electromagnetism ii 2012

8/20/2019 Electromagnetism II 2012

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MASSACHUSETTS INSTITUTE OF TECHNOLOGYPhysics Department

Physics 8.07: Electromagnetism II September 5, 2012Prof. Alan Guth

PROBLEM SET 1

DUE DATE: Friday, September 14, 2012. Either hand it in at the lecture, or by 5:00pm in the 8.07 homework box.

READING ASSIGNMENT: Chapter 1 of Griffiths: Vector Analysis .

PROBLEM 1: VECTOR IDENTITIES INVOLVING CROSS PRODUCTS(20 points)

In manipulating cross products, it is useful to define εijk (the Levi-Civita antisym-metric symbol) to be:

εijk =

+1 if ij k = (123, 231, 312)−1 if ij k = (213, 321, 132)

0 otherwise .(1.1.1)

That is, εijk is nonzero only when all three indices are different; it is then equal to +1if ijk is a cyclic permutation of 123, and -1 if ijk is an anti-cyclic permutation. Notethat εijk is totally antisymmetric, in the sense that it changes sign if any two indices areinterchanged:

εijk = −εikj = εkij . (1.1.2)

With this definition, the i

th

component of the cross product of two vectors A and

B canbe written as

A × B

i= εijk Aj Bk , (1.1.3)

where we have used the summation convention that repeated indices are summed over

(that is, εijk Ajl Bkm =3

j=1

3k=1

εijk Ajl Bkm). For the rest of this problem set, we will

always assume that this summation convention is implied, unless explicitly stated other-wise.

(a) From the definition in Eq. (1.1.1), show that

εijk εinm = δ jn δ km − δ jm δ kn , (1.1.4)

where of course there is an implied sum over the i index in Eq. (1.1.4), but theindices j , k, n, and m are free.

(b) Using Eqs. (1.1.3) and (1.1.4), show that for any vectors A, B, and C ,

A × (B × C ) = B (A · C ) − C (A · B) (1.1.5)


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8.07 PROBLEM SET 1, FALL 2012 p. 2

(c) Using Eqs. (1.1.3) and (1.1.4), show that for any vectors A and B,

∇ · ( A× B) = B · ( ∇× A) − A · ( ∇× B) . (1.1.6)

(d) Using Eqs. (1.1.3) and (1.1.4), show that for any vector A,

A× ∇× A = 12

∇A2 −

A · ∇

A . (1.1.7)

(e) Using Eqs. (1.1.3) and (1.1.4), show that for any vectors A and B,

∇×

A× B

=

B · ∇

A −

A · ∇

B + A

∇ · B− B

∇ · A

. (1.1.8)

PROBLEM 2: TRIPLE CROSS PRODUCTS (10 points)

Griffiths Problem 1.2 (p. 4), Griffiths Problem 1.6 (p. 8).

PROBLEM 3: PROPERTIES OF THE ROTATION MATRIX R (15 points)

Griffiths Eq. (1.31), p. 11, is

Ā i =3

j=1

Rij Aj .

If we use the convention that repeated indices are summed over, then this can be writtenas

Ā i = Rij Aj . (1.2.1)

(a) Show that the elements (Rij) of the three-dimensional rotation matrix must satisfythe constraint

Rij Rik = δ jk (1.2.2)

in order to preserve the length of A for all A. Matrices satisfying Eq. (1.2.2) arecalled orthogonal . Here δ jk is the Kronecker delta (δ jk is 1 if j = k and 0 otherwise),and we use the summation convention above.

(b) Using the orthogonality constraint (1.2.2), show that

Ai = Rji Ā j . (1.2.3)

Note that we can now show that RjiRki = δ jk using this relation, in a manner similarto the procedure in (a) (you do not have to show this).

(c) Using the chain rule for partial differentiation and the results of (b), show that if f is scalar function of r ≡ (x1, x2, x3), then ∇f (r) transforms as a vector; i.e., showthat if

f̄ (x̄1, x̄2, x̄3) = f (x1, x2, x3) , (1.2.4)

where x̄i = Rij xj , then∂ f̄

∂ ̄xi= Rij

∂f

∂xj. (1.2.5)


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PROBLEM 4: USE OF THE GRADIENT (10 points)

Griffiths Problem 1.12 (p.15), Griffiths Problem 1.13 part (a) only (p.15).

PROBLEM 5: THE DIRAC DELTA FUNCTION AND∇2(1/4πr) (20 points)

One of the most used identities in this course is be the relation

−∇2 14πr

= − ∇ ·

∇ 14πr

= ∇ ·

r̂

4πr2

= δ 3(r) = δ (x) δ (y) δ (z) . (1.5.1)

It turns out of course (see Griffiths 1.5.1, p. 45) that

−∇2

1

4πr

is zero everywhere except at the origin, and ill-defined there. To get a better feel for thefact that

−∇2 14πr

is a delta function, let’s look at a different function which approaches −(1/4πr) in somelimit, but which is well-behaved everywhere. The function is

f a(r) = − 14π

1√ r2 + a2

. (1.5.2)

For a nonzero, f a(r) is well-behaved everywhere, and

lima→0

f a(r) = − 14πr

(1.5.3)

(a) Calculate ga(r) = ∇2f a(r) and show that it is also well behaved for all r. Sketchga(r) for some value of a as a function of r/a.

(b) Show that

all space

ga(r) d

3

x = 1 . (1.5.4)

(c) Show thatlima→0

ga(r) = 0 if r = 0 . (1.5.5)

Thus in the limit that a goes to zero, our well-behaved function ga(r) exhibits theproperties we expect of a three-dimensional delta function.


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PROBLEM 6: EXERCISES WITH δ-FUNCTIONS (10 points)

(a) A charge Q is spread uniformly over a spherical shell of radius R. Express the volume

charge density using a delta function in spherical coordinates. Repeat for a ring of radius R with charge Q lying in the xy plane.

(b) In cartesian coordinates, we can write δ 3(r − r) = δ (x− x)δ (y − y)δ (z − z). Howwould one express δ 3(r − r) in cylindrical coordinates (s, φ, andz).

(c) A charge λ per unit length is distributed uniformly over a cylindrical surface of radiusb. Give the volume charge density using a delta function in cylindrical coordinates

(d) What is ∇2 ln r in two dimensions? (Here r is the radial coordinate, r =

x2 + y2.)

PROBLEM 7: COROLLARIES OF THE FUNDAMENTAL INTEGRAL

THEOREMS (15 points)

This problem is closely related to Problem 1.60, p. 56 of Griffiths. You will finduseful hints there– but try without hints first!!. Show that:

(a)

V ∇ψ d3x =

S ψ da, where S is the surface bounding the volume V . Show that as

a consequence of this,

S da = 0 for a closed surface S .

(b)

V ∇× A d3x = −

S A × da, where S is the surface bounding the volume V .

(c)

S ∇ψ × da = −

Γ ψ d l, where Γ is the boundary of the surface S .

(d) For a closed surface S , one has S

(

∇× A)

·da = 0.


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8.07 PROBLEM SET 1 SOLUTIONS, FALL 2012 p. 2

n, m, which in turn means that either j = n and k = m, or j = m and k = n. In thefirst case ( j = n, k = m) the two ε factors on the LHS are identical, so the value of the LHS is +1. In the second case the ordering of the indices on the ε factors differs

by an interchange of the 2nd two indices, and so one factor is the negative of theother, and the LHS is -1. The delta-functions on the RHS also give +1 in the firstcase and -1 in the second, so the equality holds.

If p = q , then the LHS is zero, because there is no value of i for which both factorsare nonzero. The RHS is also zero, since p = q implies that the pair of indices

j, k does not match the pair n, m. Thus at least one of the indices j or k must bedifferent from both n and m. If it is j , then δ jn and δ jm vanish, so the RHS vanishes.Similarly, if it is k that is different from both n and m, then δ km = δ kn = 0, so againthe RHS vanishes. So, if p = q , then LHS = RHS = 0. The formula has now beenshown to hold in all cases.

(b)

A × Bi

= εijkAjBk, so for any vectors A, B, and C ,

A ×

B × C

i

= εijkAj

B × C

k

, (1.2)

but B × C

k

= εkmnBmC n . (1.3)

So A ×

B × C

i

= εijkAjεkmnBmC n = εijkεkmnAjBmC n . (1.4)

But εijk = −εikj = εkij , so using Eq. (1.1),εijkεkmn = εkijεkmn = δ imδ jn − δ inδ jm . (1.5)

So, Eq. (1.4) can be rewritten as A×

B × C

i

= (δ imδ jn − δ inδ jm) AjBmC n= δ imδ jnAjBmC n − δ inδ jmAjBmC n= AnBiC n − AjBjC i = ( A · C )Bi − ( A · B)C i ,

(1.6)

Which is what we want, that is,

A × ( B × C ) = B ( A · C ) − C ( A · B) . (1.7)

(c) Using ∂ i to denote ∂/∂xi, we can write

∇ · ( A× B) = ∂ i(εijkAjBk) = εijk∂ i (AjBk) = εijkAj (∂ iBk) + εijkBk (∂ iAj)


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= −εjikAj (∂ iBk) + εkijBk (∂ iAj) = B · ( ∇× A) − A · ( ∇× B) . (1.8)

(d) For any vector A, show that A× ∇× A = 12 ∇A2 − A · ∇ A.

