarization~P (~r) = −k r, where r denotes the distance from its axis, and k is a constant.

(a) [4 points] Find the bound charge density inside the cylinder, ρb(~r).

Ans. The volume charge density is given by

ρb(~r) = −~∇. ~P

= −1

r

∂(rPr)

∂r− 1

r

∂Pφ∂φ− ∂Pz

∂z

= −1

r

∂(−k r)∂r

=k

r.

(b) [4 points] Find the bound charge density on the surface of the cylinder, σb(~r).

Ans. The surface charge density is given by σb(~r) = ~P .n, where n denotes the outward

normal to the surface. Here, n = r. Hence, σb(~r) = −k r.r = −k.

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(c) [6 points] Find the net electric field, ~E(~r), in each of the following regions:

(i) inside the cylinder, r < R

Ans. By symmetry, the electric field must be a function of r alone, and in the radially

outward direction. Thus we apply Gauss’ law to a closed concentric cylinder of

length l and radius r < R. The total enclosed charge is given by

Qencl =

∫ r

0

(k/r′)(2πr′ldr′) = 2πkrl .

Hence, E(2πrl) = 4π(2πkrl), or E = 4πk. =⇒ ~E(~r) = 4πk r.

(ii) outside the cylinder, r > R

Ans. Here we consider a cylinder of radius r > R. The total enclosed charge is given

by

Qencl = 2πkRl + (−k)(2πRl) = 0 .

Hence, E = 0.

Note: The result Qencl = 0 for a closed surface enclosing the dielectric is true

more generally, as the dielectric as a whole is electrically neutral.

(d) [6 points] Plot the magnitudes of the electric field ( ~E) and the displacement ( ~D) as a

function of r. Show that they satisfy the required boundary conditions at r = R.

Ans. By definition, the displacement is given by ~D = ~E + 4π ~P , which is zero both inside

and outside the cylinder. The plots are shown below.

They satisfy the boundary conditions ~E⊥r→R+ − ~E⊥r→R− = 4πσ = −4πk, and ~D⊥r→R+ −~D⊥r→R− = 0. Note that the parallel components are identically zero.

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Question 4: Dipolar hemispheres [25 points]

A conducting sphere of radius a is made up of two hemispherical shells separated by a thin

insulating ring. The two hemispheres are held at potentials +V and −V (see figure).

(a) [4 points] Roughly sketch the electric field lines you would expect for this system.

Focus on qualitative features, such as the direction of the field lines, and the angle

they make with the spherical surface.

y

z

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Suppose Φin(r, θ) and Φout(r, θ) describe the electric potential inside and outside the sphere,

where r denotes the distance from the origin, and θ denotes the angle from the z axis.

(b) [5 points] Write the boundary conditions on Φin and Φout at r = 0, r = a, and r →∞.

Ans. The boundary conditions are given by

(i) Φin(r, θ) is finite at r = 0.

(ii) Φout(r, θ)→ 0 as r →∞.

(iii) Φin(a, θ) = Φout(a, θ) =

+V θ < π/2

−V θ > π/2.

(c) [5 points] Write down the most general solution for Φin(r, θ) and Φout(r, θ) which satisfy

the boundary conditions at r = 0 and r →∞ respectively.

Ans. The most general solution to Laplace’s equation consistent with the boundary condi-

tions (i) and (ii) are given by

Φin(r, θ) =∞∑l=0

AlrlPl(cos θ) ,

Φout(r, θ) =∞∑l=0

Bl

rl+1Pl(cos θ) .

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(d) [8 points] Apply the boundary conditions at r = a to determine Φin(r, θ) and Φout(r, θ).

You may need the following property of Legendre polynomials: Pl(−x) = (−1)lPl(x),

and you may express your answer in terms of Cl ≡∫ 1

0Pl(x)dx.

Hint: If you’re stuck, try the substitution x = cos θ.

Ans. The boundary conditions at r = a read

∞∑l=0

Bl

al+1Pl(cos θ) =

∞∑l=0

AlalPl(cos θ) =

+V θ < π/2

−V θ > π/2

From the first equality we get Bl = a2l+1Al. Next we apply the orthogonality of

Legendre polynomials on the second equality to obtain the coefficients Al. To this

end, we multiply both sides by dθ sin θPl′(cos θ) and integrate over θ from 0 to π,

yielding

∞∑l=0

Alal

∫ π

0

dθ sin θPl(cos θ)Pl′(cos θ) =

∫ π

0

dθ sin θPl′(cos θ)×

+V θ < π/2

−V θ > π/2

or,2Al′a

l′

2l′ + 1= V

[ ∫ 1

0

dxPl′(x)−∫ 0

−1

dxPl′(x)

].

Here we have used the orthogonality relations of the Legendre polynomials in the left-

hand side, and made the substitution x = cos θ in the right-hand side. The right-hand

side can be further simplified as∫ 1

0

dxPl′(x)−∫ 0

−1

dxPl′(x) =

∫ 1

0

dx(Pl′(x)− Pl′(−x)) = (1− (−1)l′)

∫ 1

0

dxPl′(x) ,

which vanishes for all even values of l′. For odd l′, we get

2Al′al′

2l′ + 1= 2V

∫ 1

0

dxPl′(x) = 2V(−1)

l′−12

l′(l′ + 1)

l′!!

(l′ − 1)!!

or, Al = (−1)l−12

2l + 1

l(l + 1)

l!!

(l − 1)!!

V

al.

Therefore,

Φin(r, θ) = V∑odd l

(−1)l−12

2l + 1

l(l + 1)

l!!

(l − 1)!!

(ra

)lPl(cos θ) ,

Φout(r, θ) = V∑odd l

(−1)l−12

2l + 1

l(l + 1)

l!!

(l − 1)!!

(ar

)l+1

Pl(cos θ) . [using Bl = a2l+1Al.]

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(e) [3 points] Show that the potential Φ(~r) has the symmetry Φ(x, y,−z) = −Φ(x, y, z),

i.e., Φin(r, π − θ) = −Φin(r, θ), and Φout(r, π − θ) = −Φout(r, θ).

Hint: Use the properties cos(π − θ) = − cos θ, and Pl(−x) = (−1)lPl(x).

Ans. Both Φin and Φout are functions of θ of the form

f(θ) =∑odd l

αlPl(cos θ),

where αl are θ independent coefficients. Therefore,

f(π − θ) =∑odd l

αlPl(cos(π − θ))

=∑odd l

αlPl(− cos θ)

=∑odd l

(−1)lαlPl(cos θ)

= −∑odd l

αlPl(cos θ)

= −f(θ).

Thus they have the symmetry f(x, y,−z) = −f(x, y, z), as expected from the geometry

of the problem.

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Question 5: Charged spherical shell [35 points]

A spherical shell of radius a has a total charge Q distributed uniformly on its surface.

(a) [6 points] Find the energy of the charge distribution by integrating the energy density

in the electric field produced by it over all space.

Ans. A simple application of Gauss’ law yields that the electric field is given by

~E(~r) =

0 r < a

(Q/r2) r r > a.

The energy density in the electric field is u = E2/(8π). Thus the charge distribution

has a total energy

U =

∫ ∞a

(4πr2dr)

(Q2

8πr4

)=

∫ ∞a

Q2

2r2dr

=Q2

2a.

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Next we’ll calculate the net force on the upper hemisphere using the Maxwell stress tensor

T. For this, we need to integrate T.d~a over a surface enclosing the hemisphere. Figure

shows such a surface, S, which encloses the upper hemisphere for any R > a. In particular,

we can consider the limit R→∞, when the bottom surface spans the entire x-y plane.

(b) [4 points] Show that in the limit R → ∞, the hemispherical part of S does not

contribute to the force.