∇× Ai

= εijk∂ jAk where ∂ j ≡ ∂ ∂xj

. (1.9)

So A ×

∇× A

i

= εijkAj

∇× Ak

= εijkAj (εkmn∂ mAn)

= εijkεkmnAj∂ mAn .

(1.10)

Following the same steps that led to Eq. (1.6), we have

A×

∇× A

i

= εkijεkmnAj∂ mAn = (δ imδ jn − δ inδ jm) Aj∂ mAn

= (An∂ iAn − Am∂ mAi) = 12

∂ i (AnAn) − (Am∂ m) Ai ,(1.11)

or, as desired,

A× ∇× A = 1

2 ∇A2 − A · ∇ A . (1.12)

(e) Show that ∇×

A × B

=

B · ∇

A −

A · ∇

B + A

∇ · B− B

∇ · A

.

Using

A× Bi

= εijkAjBk, for any vectors A and B,

∇×

A× B

i

= εijk∂ j

A× Bk

= εijk∂ j(εkmnAmBn)

= (δ imδ jn − δ inδ jm) ∂ j (AmBn) = ∂ n (AiBn) − ∂ m (AmBi)

= Ai (∂ nBn) + (Bn∂ n) Ai − Bi (∂ mAm) − (Am∂ m) Bi , (1.13)

or

∇×

A × B

= A

∇ · B

+

B · ∇

A − B

∇ · A−

A · ∇

B . (1.14)


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PROBLEM 2: TRIPLE CROSS PRODUCTS (10 points)

Griffiths Problem 1.2 (p. 4): Simple counterexample: (ex×ey)×ey = ez×ey = −ex whileon the other hand ex × (ey × ey) = 0. More generally, the proposed associativitycould hardly be true given that ( A × B) × C lies on the plane defined by A and Bwhile A × ( B × C ) lies on the plane defined by C and B.

Griffiths Problem 1.6 (p. 8): We expand

A × ( B × C ) + B × ( C × A) + C × ( A× B)

= [ B( A · C ) − C ( A · B)] + [ C ( A · B) − A( B · C )] + [ A( B · C ) − B( A · C )] = 0as one can see that the terms cancel in pairs. This type of identity is a Jacobi identity satisfied also by commutators.

In order to find out when A× ( B × C ) = ( A× B) × C , we rewrite it as A× ( B × C ) + C × ( A × B) = 0 .

Comparing with the Jacobi identity we just proved, we see that this requires

B × ( C × A) = 0 .

A cross product of two vectors vanishes if they are parallel or either one is zero.With non-zero vectors A, B and C this requires that either (1) B is parallel to

C × A (equivalently B is orthogonal to both A and C ) or, (2) that C × A vanish(equivalently, A is parallel to C .).

PROBLEM 3: PROPERTIES OF THE ROTATION MATRIX R (15 points)

(a) Show that the elements (Rij) of the three-dimensional rotation matrix must satisfythe constraint

RijRik = δ jk . (3.1)

Griffiths equation (1.31), page 11, is Ā i = RijAj , so

Ā2 = A2 = AkAk = δ jkAjAk , (3.2)

but we also have

Ā2 = Ā i Ā i = (RijAj) (RikAk) = (RijRik) AjAk . (3.3)

In order for Eqs. (3.2) and (3.3) to hold for all Ak, we must have

(RijRik − δ jk)AjAk = 0 for all Aj , (3.4)


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which implies that

RijRik = δ jk , (3.5)

as desired.

[The implication of Eq. (3.5) from Eq. (3.4) may seem obvious, and full credit willbe given to any student who treats it as obvious. Note, however, that if RijRikwere not manifestly symmetric in j and k, then the antisymmetric part of RijRikwould be unconstrained. If we wanted to actually prove Eq. (3.5), we could defineS jk ≡ RijRik− δ jk , so S jkAjAk = 0 for all Aj . Then we can choose Aj = δ j1, whichimplies that S 11 = 0. We can similarly show that S 22 = S 33 = 0. Then we canchoose Aj = δ j1 + δ j2. Given what we already know that the diagonal entries of S jkvanish, we would find that S 12 + S 21 = 0. Using the fact that S jk is symmetric, thisshows that S 12 = 0, and similarly we can show that any matrix element with twodifferent indices must vanish.]

(b) Using Eq. (3.1) show that Ai = Rji Ā j.

Take the equation Ā i = RijAj , multiply by Rik, and sum over i, giving

Rik Ā i = RikRijAj = δ kjAj = Ak . (3.6)

Renaming indices, Eq. (3.6) is

Ai = Rji Ā j , (3.7)

as desired.

(c) Show that ∇f transforms as a vector.The chain rule for partial differentiation is

∂ f̄

∂ ̄xi=

∂f

∂xj

∂xj∂ ̄xi

, (3.8)

and since x j = Rkj x̄k,

∂xj∂ ̄xi

= ∂ ∂ ̄xi

(Rkj x̄k) = Rkj∂ ̄xk∂ ̄xi

= Rkjδ ik = Rij . (3.9)

Thus we have∂ f̄

∂ ̄xi= Rij

∂f

∂xj, (3.10)

as desired.


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PROBLEM 4: USE OF THE GRADIENT (10 points)

GRADING: This problem was graded, with a maximum number of points equal to 30

(3× 10), distributed as shown below. (See the note at the beginning of the problemset for a description of the spot-grading system that we are using.)Griffiths Problem 1.12 (p. 15): The height function is given by

h(x, y) = 10(2xy − 3x2 − 4y2 − 18x + 28y + 12)

(a) (8 points) The top of the hill must correspond to an extremal point of h(x, y), in otherwords a point where its partial derivatives vanish, or briefly, the gradient vanishes.Thus we have the conditions

∂h

∂x

= 10(2y

−6x

−18) = 0 ;

∂h

∂y

= 10(2x

−8y + 28) = 0 .

These have just one solution at (x, y) = (−2, 3), so the hilltop is located2 miles west and 3 miles north of S. Hadley.

You were not asked to do so, but a thorough analysis should include a determinationof whether this stationary point (i.e., point of vanishing gradient) is a local maximum,minimum, or saddle point. This is determined by the eigenvalues of the Hessianmatrix, the 2 × 2 matrix of partial derivatives,

H ij ≡ ∂ 2h

∂xi ∂xj=

−60 20

20

−80

.

The eigenvalues can be found by solving the characteristic equation

Det(H − λI ) = 0 ,which expands to λ2+140λ+4400 = 0. The roots are λ = −70±10√ 5 = −92.36 and−47.74. Since all roots are negative, the stationary point is indeed a local maximum,as the wording of the problem implied.

(b) (6 points) The height there is h(−2, 3) = 720 feet.(c) (8 points) For this we just need to evaluate the gradient

∇h = 10(2y − 6x− 18 , 2x− 8y + 28)at (x, y) = (1, 1). This gives ∇h = (−220, 220), a vector pointing NW, and corre-sponds to the direction of steepest slope. Since |∇h| = 220√ 2 = 311.13 the slope atthat point is 311.1 feet per mile.

Griffiths Problem 1.13 part (a) (p. 15) (8 points): We have 2 = (x − x)2 + (y − y)2 +(z − z)2. Then ∇ 2 = 2(x − x)êx + 2(y − y)êy + 2(z − z)êz = 2(r − r ) = 2 .


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PROBLEM 5: THE DIRAC DELTA FUNCTION AND∇2(1/4πr) (20 points)

The function f a(r) is given by

f a(r) = − 14π

1√ r2 + a2

. (5.1)

(a) To calculate ∇2f a(r), use Eq. (1.73) (p. 42) of Griffiths for the Laplacian in sphericalcoordinates. For functions of only r , this gives

ga(r) = ∇2f a(r) = 1r2

∂

∂r

r2

∂f a(r)

∂r

=

−

1

4π

1

r2

∂

∂rr2 ∂

∂r

1

√ r2

+ a2 = 1

4π

1

r2

∂

∂r r

3

(r2

+ a2

)3/2

= 1

4π

1

r2

3r2

(r2 + a2)3/2

− 3r

4

(r2 + a2)5/2

= 1

4π

3

(r2 + a2)5/2

r2 + a2

− r2

= 3

4π

a2

(r2 + a2)5/2

. (5.2)

To sketch ga(r) as a function of r/a, rewrite it as

ga(r) = 3

4πa31

1 + r2

a2

5/2 . (5.3)Thus, the value of a is affects only the coefficient, and not the shape of the curve.Choosing a = 1

10, the graph looks like


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(b) We must show that the integral of ga(r) over all space is 1. Denoting the integral byI ,

I = 4π ∞0 ga(r) r

2

dr = 4π ∞0

3

4πa3

11 + r

2

a2

5/2 r

2

dr

= 3

∞

0

η2 dη

(1 + η2)5/2 ,

(5.4)

where η = r/a. To carry out the integration use the substitution η = tan θ, so

dη = dθ

cos2 θ

and

1 + η2 = 1cos2 θ

.

Then

I = 3

π/20

tan2 θ cos5 θ dθ

cos2 θ = 3

π/20

sin2 θ cos θ dθ .

Now let u = sin θ, du = cos θ dθ, so

I = 3

1

0

u2 du = 3

1

3u31

0

= 1 , (5.5)

as desired.

(c) Looking at Eq. (5.2), it is obvious that as a → 0, the numerator vanishes and thedenominator approaches r5, so

lima→0

ga(r) = 0 for r = 0. (5.6)

PROBLEM 6: EXERCISES WITH δ-FUNCTIONS (10 points)

Here I will write the answers and confirm the normalizations (of course, when doingthis for the first time, one typically puts an undetermined constant which is fixed by thenormalization condition).