Hint: You’ll need to argue that Tij fall off faster with r than the rate at which the

area of hemispherical part grows.

Ans. The elements of the stress tensor are given by Tij = 14π

(EiEj +BiBj − 1

2(E2 +B2)δij

).

In our problem, there is no magnetic field. Thus it reduces to Tij = 14π

(EiEj − 1

2E2δij

).

We see that Tij involve products of electric fields. Since E falls off as 1/r2 with distance,

Tij will fall off as 1/r4.

On the other hand, the area of the hemispherical part grows as r2 with distance.

Therefore, the magnitude of∫

T.d~a will be of order O(1/R2), which vanishes in the

limit R → ∞. Thus the hemispherical part of S does not contribute to the force in

this limit.

(c) [6 points] Express the Cartesian components of the electric field (Ex, Ey, and Ez) on

the x-y plane in terms of r and φ, where r and φ denote plane polar coordinates.

Ans. The electric field on the x-y plane is zero for r < a, and is radially outward for r > a.

Thus Ez = 0, and

Ex =

0 r < a

(Q/r2) cosφ r > a, Ey =

0 r < a

(Q/r2) sinφ r > a.

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(d) [6 points] Evaluate all elements of the stress tensor (Tij) using the electric field com-

ponents you found in (d).

Ans. Using the relation Tij = 14π

(EiEj − 1

2E2δij

), we find Tij = 0 for r < a.

For r > a,

Txx =1

8π

(E2x − E2

y

)=

Q2

8πr4(cos2 φ− sin2 φ) =

Q2

8πr4cos 2φ .

Txy =1

4πExEy =

Q2

4πr4sinφ cosφ =

Q2

8πr4sin 2φ .

Txz = 0 .

Tyx = Txy =Q2

8πr4sin 2φ .

Tyy =1

8π

(E2y − E2

x

)=

Q2

8πr4(sin2 φ− cos2 φ) = − Q2

8πr4cos 2φ .

Tyz = 0 .

Tzx = 0 .

Tzy = 0 .

Tzz = − 1

8π

(E2x + E2

y

)= − Q2

8πr4.

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(e) [8 points] Integrate T.d~a over the x-y plane to find the net force ~F on the upper

hemisphere. Be careful to exclude regions where T = 0.

Hint: The area vector points in the direction of the outward normal of a closed surface.

Ans. The x-y plane forms the bottom surface of S in the limit R→∞. Hence the outward

normal points in the −z direction, i.e., d~a = −da z.

We can calculate the x, y, and z components of the force as

Fx =

∫Txxdax + Txyday + Txzdaz = 0 .

Fy =

∫Tyxdax + Tyyday + Tyzdaz = 0 .

Fz =

∫Tzxdax + Tzyday + Tzzdaz

= −∫Tzzda

=

∫ ∞a

(Q2

8πr4

)(2πrdr)

=Q2

8a2.

As expected from symmetry arguments, the force on the upper hemisphere is along

the +z direction. The two similarly charged hemispheres repel one another.

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Now consider a more general charge case where the total charge Q, instead of being all on

the surface, is distributed according to a function q(r) which gives the total charge inside

radius r. A possible form of q(r) is shown in the figure below.

You can convince yourself that the electric field produced by such a charge distribution is

given by ~E(~r) =(q(r)/r2

)r.

(f) [5 points] Show that the energy of any such distribution will be larger than the case

where all charges reside on the surface. Explain why this is relevant for conductors.

Hint: Write the energy as an integral over space, and compare with (a).

Ans. We can calculate the energy by integrating the energy density in the electric field,

u = E2/(8π), as in part (a). This gives

U =

∫ ∞0

(4πr2dr)(q(r))2

8πr4

=

∫ ∞0

(q(r))2

2r2dr

=

∫ a

0

(q(r))2

2r2dr +

∫ ∞a

Q2

2r2dr

=

∫ a

0

(q(r))2

2r2dr +

Q2

2a.