(a) For the sphere

ρ = Q

4πR2δ (r − R) . (6.1)


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This clearly has the right units Q/L3 (an L2 from the R’s and the other from theunits of the delta function). Moreover with spherical symmetry d3x = 4πr2dr andthus

d3xρ(x) =

4πr2dr Q4πR2

δ (r −R) = Q , (6.2)as should be the case.

For the charge distributed over the circle, the angle θ = π/2, and we write

ρ = Q

2πR2δ (r − R)δ

θ − π

2

=

Q

2πR2δ (r − R)δ (cos θ) . (6.3)

Make sure you understand why δ (cos θ) = δ θ − π2 . To confirm the normalization

we integrate again: ∞

0

(r2dr)

1

−1

d(cos θ)

2π

0

dφ Q

2πR2δ (r − R)δ (cos θ)

= R2 · 1 · 2π Q2πR2

= Q .

(6.4)

Similarly, we could write the charge distribution in cylindirical coordinates as

ρ = Q

2πRδ (r − R)δ (z − z) . (6.5)

The normalization is assured by the integration with volume element d3x =dr(rdφ)dz.

(b) The volume element here is d3x = ds(sdφ)dz and we thus guess

δ (x − x) = 1s

δ (s− s)δ (φ − φ)δ (z − z) , (6.6)

which is confirmed by computing

d3x δ (x− x) =

∞0

sds 2

π

0

dφ ∞−∞

dz 1s

δ (s − s)δ (φ − φ)δ (z − z) = 1 .

(c) We write

ρ = λ

2πbδ (s − b) , (6.7)


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which we confirm by integrating over a finite cylinder CL bounded by the planesz = ±L:

CL d

3

x ρ = ∞0 sds

2π0 dφ

L−L dz

λ

2πb δ (s− b) = 2Lλ , (6.8)which is the expected result.

(d) Away from the origin,

∇2 ln r = ∇ · ( ∇ ln r) = ∇ · r̂

r

= ∇ ·

rr2

= ∇

1r2

· r + 1

r2 ∇ · r .

In two dimensions, ∇ · r = 2 and thus we get

∇2

ln r = −2r̂

r2 · r + 2

r2 = 0 . (6.9)

Since we could be missing a delta function at the origin we use the analog of the di-vergence theorem, which in two dimensions relates the area integral of the divergenceto the flux across the boundary line:

r


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Since c was arbitrary this means

V

∇ψd3x =

S

ψda . (7.1)

(b) (12 points) This time choose F = A× c, so ∇ · F = c · ∇× A. Then

c · V

∇× A

d3x =

V

∇·

A × c

d3x =

S

A× c

· da = −c ·

S

A×da . (7.2)

(c) (12 points) Apply Stokes theorem to ψ b, where b is a constant vector:

S

∇× (ψ b) · da = Γ

ψ b · d l .

Using the identity ∇× (ψ b) = ∇ψ × b, S

( ∇ψ × b) · da = b · S

(da × ∇ψ) = b · Γ

ψ d l .

Since this equation holds for all b, we can write

S

∇ψ×

da =− Γ ψ d l , (7.3)

where we have reordered the cross product to match the identity in the problemstatement.

(d) (9 points) The boundary ∂S of a closed surface S is zero, therefore by Stokes’ theorem

we have S (

∇× A) ·da = ∂S A ·d l = 0. Alternatively, we can assume that the closedsurface S is non-self-intersecting, so that Klein bottle surfaces are not allowed. ThenS will necessarily be the boundary of some volume V . In that case we can use thedivergence theorem and the fact that ∇ · ( ∇× A) = 0 for all A to show

S

( ∇× A) · da = V

∇ · ( ∇× A) d3x = 0 . (7.4)


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PROBLEM SET 2

DUE DATE: Monday, September 24, 2012. Either hand it in at the lecture, or by 6:00pm in the 8.07 homework boxes.

READING ASSIGNMENT: Chapter 2 of Griffiths: Electrostatics .

GRADING OF THIS PROBLEM SET: The problem set is worth 85 points plus20 points extra credit. It is therefore time to clarify the operational definition of “extra credit”. We will keep track of the extra credit grades separately, and at theend of the course I will first assign provisional grades based solely on the regularcoursework. I will consult with Prof. Chen and Ahmet Demir, and we will try tomake sure that these grades are reasonable. Then I will add in the extra credit,

allowing the grades to change upwards accordingly. Finally, we will look at eachstudent’s grades individually, and we might decide to give a higher grade to somestudents who are slightly below a borderline. Students whose grades have improvedsignificantly during the term, and students whose average has been pushed down bysingle low grade, will be the ones most likely to be boosted.

The bottom line is that you should feel free to skip the extra credit problems,and you will still get an excellent grade in the course if you do well on the regularproblems. However, if you are the kind of student who really wants to get themost out of the course, then I hope that you will find these extra credit problemschallenging, interesting, and educational. (As described in the solutions to ProblemSet 1, the problem sets in the course will not be graded in full, but instead only

selected problems will actually be graded. The extra credit problems will never beamong the graded problems.)

PROBLEM 1: THE LAPLACIAN AS THE ANTI-LUMPINESS OPERA-TOR (15 points)

In this problem you will prove a relation that was stated in lecture. Le ϕ(r) be anyscalar function of position r. We are interested in relating the value of ϕ at an arbitrarypoint r0 to the average value of ϕ on a sphere that is centered at r0. While the point r0is arbitrary, we can simpify our notation by choosing a coordinate system so that r0 isthe origin 0. Then the relation to be proved can be written

ϕ( 0)− ϕ̄(R) = − 14π

r


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PROBLEM 3: THE ELECTRIC FIELD, POTENTIAL, AND ENERGY OFA UNIFORM SPHERE OF CHARGE (15 points)

(a) A uniformly charged sphere of charge has radius R and total charge Q. Using Gauss’slaw, calculate the electric field E (r) everywhere.

(b) Using the electric field you calculated in part (a), find the electric potential V (r)everywhere.

(c) Using the expression

W = 1

20

all space

| E |2 d3x , (3.1)

for the total work needed to assemble the charge configuration, calculate W usingyour expressions above.

(d) Using the expression

W = 1

2

all space

ρV d3x , (3.2)

calculate W again using your expressions above.

PROBLEM 4: CALCULATING FORCES USING VIRTUAL WORK (10 points)

Use “virtual work” to calculate the attractive force between conductors in the parallelplate capacitor (area A, separation d). That is, use conservation of energy to determine

how much work must be done to move one plate by an infinitesimal amount, and thenuse the value of the work to determine the force. Do your virtual work computations intwo ways:

(a) keeping fixed the charges on the plates, and,

(b) keeping a fixed the voltage between the plates.

PROBLEM 5: MUTUAL CAPACITANCE (15 points)

In lecture we discussed relations of the form

Qi =

nj=1

C ijV j , i, j = 1, 2, . . . , n . (5.1)

governing the potentials and charges of n conductors (with the potential taken to be zeroat spatial infinity).

(a) Prove that C ij = C ji . [Hint: Consider how much energy is needed to start with thesystem uncharged, then add charge Qi to conductor i, and then add charge Qj to


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conductor j . Then consider starting again with the system uncharged, and perform-ing these operations in the opposite order. That is, add charge Qj to conductor j,and then Qi to conductor i. Then think about how to use your answers to prove the

desired result.]

(b) Consider a two-conductor configuration. Calculate the conventional capacitance C in terms of C 11, C 12, C 21, and C 22.

(c) Consider two concentric spherical conducting shells of radii a and b with a < b.Call the inner shell conductor 1, and the outer shell conductor 2. Calculate thematrix of capacitances C ij and use your result from part (b) to infer the conventionalcapacitance C . Compare your answer with Example 2.11 in Griffiths, p. 105.

PROBLEM 6: SPACE CHARGE, VACUUM DIODES, AND THE CHILD-

LANGMUIR LAW (20 points)

Griffiths Problem 2.48 (p. 107). Challenging! For part (e), you can solve the dif-ferential equation either by guessing a solution and showing that it works, or by findinga first integral by the same method that is used in mechanics to go from Newton’s 2ndorder equation of motion to the first order equation for the conservation of mechanicalenergy.

PROBLEM 7: ∇2(1/r) IN THE LANGUAGE OF DISTRIBUTIONS (20 points extra credit)

This problem will have a longwinded pedagogical introduction, since it concerns anapproach which was discussed in lecture, but is not discussed in the textbook.

In Problem 5 of Problem Set 1, you evaluated −∇2(1/4πr) by replacing 1/r by1/√

r2 + a2. After calculating ga(r) ≡ −∇2(1/4π√

r2 + a2), you showed that its integralover all space is 1, and that for any r = 0 it approaches 0 as a → 0. This exercise wasintended to convey a useful intuition about δ -functions, and about the relation

−∇2 14πr

= δ 3(r) . (7.1)

However, from the standpoint of a mathematically rigorous treatment, there is a short-coming to this and all similar treatments of the δ -function as a limit of a sequence of functions. While the sequence of functions leads to reliable intuition, the precise math-ematical picture is complicated by the ordering of limits. That is, you showed in yourproblem set solutions that

all space

ga(r) d3x = 1 for any a > 0, and hence

lima→0

all space

ga(r) d3x = 1 . (7.2)


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However, if we had taken the limit first, we would have found

lima→0 ga(r) = 0 if r

= 0

∞ if r = 0 , (7.3)and we showed in lecture that the integral of this function, defined as the area underthe curve, is in fact zero. So we cannot quite say that ga(r) approaches a δ -function asa → 0. Instead, we have to keep in mind the slightly more complicated picture in whichga(r) acts like a δ -function when a is very very small, and behaves exactly as a δ -functionif we take the limit a → 0 after any integrations have been carried out.

Since the integral of the function described in Eq. (7.3) vanishes, there is no normalfunction that behaves as a Dirac δ -function. Thus the δ -function is technically not afunction, but rather what the mathematicians call a generalized function , or a distribution .

It is really the concept of integration that is being generalized, and a distribution isthe integrand of a generalized integral. Starting with functions of one variable, we canconsider an arbitrary function ϕ(x). Its integral,

∞

−∞

ϕ(x) dx , (7.4)

maps the function ϕ(x) into a single real number, the value of its integral. It is a linearmap, in the sense that

∞−∞

[ϕ1(x) + λϕ2(x)] dx = ∞−∞

ϕ1(x) dx + λ ∞−∞

ϕ2(x) dx , (7.5)

where λ is a constant. A distribution defines a generalized integral, which is an arbitrary linear map from the space of smooth “test” functions ϕ(x) to real numbers. These testfunctions are required not only to be smooth, but also to fall off rapidly at large valuesof |x|.* The distribution that corresponds to a δ -function is the map which takes thefunction ϕ(x) to ϕ(x0), its value at some particular point x0. While there is no functionthat behaves as a Dirac δ -function, it is perfectly clear that this map from functions toreal numbers is well-defined. Thinking of this map as a generalization of integration, wecan write it as

ϕ(x) δ (x− x0) dx ≡ ϕ(x0) . (7.6)

* Various choices can be made for the precise restrictions on the space of test func-tions. A frequently used choice is the space of Schwartz functions, which are infinitelydifferentiable, and which have the property that the function and all its derivatives falloff faster than any power at large |x|. The distributions associated with this definition of smoothness are called tempered distributions.


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But remember that the integral sign here does not describe the area under a curve; insteadit denotes a linear map from the function ϕ(x) to a real number, and the symbol δ (x−x0)indicates the particular linear map which maps ϕ(x) to its value at x0, namely ϕ(x0).

Mathematically, Eq. (7.6) defines the δ -function, which is defined solely as a prescriptionfor a generalized type of integration.

The derivative of a distribution is defined so that generalized integration is consistentwith the usual procedure of integration by parts:

ϕ(x)

d

dxδ (x− x0) dx ≡ −

dϕ(x)

dx δ (x− x0) dx ≡ −ϕ(x0) , (7.7)

where ϕ(x) ≡ dϕ(x)/dx. Note that we do not include any boundary terms, as ϕ(x) isrequired to fall off at large

|x

| fast enough to cause any boundary terms to vanish.

We are now ready to evaluate ∇2(1/r) in the language of distributions. Note thatin the language of functions ∇2(1/r) is ill-defined, because 1/r is not differentiable atr = 0. But we can promote 1/r to a distribution by defining it as a mapping from testfunctions ϕ(r) to numbers, where the mapping is given by the (ordinary) integral

ϕ(r)

1

r d3x . (7.8)

Note that even though 1/r is singular at r = 0, this integral is perfectly well defined,since in spherical polar coordinates we have

1

r d3x = r dr sin θ dθ dφ . (7.9)

By defining the derivative of a distribution by integration by parts, as in Eq. (7.7), wecan write the distribution corresponding to ∇2(1/r), which I will call F [ϕ(r)] for futurereference:

ϕ(r)∇2

1

r

d3x =

ϕ(r) ∂ i∂ i

1

r

d3x (7.10a)

= − ∂ iϕ(r) ∂ i1r d3x (710b)=

∇2ϕ(r)

1

r

d3x , (7.10c)

so

F [ϕ(r)] ≡ ∇2ϕ(r)

1

r

d3x . (7.11)


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Note that the square brackets used for the argument of F is a common notation for afunctional, i.e., a function of a function. Here F maps the function ϕ(r) to the numbergiven by the (ordinary) integral on the right-hand side of Eq. (7.11).

AT LAST: THE HOMEWORK PROBLEM:

(a) Evaluate F [ϕ(r)] (as defined by Eq. (7.11)) for an arbitrary smooth test functionϕ(r) which falls off rapidly for large |r|. Show that

F [ϕ(r)] = −4πϕ( 0) . (7.12)

Since ϕ(r) δ 3(r) = ϕ( 0) , (7.13)

Eq. (7.12) is equivalent to writing

∇2

1

r

= −4πδ 3(r) (7.14)

in the sense of distributions, which is the result we seek. [Hint: Although Eq. (7.11)is the defining equation, there is nothing that prevents you from integrating by partsonce to retrieve the integral in the form of Eq. (7.10b). Write this in spherical polarcoordinates, and then try to evaluate it.]

(b) Use the language of distributions to evaluate ∇2 ln r in two dimensions. (See Problem

6(d) of Problem Set 1.)


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PROBLEM SET 2 SOLUTIONS

PROBLEM 1: THE LAPLACIAN AS THE ANTI-LUMPINESS OPERA-TOR (15 points)

(a) Using divergence theorem, r


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Putting this result into Eq. (1.3),

4π ϕ( 0)− ϕ̄(R) + r


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PROBLEM 3: THE ELECTRIC FIELD, POTENTIAL, AND ENERGY OFA UNIFORM SPHERE OF CHARGE (15 points)

(a) A uniformly charged sphere of charge has radius R and total charge Q. Using Gauss’slaw, calculate the electric field E (r) everywhere.

Consider a Gaussian sphere of radius r < R. The charge enclosed Q(r) = 43

πr3ρ.By Gauss’ law,

4πr2E (r) = Q(r)

0=

1

0

4

3πr3ρ =⇒ E (r) =

ρ r

30, for r < R .

For a Gaussian sphere of radius r > R, the charge enclosed is Q, so

4πr2E (r) = Q

0=⇒ E (r) =

Q

4π0r2 .

Thus,

E =

Qr

4π0R3 r̂ for r < R

Q

4π0r2 r̂ for r > R .

(b) Using the electric field you calculated in part (a), find the electric potential V (r)

everywhere.

Outside of the sphere, r > R,

V (∞)− V (r) = −

∞

r

E (r)dr = −

∞

r

Q

4π0r2dr = −

Q

4π0r .

Inside the sphere, r < R,

∆V = V (R)− V (r) = − R

r

E (r)dr = − R

r

Qr

4π0R3

dr = −Q(R2 − r2)

8π0R3

.

Putting these results together,

V (r) =

Q

3R2 − r2

8π0R3 if r < R

Q

4π0r if r > R.


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(c) Using the expression

W = 1

20 all space |

E |2 d3x ,

for the total work needed to assemble the charge configuration, calculate W usingyour expressions above.

W = 1

20

all space

| E |2 d3x = 2π 0

R0

r2dr

Qr

4π0R3

2+

∞

R

r2dr

Q

4π0r2

2

= Q2

8π0

1

5R +

1

R

=

3

5

Q2

4π0R .

(d) Using the expressionW =

1

2

all space

ρV d3x ,

calculate W again using your expressions above.

W = 1

2

all space

ρV d3x = 4π

2

R0

r2dr

3Q

4πR3

Q

3R2 − r2

8π0R3

= 3Q2

16π0R

1−

1

5

=

3

5

Q2

4π0R .

PROBLEM 4: CALCULATING FORCES USING VIRTUAL WORK (10 points)

The capacitance is C = 0Ad and the energy stored in the capacitor is

E = 1

2QV =

1

2CV 2 =

1

2

Q2

C .

To calculate the force, let’s separate the plates by a displacement δx. The work done byus is F δx.

(a) Fixed charges: Since the battery does no work, our work F δx must equal thechange in electrostatic energy:

F δx = ∆E = ∆

1

2

Q2

C

=

1

2Q2∆

1

C

= −

1

2

Q2

C 2∆C . (4.1)


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The change in the capacitance is

∆C =

∂C

∂d δx = −

0A

d2 δx = −

C

d δx .

Thus

F δx = 1

2

Q2

C 2C

d δx

F constQ = 1

2

Q2

0A =

σ2

20A .

(b) Fixed potential: The change in the energy stored in the capacitor is the sum of the work done by the force and the work done by the battery.

∆

1

2CV 2

= F δx + V ∆Q

1

2(∆C )V 2 = F δx + V ∆C V

F δx = −1

2(∆C )V 2 = −

1

2(∆C )

Q2

C 2

So, comparing with Eq. (4.1), we see that the force is the same as before (as it shouldbe !!).