Thus U > Q2/(2a), i.e., the energy is larger than the case where all charges reside on

the surface. The extra energy originates from having a non-zero electric field inside

the sphere. This is why all charges in a conductor reside on its surface in equilibrium

– that yields the lowest energy configuration.

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(g) Extra credit: [5 points] Show that the net force on the upper hemisphere for any

distribution q(r) will be larger than the case where all charges reside on the surface.

Hint: Argue that the stress tensor will be non-zero for r < a on the x-y plane, in such

a way that it contributes to a force in the +z direction.

Ans. The net force can be calculated by integrating the stress tensor T.d~a over the x-y

plane, as in parts (c)–(e). Again, d~a points in the −z direction, and Txz = Tyz = 0.

Therefore, Fx = Fy = 0. The z-component of the force is given by

Fz = −∫Tzzda

=1

8π

∫E2da

=1

8π

∫ ∞0

(q(r))2

r4(2πrdr)

=

∫ ∞0

(q(r))2

4r3dr

=

∫ a

0

(q(r))2

4r3dr +

∫ ∞a

Q2

4r3dr

=

∫ a

0

(q(r))2

4r3dr +

Q2

8a2.

Therefore, Fz >Q2

8a2, i.e., the net repulsive force on the upper hemisphere is larger than

the case where all charges reside on the surface. This is consistent with the result in

(g) that the latter charge configuration has a lower energy.

- - - - - END OF EXAM - - - - -

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Formulae

σb = n · P ρb = −∇ · P (D2 −D1) · n = 4πρf (E2 − E1)× n = 0

Kb

c= −n×M Jb

c= ∇×M (B2 −B1) · n = 0 (H2 −H1)× n = −4π

cKf

D = E + 4πP P = χeE D = εE

H = B − 4πM M = χmH B = µH

Φ(r) =

∫V

dv′ρ(r′)

|r − r′|E = −∇Φ A(r) =

∫V

dv′J(r′)

|r − r′|B = ∇× A

Q =∑α

qα Φ(1) =Q

rp =

∑α

qαr′α Φ(2) =

p · rr3

Qij =∑α

qα(3x′iαx′jα − r′2α δij) Φ(4) =

1

6

∑ij

Qij

(3xixj − r2δij

r5

)

Φ(x, y, z) =∑α,β,γ

∣∣∣∣α2+β2+γ2=0

e±αxe±βye±γz Φ(r, θ) =∞∑l=0

[Alr

l +Bl

rl+1

]Pl(cos θ)

Φ(r, φ, θ) =∞∑l=0

l∑m=−l

[Aml r

l +Bml

rl+1

]Y ml (θ, φ)

Φ(r, φ, z) = A0 +B0 log r +∞∑n=1

([Anr

n +Bn

rn

]cos(nφ) +

[Cnr

n +Dn

rn

]sin(nφ)

)+∑m,n

[AmnJn(kmr) +BmnNn(kmr)

]e±inφe±kmz

∇ · A =∂Ax∂x

+∂Ay∂y

+∂Az∂z

=1

r

∂(rAr)

∂r+

1

r

∂Aφ∂φ

+∂Az∂z

=1

r2

∂(r2Ar)

∂r+

1

r sin θ

[∂(sin θAθ)

∂θ+∂Aφ∂φ

]∫ L

0

dx sinnπx

Lsin

mπx

L=L

2δmn

∫ π

0

dθ sin θPl(cos θ)Pm(cos θ) =2

2l + 1δlm

∫4π

Y ml (θ, ϕ)Y n ∗

k (θ, ϕ)dΩ = δmnδlk

∫ ρ

0

rdrJn(kmr)Jn(klr) =ρ2

2J2n+1(kmρ)δml

Tij =1

4π

(EiEj +BiBj −

1

2(E2 +B2)δij

)

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