PROBLEM 5: MUTUAL CAPACITANCE (15 points)

(a) To add charge Qi to conductor i, we need to gradually change the charge q i onconductor i from 0 to Qi. At any point during the change, the potential of conductori will be

V i =n

k=1

P ikq k = P iiq i , (5.1)

where the sum collapses because only q i = 0. (Note that we are not using a summa-

tion convention here — if there is no

sign, there is no sum.) The work needed tomove each charge element dq i from infinity (V = 0) to the conductor is then

dW = V i dq i = P ii q i dq i , (5.2)

so the total work is

W =

Qi0

P ii q i dq i = 1

2P iiQ

2i . (5.3)


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Now we hold fixed the charge on conductor i, and increase the charge q j on conductor j from 0 to Qj . The potential of conductor j is

V j =n

k=1

P jkq k = P jiQi + P jjq j . (5.4)

The work to move a charge element dq j from infinity to conductor j is then

dW = V j dq j = [P jiQi + P jjq j ] dq j , (5.5)

so the total work for this operation is

W = Qj

0

[P jiQi + P jjq j ] dq j = P jiQiQj + 1

2P jjQ

2

j

. (5.6)

The total work done to charge the two conductors is then

W tot = 1

2P iiQ

2i + P jiQiQj +

1

2P jjQ

2j . (5.7)

If we had charged the capacitors in the opposite order, we would have obtained asimilar expression, except that we would have P ij appearing instead of P ji . Butthese two sequences both started with the charges at infinity, and ended with Qi onconductor i and Qj on conductor j . Since electrostatics is known to be conservative,

the work done must be the same in both cases, and hence P ij = P ji .Finally, the matrix C ij is the matrix inverse of P ij . It can be easily proven that if a symmetric matrix is invertible then its inverse is symmetric also. One way to seethis is to look at the explicit expression for the matrix inverse given by Cramer’srule.

(b) The matrix equations are

Q1Q2

=

C 11 C 12C 21 C 22

V 1V 2

. (5.8)

Inverting we have V 1V 2

=

1

detC

C 22 −C 12−C 21 C 11

Q1Q2

. (5.9)

If we set Q1 = Q and Q2 = −Q then, the capacitance C is defined from

Q = C (V 1 − V 2) . (5.10)


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Inverting, this givesQ1 = C̃ (V 1 − V 2) , (5.18)

where for convenience we have defined

C̃ ≡ 4π0ab

b− a . (5.19)

Having found Q1, we can find Q2 from Eq. (5.16):

Q2 = b

4π0V 1 −

Q1a

= 4π0b

V 1 −

b

b− a(V 1 − V 2)

= 4π0b

bV 2 − aV 1

b− a

= −C̃V 1 +

b

aC̃V 2 .

(5.20)

The C ij are defined so that

Qi =n

j=1

C ijV j , (5.21)

so comparing with Eqs. (5.18) and (5.20) we see that

C =

C̃ −C̃ −C̃ ba

C̃

. (5.22)Substituting these matrix elements into Eq. (5.13), we find

C =b

a˜

C 2

− ˜C

2

ba C̃ −

C̃ = C̃ . (5.23)

This agrees with Griffiths’ example 2.11 (p. 105).

PROBLEM 6: SPACE CHARGE, VACUUM DIODES, AND THE CHILD-LANGMUIR LAW (20 points)

(a) In general Poisson’s equation is ∇2V = −ρ/0. In this case we know that V and ρdepend only on x, so we can write

d2V (x)

dx2 = −ρ(x)

0 . (6.1)

(b) By energy conservation we know that the change of electrostatic energy ∆W elec of the electron, as it travels from the cathode to a point x, goes into a change of kineticenergy ∆K . Letting v(x) denote the velocity of the electron at x, we have

∆W elec + ∆K = 0 =⇒ (−e)[V (x)− V (0)] + 1

2m[v2(x)− v2(0)] = 0 .


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But the potential at the cathode is zero, V (0) = 0, and the velocity of the electronsat the cathode is also zero, v(0) = 0. Therefore

12

mv2(x) = eV (x) =⇒ v(x) =

2eV (x)

m . (6.2)

(c) If a fluid has velocity v and charge density ρ, then the current I moving across anarea A normal to the velocity is I = ρvA (note that since ρ is negative the currentI is also negative). Thus,

I

A = ρv =⇒ ρ(x) v(x) =

I

A . (6.3)

The above equation, with I independent of x, is the requested relation between ρ(x)and v(x). They are both x-dependent, but their product is not!

(d) From Eqs. (6.1) and (6.3) we have

d2V

dx2 = −

ρ

0= −

I

0A v(x) ,

and using Eq. (6.2) we find

d2V

dx2 = −

I

A0 m

2e [V (x)]−1/2 .

This is the desired differential equation. Calling

α ≡ − I

A0

m

2e , (6.4)

the equation readsd2V

dx2 = α V −1/2 . (6.5)

(e) One method is to write Eq. (6.5) as V = αV −1/2 and multiply by V to find

V V = αV −1/2V =⇒ 1

2

d

dxV 2 = 2α

d

dxV 1/2 =⇒

d

dx

V 2−4αV 1/2

= 0 .

This shows that the expression in parentheses is a constant:

V 2 − 4αV 1/2 = E .


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(Note that this method is also used in Newtonian mechanics to show that mẍ =−V (x) implies that mẋẍ = −ẋV (x), which implies that d

dt [12

mẋ2 + V (x)] = 0.)The constant E can be evaluated using the boundary conditions at x = 0. Since

V (0) = 0 and V (0) = 0 we have E = 0. Thus the differential equation becomesquite simple:

V 2 = 4αV 1/2 =⇒ dV

dx = 2α1/2V 1/4 . (6.6)

Integrating we find

dV

V 1/4 = 2α1/2x =⇒

4

3V 3/4 = 2α1/2x + F ,

where F is another constant of integration. Again, V (0) = 0 implies that the constant

of integration is zero, so

V 3/4 = 3

2α1/2x =⇒ V (x) =

9α

4

2/3x4/3 . (6.7)

As an alternative method, we can try to invent a trial solution. If it satisfiesthe differential equation (6.5) and boundary conditions V (0) = V (0) = 0, then wewould have the desired solution. Since the right-hand side of Eq. (6.5) is a power of V , it is natural to try a power-law solution of the form

V = Cxβ , (6.8)

where C and β > 0 are constants. Inserting this trial solution into Eq. (6.5), we find

Cβ (β − 1)xβ−2 = αC −1/2x−β/2.

Equating the powers of x we find β = 4/3 and then

C 4

9 = αC −1/2 =⇒ C =

9α

4 2/3

.

We therefore find

V (x) =

9α

4

2/3x4/3 , (6.9)

in agreement with Eq. (6.7). Since β > 1 this solution satisfies not only V (0) = 0 butalso the condition that the field at the cathode vanishes, −V (0) ≡ −dV

dx |x=0 = 0, sothis is the solution we wanted.


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The problem tells us that V (d) = V 0, so using the fact that V (x) ∝ x4/3 we can

write

V (x) = V 0

xd4/3

. (6.10)

This result is consistent with Eqs. (6.7) or (6.9), since α depends on the unknownI and is therefore not yet determined. For this part, however, it suffices to useEq. (6.10). ρ(x) can be found directly from Poisson’s equation in the form (6.1),with Eq. (6.10):

ρ(x) = −0d2

dx2 V 0 x

d4/3

= − 40V 0

9(d2 x)2/3 . (6.12)

And v(x) can be found directly from Eqs. (6.2) and (6.10):

v(x) =

2eV (x)

m =

2eV 0

m

xd

2/3. (6.13)

The potential V (x) looks like in this figure. The straight line represents the potential

without space charge.

(The above graph was drawn accurately, but for purposes of grading we will acceptany graph with the same qualitative features.)

(f) Comparing either Eqs. (6.7) or (6.9) with Eq. (6.10), we determine the value of α:

9α

4

2/3= V 0 =⇒ α =

4V 3/20

9d2 .


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Then using Eq. (6.4), we can determine the value of I :

I = −49

0AV 3/20

d2

2em

. (6.14)

Thus we can write I = KV 3/20 , with

K = −4

9

0AV 3/20

d2

2e

m . (6.15)

PROBLEM 7: ∇2(1/r) IN THE LANGUAGE OF DISTRIBUTIONS (20 Extra points)

(a) We want to evaluate F [ϕ(r)], which is defined by

F [ϕ(r)] ≡

∇2ϕ(r)

1

r

d3x . (7.1)

As pointed out in the statement of the problem, we can begin by integrating byparts, which takes us back to Eq. (7.10b) of the problem statement,

F [ϕ(r)] = −

∂ iϕ(r) ∂ i

1r

d3x . (7.2)

But

∂ i

1

r

= −

1

r2 r̂i , (7.3)

so

F [ϕ(r)] = +

1

r2 r̂ · ∇ϕ(r) d3x . (7.4)

If we write this in polar coordinates, the radial component of the gradient is just∂/∂r, and the volume element is d3x = r2 dr sin θ dθ dφ, so

F [ϕ(r)] =

∞

0

dr

π0

sin θ dθ

2π0

dφ ∂ϕ(r,θ,φ)

∂r . (7.5)

The integral over r is an integral of a derivative, so if we integrate first over r wefind

F [ϕ(r)] =

π0

sin θ dθ

2π0

dφ [ϕ(∞, θ , φ)− ϕ(0, θ , φ)] . (7.6)


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But the test function ϕ(r) is required to approach zero at r = ∞, and continuityrequires that ϕ(0, θ , φ) is independent of θ and φ. The integral over angles then givesa factor of 4π, and we have

F [ϕ(r)] = −4πϕ( 0) , (7.7)

which is exactly what we were asked to prove.

(b) To evaluate ∇2 ln r as a distribution in two dimensions, we introduce a test functionϕ(x, y), and evaluate

F [ϕ(x, y)] =

∇2ϕ ln

1

r d2x , (7.8)

where we used the fact that derivatives of a distribution are defined by integratingby parts. From Eq. (7.8) we can integrate by parts, finding

F [ϕ(x, y)] = −

∂ iϕ ∂ i ln

1

r d2x = −

∂ iϕ

1

rr̂i

d2x . (7.9)

Using polar coordinates r̂i∂ iϕ = ∂ϕ/∂r, and d2x = r dr dϕ, so

F [ϕ(x, y)] = −

2π0

dφ

∞

0

r dr 1

r

∂ϕ

∂r

= − 2π

0

dφ ϕ(r=∞, φ)− ϕ(r=0, φ) = 2πϕ( 0) ,(7.10)

where as in part (a) we used the fact that ϕ is required to approach zero as r →∞,and that it cannot depend on angle at r = 0. Since

ϕ(r) δ 2(r) d2x ≡ ϕ( 0) , (7.11)

and since distributions are defined solely in terms of the generalized integral thatthey produce, we see that

∇2 ln r = 2πδ 2(r) , (7.12)

in the sense that the two sides of the equation represent the same distribution.


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PROBLEM SET 3

DUE DATE: Friday, September 28, 2012. Either hand it in at the lecture, or by 6:00

pm in the 8.07 homework boxes.

READING ASSIGNMENT: Chapter 3 of Griffiths: Special Techniques , Secs. 3.1–3.3.

PROBLEM 1: SPHERES AND IMAGE CHARGES (10 points)

Griffiths Problem 3.8 (p. 126).

PROBLEM 2: IMAGE CHARGES WITH A PLANE AND HEMISPHERI-

CAL BULGE (15 points)

Consider a conducting plane that occupies the x-y plane of a coordinate system, butwith the circular disk x2+y2 < a2 removed. The circular disk is replaced by a conducting

hemisphere of radius a, described by the equation

x2 + y2 + z2 = a2 , z > 0 . (2.1)

A charge q is placed on the z-axis at (0, 0, z0), with z0 > a. Find a suitable set of imagecharges for this configuration. Show that the charge is attracted toward the plate with a

force

| F | = 1

4π0

q 2

4z2 +

4q 2a3z3

(z4 − a4)2

. (2.2)

PROBLEM 3: IMAGES FOR A CONDUCTING CYLINDER (15 points)

This problem is based on Problem 2.11 of Jackson: Classical Electrodynamics, 3rd

edition.A line of charge with linear charge density λ is placed parallel to, and at a distance

R away from, the axis of a conducting cylinder of radius b held at fixed voltage so that

the potential vanishes at infinite distance from the cylinder.

(a) Find the magnitude and position of the image charge(s).

(b) Find the potential V 0 of the cylinder in terms of R, b, and λ.


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PROBLEM 4: CAPACITANCE OF A SINGLE CONDUCTOR (20 points)

(a) Consider a single conductor, and define its capacitance by Q = C V , where Q is the

charge on the conductor, and V is the potential of the conductor defined so thatV = 0 at infinity. Show that C can be expressed as

C = 0

V 20

V

| ∇V |2 d3x , (4.1)

where V is the space outside the conductor, and V (r) is the solution for the potentialwhen the conductor is held at V = V 0.

(b) Show that the true capacitance C is always less than or equal to the quantity

C [Ψ(r)] = 0

V 20

V | ∇Ψ|

2

d3

x , (4.2)

where Ψ(r) is any trial function satisfying the boundary condition Ψ = V 0 at theconductor, and Ψ = 0 at infinity. (Note that Ψ is not required to satisfy Laplace’sequation, or any other equation.)

(c) Prove that the capacitance C of a conductor with surface S is smaller than thecapacitance C of a conductor whose surface S encloses S .

(d) Use part (c) to find upper and lower limits for the capacitance of a conducting cubeof side a. Write your answer in the form: α(4π0a) < C cube < β (4π0a) and find the

constants α and β . A numerical calculation* gives C 0.661(4π0a). Compare thisanswer with your limits.

PROBLEM 5: LAPLACE’S EQUATION IN A BOX (15 points)


* C.-O. Hwang and M. Mascagni, Journal of Applied Physics 95, 3798 (2004).


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Physics 8.07: Electromagnetism II October 3, 2012Prof. Alan Guth


PROBLEM 1: SPHERES AND IMAGE CHARGES (10 points)

Griffiths 3.8. We use the same notation we had in class: sphere of radius a and acharge Q at a distance R > a from the center. The image is q = −aQ/R at a distancer from the center, with Rr = a2 and such that the image is on the segment joining thecenter and the outside charge.

The above charges produce zero potential on the surface of the sphere. To produce aconstant potential V 0 at the sphere we need an extra charge

q at the center, of magnitude

defined by

V 0 = 14π0

q a

=⇒ q = (4π0) aV 0 .For a neutral conducting sphere we need an extra image at the center of the sphere withcharge q neutral = −q = aQ/R, so the total image charge is zero. The force due to thefirst image is attractive, and the force due to the second image is repulsive but lower inmagnitude since the second image is further away. Thus, the net force felt by the outsidecharge is attractive and of magnitude:

F = 1

4π0 |Qq

| 1

(R− r)2

− 1

R2

= Q2

4π0

a

R

1

(R− r)2 − 1

R2

=

Q2

4π0

a3

R32R2 − a2

(R2 − a2)2 .

Note that as expected the force does not change sign since R > a. Finally, for large Rthe force falls of like 1/R5.

PROBLEM 2: IMAGE CHARGES WITH A PLANE AND HEMISPHERI-

CAL BULGE (15 points)

GRADING: This problem was graded, with a maximum number of points equal to 45(3× 15). 10 points were assigned to each of the three image charges, and 15 pointswere allocated to the force calculation. (See the note at the beginning of ProblemSet 1 Solutions for a description of the spot-grading system that we are using.)

We have a charge q = q 1 at z = z1. Let’s consider the two related problems:


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Charge q1 and V = 0 at z = 0:

The suitable image is

q 2 = −q 1 , at z2 = −z1 .Charge q1 and V = 0 on a sphere of radius a: Here the suitable image is

q 3 = − az1

q 1 , at z3 = a2

z1.

Now let’s consider the configuration with these three charges and a fourth one

q 4 = − a|z2|q 2 = a

z1q 1 at z4 =

a2

z2= −a

2

z1.

We thus have:

At the surface of the bulge:

V (q 1) + V (q 3) = 0 , V (q 2) + V (q 4) = 0 .

At z = 0 and x2 + y2 ≥ a2:

V (q 1) + V (q 2) = 0 , V (q 3) + V (q 4) = 0 .

Since q 2, q 3 and q 4 are outside the volume of interest, they form a suitable set of images.The force on q 1 is

F tot = kq 1q 2êz

|z1 − z2|2 +

kq 1q 3êz

|z1 − z3|2 +

kq 1q 4êz

|z1 − z4|2

= −kq 2êz

(2z)2 −

az

kq 2êzz − a2

z

2 + az kq 2êzz + a

2

z

2= −kq

2êz4z2

+ kq 2a3z3êz

1

z4a2

1

z + a2

z

2 − 1z − a2

z

2

= −kq 2êz

4z2 + kq 2a3z3êz

1

z2a2

z4 + a4 − 2a2z2 − z4 − a4 − 2a2z2

[(z2 + a2)(z2 − a2)]2

= −kq 2êz4z2

− 4kq 2a3z3êz(z4 − a4)2 ,

where k = 14π0

. The force is attractive towards the plane:

F tot = −14π0

q 2

4z2 +

4q 2a3z3

(z4 − a4)2

êz .


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PROBLEM 3: IMAGES FOR A CONDUCTING CYLINDER (15 points)

GRADING: This problem was graded, with a maximum number of points equal to 45

(3× 15), distributed as shown below. (See the note at the beginning of Problem Set1 Solutions for a description of the spot-grading system that we are using.)(a) (35 points) Looking to the symmetry of the cylinder, we can assume a image line

with charge λ at position R along the line connecting the origin to the charge λ.

The electric field of a straight line charge is simply λ2π0r

r̂, where r is the distancefrom the line and r̂ is a unit vector pointing outward from the line, as can be foundfrom Gauss’s law. Then the integral yields a logarithmic potential which is infinitein magnitude at both r = 0 and r = ∞. It is therefore necessary to pick a referencepoint at some arbitrary distance.

We choose an arbitrary reference point for the potential for each of the line sources,and then we can superimpose them to find an expression for the total potential:

V (ρ, φ) =

1

2π0 λ ln r0r + λ ln r0r + V c=

1

4π0

λ ln

r2

0

ρ2 + R2 − 2ρR cos φ

+ λ ln

r20


+ V c ,

where r and r are the distances from λ and λ respectively, and V c is an arbitraryconstant, which gives us the most general possible expression for the potential of thetwo charges. (For this problem we will find that V c = 0, but it would not be zero if the problem had specified that the potential at infinity was 10 volts.) Since we aregiven that the potential vanishes at infinity, we require limρ→∞ V (ρ, φ) = 0. Theleading terms are

V (ρ →∞) ∼ −λ + λ

2π0 ln(ρ/r0) + V c + O(R/ρ) ,

where the order of the correction can be seen by expanding

ln


r20

= 2 ln

ρ

r0+ ln

1− 2 R

ρ cos φ +

R2

ρ2

= 2 ln

ρ

r0− 2

R

ρ

cos φ + O(R/ρ)2 .


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We must choose λ = −λ and V c = 0 so that the potential vanishes at infinity. Usingthis, we can rewrite the expression for the potential as

V (ρ, φ) = λ

4π0lnρ2 + R2 − 2ρR cos φ


.

To determine the location R, we impose the constant voltage boundary conditionV (b, φ) = V 0. This implies

b2 + R2 − 2bR cos φb2 + R2 − 2bR cos φ = c .

Rearranging the terms, we get

(1− c)b2 + R2 − cR2 = 2b(R − cR)cos φ .

Since we need this equation to hold for any angle φ along the cylinder, we see that

both sides have to independently vanish. The RHS of the equation gives R = cR.Putting c = R/R into the equation we get a quadratic equation for R with twosolutions:

R1 = b2

R , R2 = R .

Both solutions give V = constant on the surface of the cylinder, which is the re-

quirement that we used in deriving the quadratic equation. The second solution

(R = R), however, puts the second line charge on top of the first, canceling theelectric field entirely! This is not the image charge that we seek, which must always

lie outside the region in which we are trying to determine the fields. Therefore only

the R1 = b2/R solution is relevant to our problem.

This problem could alternatively be solved using the electric fields from the two line

charges. Since the cylinder is a conductor, the electric field just outside its surface

must be normal to the surface. Equating the tangential electric field E

to zero

yields the same results.

(b) (10 points) The potential V 0 is given as

V (b, φ) = V 0 = λ

4π0ln

b2 + R2 − 2bR cos φb2 + R2 − 2bR cos φ

=

λ

4π0ln c =

λ

2π0ln

b

R.


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PROBLEM 4: CAPACITANCE OF A SINGLE CONDUCTOR (20 points)

(a) The potential energy W of the capacitor is W = 12

QV 0 = 1

2CV 2

0 . This energy is elec-

trostatic energy given by the expression: W = 02 V | E |2d3x = 02 V | ∇V (r)|2d3x ,

where V (r) is the potential on the volume V outside the conductor. Comparing thetwo expressions gives:

C = 0V 20

V | ∇V |2d3x . (4.1)

(b) Write the trial function Ψ asΨ = V + δV . (4.2)

Here Ψ satisfies the boundary conditions, thus δV = 0 on the boundary S of theconductor, and also at infinity.

C [Ψ] = 0V 20

V | ∇Ψ|2d3x = 0

V 20

V

| ∇V |2 + 2 ∇V · ∇δV + | ∇δV |2

d3x

= C + 0V 20

V

∇δV |2d3x + 20V 20

V

∇ · (δV ∇V ) − δV ∇2V

d3x

I 1

. (4.3)

To deal with the last integral above, denoted by I 1, note that V (r) is the potentialoutside the conductor when the conductor is held at potential V 0, so ∇2V = 0. Theremaining integral can be turned into a surface integral by use of the divergence

theorem:I 1 =

20V 20

S

δV ∇V · da + 20V 20

∞

δV ∇V · da , (4.4)

where the second term is an integral over a large spherical surface at infinity. The firstterm clearly vanishes, since δV = 0 on the boundary of the conductor S . δV = 0at infinity also, but we need to be more careful, since the integration is over aninfinity area. If we let R denote the radius of the sphere, which will be taken toinfinity, then the area grows as R2. But ∇V is the negative of the electric field, soits magnitude falls off as 1/R2, canceling the infinity of the area. In fact we know

that

∞ ∇V ·da = −Q/0, where Q is the charge on the conductor. Since the infinity

of the surface area is canceled by the falling off of ∇V , the integral will vanish if δV → 0 as R →∞, no matter how slowly. Thus I 1 = 0, and Eq. (4.4) becomes

C [Ψ] = C + 0V 20

V | ∇δV |2d3x ≥ C . (4.5)

The integrand above is nonnegative, so C [Ψ] can equal C only if ∇δV = 0 every-where. Since δV = 0 on S , C [Ψ] is in fact strictly bigger than C whenever δV is


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nonzero. In other words, C [Ψ] > C whenever Ψ(x) is not equal to the true potentialfunction.

(c) Let

V be the volume outside of the surface S of the

smaller conductor, which we imagine fits entirely insidethe larger conductor whose boundary is S . Assume thatwe hold the smaller conductor at potential V 0. We willuse δ V to denote the volume enclosed between S and S ,so δ V = V −V , as shown in the diagram at the right. Tofind an upper bound for C , let’s use the trial function:

Ψ(x) =

V (x) for x ∈ V V 0 for x ∈ δ V .

(4.6)

Note that for the region outside the surface S of the larger conductor, this trialfunction uses the true potential for the larger conductor, when the conductor is heldat potential V 0. In the region between S

and S , it has the constant value V 0. Thusthe trial function Ψ(x) satisfies the correct boundary condition (Ψ = V 0) at S

, andit is continuous at S and throughout the region V outside of the smaller conductor.Using this trial function we get

C ≤ C [Ψ] ≡ 0V 20

V | ∇Ψ|2d3x = 0

V 20

δV | ∇Ψ|2d3x + 0

V 20

V | ∇Ψ|2d3x . (4.7)

Since ∇Ψ = 0 in δ V we getC

≤

0

V 2

0 V | ∇V |2d3x = C . (4.8)

Given the remarks at the end of part (b), whenever S is different from S (but stillcontained inside it), we can conclude that C < C , i.e., C will be strictly smallerthan C . To prove this, we need only show that the test function Ψ(x) is not thetrue potential function V (x) for the smaller conductor. While this seems obvious,we will nonetheless give a proof, or maybe even two proofs. Suppose first that S is completely in the interior of S , as shown in the diagram above, so that S iscompletely surrounded by the region we called δ V . In this region ∇Ψ = 0, so if Ψwere the potential we would have E (x) = 0 over the surface S , and then by Gauss’slaw the total charge on the (smaller) conductor would be zero.

More generally, we can consider trickier cases

where the smaller conductor lies completely inside thelarger conductor, but where their surfaces might insome places coincide. The smaller conductor might,for example, be a sphere, and the larger conductor canbe the same sphere, but with a bump sticking out,as shown in the diagram at the right. We assume of course that the two surfaces are not identical, so theregion δ V is nonempty. For this situation it is harder to


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prove that Ψ(x) cannot be equal to the true potential V (x), but we can do it by using theLaplacian mean value theorem: the statement that for any solution to ∇2V = 0, the valueof V at any point is equal to the average value of V on any sphere centered on that point.

From this theorem, we know that the maximum of V (x) (taking V 0 > 0) must be on theconductor, the boundary of the region where ∇2V = 0, because the theorem preventsthe function from having a maximum or minimum in the interior. Thus V (x) ≤ V 0. If Ψ(x) is anywhere greater than V 0, then we know that Ψ(x) = V (x), and we are done. Sowe can focus on the case where Ψ(x) ≤ V 0. The trial solution Ψ( x) = V 0 in the nonemptyregion δ V . Let V be the region for which Ψ(x) = V 0, which includes δ V but whichcould conceivably be larger. In the diagram at theright, V is shown as the hatched region. Now con-sider a point P on the outer boundary of V , andconsider a sphere that is centered at P , but smallenough to lie outside S . Ψ(P ) = V 0, but the spherecontains points inside V with Ψ = V 0, but alsopoints outside V for which Ψ < V 0. Thus the av-erage value of Ψ on the sphere is less than V 0, so Ψcannot satisfy ∇2Ψ = 0.(d) A cube of side a can be contained in a sphere of radius R1 =

√ 3

2 a and it can contain

a sphere of radius R2 = 1

2a. Moreover, the capacitance of a conducting sphere is

given by C = 4π0R (this is a simple exercise). Thus

4π0a√

3

2 ≥ C cube ≥ 4π0a 1

20.866 (4π0a) ≥ C cube ≥ 0.5 (4π0a)

The average of the two bounds is 0.683(4π0a). This is about 3.3% off the precisenumerical estimate for the capacitance 0.661(4π0a).


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PROBLEM SET 4 REVISED∗

DUE DATE: Friday, October 5, 2012. Either hand it in at the lecture, or by 6:00 pmin the 8.07 homework boxes.

READING ASSIGNMENT: Chapter 3 of Griffiths: Special Techniques , Secs. 3.3-3.4.

PROBLEM 1: LAPLACE’S EQUATION IN A BOX (15 points)


PROBLEM 2: A SPHERICAL CONDUCTOR AND A CONDUCTINGPLANE (25 points)

Consider a solid spherical conductor of radius R

, with center on the positive z

-axisat z = z0, with z0 > R. Suppose that the x-y plane is conducting, and is held at potentialV = 0, while the sphere is held at potential V 0.

To first approximation, we can think of the field as that of a point charge q 0 at thecenter of the sphere, with q 0 related to V 0 by

V 0 = q 0

4π0R . (2.1)

The field due to this charge gives a potential V 0 on the surface of the sphere, as desired.But now the potential on the x-y plane is not zero.

(a) The potential on the x-y plane can be restored to zero by placing an image charge

below the x

-y

plane (i.e., at negative z

). What charge q

should this image have,and where should it be placed?

(b) The potential on the surface of the spherical conductor is now no longer constant,but it can be made constant by adding another image charge q . The potential onthe x-y plane can be restored to zero by adding another image charge q , and thepotential on the sphere can be restored to a constant by adding yet another imagecharge q . The series will continue forever, but it does converge fairly quickly.Calculate the positions and charges of the image charges q , q , and q .

(c) After all the image charges are added through q , what is the potential V of thespherical conductor?

(d) What is the total potential energy of this configuration? Express your answer asthe first terms of an infinite series, showing those terms corresponding to the imagecharges through q .

(e) Would the fields outside the conductors be different if the solid spherical conductorwere replaced by a spherical conducting shell, with the same outer radius?

∗ The wording of Problem 2(d) has been changed since the original version of Septem-ber 29, 2012.


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PROBLEM 3: ELECTROSTATICS INSIDE A SPHERICAL CAVITY (15 points)

(Patterned after Jackson Problem 3.5.)A hollow sphere of inner radius a has the potential specified on its surface to be

V = V 0(θ, φ). Show that the potential inside the sphere can be written as

V (r) =∞l=0

lm=−l

Alm

ra

lY lm(θ, φ) , (3.1)

where you are asked to find the expressions for Alm in terms of V 0(θ, φ).

PROBLEM 4: A CHARGED METAL SPHERE IN A UNIFORM FIELD (15 points)


PROBLEM 5: A SPHERE WITH OPPOSITELY CHARGED HEMI-

SPHERES (15 points)


PROBLEM 6: AVERAGE FIELD INSIDE A SPHERE (20 points)



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PROBLEM 1: LAPLACE’S EQUATION IN A BOX(15 points)


We use separation of variables in Cartesian coordinates: V (x,y,z) = X (x)Y (y)Z (z),where Laplace’s equation becomes

1

X

d2X

dx2 = C 1 ,

1

Y

d2Y

dy2 = C 2 ,

1

Z

d2Z

dz2 = C 3 , with C 1 + C 2 + C 3 = 0. (1.1)

Since the potential is required to vanish for x = 0 and x = a, we use the functions

sinmπx

a

to describe X (x), with an analogous choice for Y (y). This implies that C 1 and C 2are negative, so C 3 ≡ γ

2 must be positive. Thus Z (z) must be expressed as a linearcombination of eγz and e−γz , or as a linear combination of sinh γz and cosh γz . Theboundary condition that V = 0 at z = 0 is most easily enforced by the latter choice, inwhich case the boundary condition simply implies that the coefficient of the cosh termmust be zero. Thus we can choose basis functions

Φmn(x,y,z) = sinmπx

a

sinnπy

a

sinh(γ mnz)sinh(γ mna)

, γ mn =

mπa

2+nπ

a

2. (1.2)

Note that we divided by a constant in the factor relevant to the z-direction, in order tohave a simple behavior at z = a.

The desired potential V (x,y,z) is a superposition of the above ones, with coefficientsV mn to be determined:

V (x,y,z) =

m,n≥1V mnΦmn(x,y,z) . (1.3)

The boundary conditions on all the faces except for the one at z = a have already beensatisfied. The boundary condition at z = a is satisfied if

V 0 =

m,n≥1

V mn sinmπx

a

sinnπy

a

, (1.4)


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where V 0 is a the constant specified in the problem. The V mn are readily calculated byusing the orthogonality relation

a0

dx sin

m

πxa

mπx

a

= a

2δ mm . (1.5)

We multiply both sides of Eq. (1.4) by sinmπxa

sinnπya

and integrate over x and y: a

0

dx sinmπx

a

a0

dy sinnπy

a

V 0

=

m,n≥1

V mn

a0

dx sinmπx

a

sinmπx

a

a0

dy sinnπy

a

sinnπy

a

= a

2

4

m,n≥1

V mnδ mmδ nn = a

2

4 V mn .

(1.6)

Thus

V mn = 4V 0

a2

a0

dx sinmπx

a

a0

dy sinnπy

a

=

16V 0

π21

mn for m, n odd

0 otherwise .

(1.7)

Therefore, the full potential is given by Eq. (1.3), with V nm given above:

V (x,y,z) = 16V 0

π2

odd m,n

1

mn sinmπx

a sinnπy

a sinh(γ mnz)

sinh(γ mna) , (1.8)

where γ mn is given in Eq. (1.2).

PROBLEM 2: A SPHERICAL CONDUCTOR AND A CONDUCTING

PLANE (25 points)

GRADING: This problem was graded, with a maximum number of points equal to 75(3× 25), distributed as shown below. (See the note at the beginning of Problem Set1 Solutions for a description of the spot-grading system that we are using.)

(a) (10 points) The potential on the x-y plane can be restored to zero by placing animage charge below the x-y plane. The image charge q = −q 0 must be placed atz = −z0:

q = −q 0, at z = −z0 . (2.1)

(b) (25 points) The potential on the surface of the spherical conductor is now no longerconstant, but it can be made constant by adding another image charge q . The


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are actually distributed over the conducting plane, but since this plane is at V = 0,there is no contribution to W . Thus,

W = 1

2V 0(q 0 + q

+ q )

= q 208π0R

1 +

R

2z0+

R2

4z20 − R2

+ . . .

.

(2.7)

It is tempting to try to find W by treating the image charges as if they were realpoint charges, and then calculating half the potential energy of the point chargeconfiguration. We found that this works for some special cases, but we never derivedit as a general result, and it is not valid in this case. By contrast, Eq. (2.6) is auniversally valid relation for electrostatics.

The force on the conducting sphere can be calculated by the method of virtual work,which relates it to the derivative of W with respect to z0. But there is a paradoxthat needs to be resolved: Eq. (2.7) implies that dW/dz0


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where dΩ = sin θ dθ dφ. Nonzero Blm’s would give infinite potential at the center of thesphere, so Blm = 0. We can determine the Alm’s as follows. The potential on the surfaceof the sphere is specified by V = V 0(θ, φ), so

V (a,θ,φ) = V 0(θ, φ) =∞l=0

lm=−l

AlmY lm(θ, φ) . (3.3)

Multiplying both sides of the equation by Y ∗lm (θ, φ) and integrating,

V 0(θ, φ)Y

∗lm(θ, φ) dΩ =

∞l=0

lm=−l

Alm

Y ∗lm(θ, φ) Y lm(θ, φ)dΩ

=

∞l=0

lm=−l

Almδ llδ mm = Alm .

(3.4)

Thus

Alm =

2πφ=0

πθ=0

V 0(θ, φ)Y ∗lm(θ, φ)sin θ dθ dφ . (3.5)

The potential inside the spherical cavity is then

V (r ) =∞

l=0l

m=−lAlm

r

al

Y lm(θ, φ) , (3.6)

where the Alm are given by Eq. (3.5).

PROBLEM 4: A CHARGED METAL SPHERE IN A UNIFORM FIELD (15 points)


GRADING: This problem was graded, with a maximum number of points equal to 45(3×15). Each of the 5 numbered equations counts 8 points, and 5 points are allocatedfor a clear explanation of where you have set the zero of potential.

We choose a coordinate system with the sphere centered at the origin, and with the z-axisaligned with the background electric field. We can then describe the background field by E 0 = E 0ẑ. The corresponding potential is

V 0 = −E 0z = −E 0 r cos θ = −E 0 rP 1(cos θ) . (4.1)

We could have added an arbitrary constant, but there is no need to. Note that thisterm is divergent as |r | → ∞, which is necessary because the electric field is constant at


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infinity. The complete potential outside the sphere must be an azimuthally symmetricsolution to Laplace’s equation, which can always be written as

V (r, θ) =∞l=0

AlrlP l(cos θ) +

∞l=0

Blrl+1

P l(cos θ) . (4.2)

Far from the sphere the potential is V = −E 0 rP 1(cos θ), which implies that Al = 0except for l = 1. Thus,

V (r, θ) = −E 0 rP 1(cos θ) +∞l=0

Blrl+1

P l(cos θ) . (4.3)

The potential on the sphere must be a constant, which implies that Bl≥2 = 0, and alsorequires that the two terms proportional to P 1(cos θ) give a total contribution of zero atr = a. This fixes −E 0a + B1/a

2 = 0, so B1 = E 0a3, resulting in

V (r, θ) = −E 0

r −

a3

r2

P 1(cos θ) +

B0r

. (4.4)

Given that the sphere has charge Q we write

V (r, θ) = −E 0 r − a3

r2P 1(cos θ) + 1

4π0

Q

r . (4.5a)

In this solution the zero of the potential is for some r > a along the positive z axis. If wewant to be able to make a clearer statement about where V = 0, we could add a constantand write

V (r, θ) = −E 0

r −

a3

r2

P 1(cos θ) +

Q

4π0

1

r −

1

a

. (4.5b)

In this form the zero of the potential is on the surface of the sphere. Either of these twoanswers is acceptable for grading purposes.

PROBLEM 5: A SPHERE WITH OPPOSITELY CHARGED HEMI-

SPHERES (15 points)


Held over to next week.


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PROBLEM 6: AVERAGE FIELD INSIDE A SPHERE (20 points)


(a) Let x denote an arbitrary point inside the sphere of radius R, and r the position of a charge q . The average electric field E q due to this charge on the volume is

E q = 14πR3

3

q

4π0

d3x

x − r

|x − r|3 .

A minor rearrangement leads to

E q = 1

4π0

d3x

−q 4πR3

3

r − x|x − r|3

.

The above has the interpretation of the field at r due to a charge −q uniformlydistributed over the volume of the sphere. Indeed, ρ = −q/ 4πR

3

3 .

(b) You showed in Problem Set 2, Problem 3 that for a un

electromagnetism ii 2012

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