Gabriel Barello - Classical Electrodynamics

1.21

a.)

For this problem, we compute ALz

, the action per unit length. We use g = 1 and will use the function

Ψ(x, y) = Ax(1− x) · y(1− y)

as the form of our approximate solution. First we compute

5Ψ = A ((1− x) · y(1− y)− x · y(1− y), (1− y) · x(1− x)− y · x(1− x), 0)

| 5Ψ |2= A2(y2(1− y)2(1− 2x)2 + x2(1− x)2(1− 2y)2

)And note that gΨ = Ψ. The integrals compute to be

∫v

gΨd2x =A

36∫v

| 5Ψ |2 d2x =A2

45

Then the minimum of the difference of these is given when A = 54 . Indeed, the second derivative is positive at this value

of A, so it is indeed a minimum (if that wasn’t obvious from the form).

b.)

Figure 1: This is a comparison of the true potential (in blue) and the approximate potential (dashed) The left plot showsy = .25 and the right shows y = .5. The vertical axis is in units of eV·4πε0

1

1.24

Since this setup is π/2 radians rotationally symmetric each of the corner points and each of the edge points are equivalent,so we really only need to compute one corner, one edge and the middle. Let us label these points {a, b, e} respectively. Recallthe best-average formula for the poisson equation

〈〈Φ〉〉p = 〈〈Φ〉〉+h2

5g +

h2

10〈g〉c

where

〈〈Φ〉〉 =4

5〈Φ〉c +

1

4〈Φ〉s

And the c and s stand for cross and square average. (the cross average is the one that makes a ”plus” sign on the grid).Note, too that h2 = 1/16. Now we are ready to get things lined up. Note the following table:

ij i+1,j i,j+1 i-1,j i,j-1 i+1,j+1 i-1,j-1 i+1,j-1 i-1,j+1a b 0 0 b 0 0 e 0b a 0 a e 0 b b 0e b b b b a a a a

After some fussing, the iterative equations we get are:

ai+1 =ei16

+4bi10

+3

160

bi+1 =2ai5

+ei5

+bi8

+1

40

ei+1 =4bi5

+a

4+

3

80

Using the jacobian method with ε0 · Φ = 1 on all sites initially gives

0. 1. 1. 1.1. 0.48125 0.75 0.68752. 0.361719 0.44875 0.4578133. 0.226863 0.317344 0.307434. 0.164902 0.216899 0.2211535. 0.119332 0.162304 0.1654856. 0.0940143 0.126118 0.1322547. 0.077463 0.104821 0.1114518. 0.0676442 0.091378 0.09879439. 0.0614758 0.0832388 0.090962310. 0.0577307 0.0781876 0.0861645

Note that my answers differ by a factor of 4π from the book’s exact solution, because I computed Φ · ε0 whereas Jacksongave 4πε0 · Φ.

2

b.)

After ten iterations of the gauss-seidel method using the same initial values we get

Iteration a b e0 1 1 11. 0.48125 0.5425 0.3748132. 0.259176 0.271445 0.2108723. 0.140508 0.157308 0.135554. 0.0901451 0.107832 0.1031695. 0.0683307 0.086445 0.08916076. 0.0589005 0.077198 0.08310437. 0.0548232 0.0731999 0.08048588. 0.0530603 0.0714713 0.07935369. 0.0522981 0.0707239 0.078864110. 0.0519686 0.0704007 0.0786524

c.)

We factored out a 1/ε0 at the begining, so we are actually reporting the value ε0Φ. The following plot compares ourapproximate solution to the exact one

Figure 2: This is a log plot of the aproximate solution using the jacobian method (blue) and the gauss seidel method (red)compared to the exact solution (lines). Unfortunately the lowest exact value is obscured by the log plot, but it is there, Ipromise. The vertical axis is in units of Φ · ε0. As you can see the gauss-seidel method converges faster and is more accurateafter 10 iterations.

3

2.1

a.)

The apropriate image charge is −q at position (−d, 0). Then the electric field at a point (0, y) is

E(0, y) =q

4πε0

1

(d2 + y2)32

(−dx+ yy − dx− yy)

= − q

2πε0

d

(d2 + y2)32

x

This is related to the surface charge by the equation σ = ε0E. So,

σ = − qd2π

(d2 + y2

)− 32

Here is a plot

Figure 3: This shows the (nondimentionalized) charge density with d = 1

b.)

The force on the charge is the same as the force that would result from the presence of the image charge.

F = − q2

4πε0

1

(2d)2= − q2

16πε0

1

d2

c.)

The total force acting on the plane can be computed by computing the integral

Fp =(qd)2

8π2ε0

∫ 2π

0

∫ ∞0

(d2 + r2)−3rdrdθ u ≡ d2 + r2

=(qd)2

8πε0

∫ ∞d2

u−3du

= − 1

16πε0

q2

d2

4

d.)

The work is given by the integral

W =

∫F · dl = − q2

16πε0

∫ ∞d

x−2dx

=1

16πε0

q2

d

e.)

The potential energy is given by the integral

PE =

∫ 2d

∞F · dl = − q2

4πε0

∫ ∞2d

x−2dx

= − 1

8πε0

q2

d

This is a factor of two larger, of course this is because as the charge moves away from the plane the charge on the planedisperses, as if the image charge were moving away at the same rate. We usually define potential energy by holding onecharge fixed while moving the other charge in from infinity. In this case, it is as if we are moving both charges in from infinityat the same time, resulting in a factor of two difference. One could also think of it in terms of the fields. Potential energyis stored in the field, but in the current situation, only half the field actually exists (the other half is cancelled out by theconductor) so there is only half as much energy stored in the field as in the corresponding situation with a real charge insteadof a conducting plane.

f.)

q = e = −1.602 · 10−19C d = 1A = 1 · 10−10m ε0 = 8.854 · 10−12F/m → W = 3.59959 eV.

2.9

The reaction of the sphere (radius a) to a uniform electric field E = Ez can be modeled using image charges of charge ±QaRat positions ±a

2

R with Q2πε0R2 = E and taking the limit R → ∞. In the limit, this becomes an electric dipole, with dipole

moment p = − 2Qa3

R2 z = −4πε0a3Ez. Which gives a surface charge density

σ0 = −3ε0E cos(θ) Uncharged (1)

σq = σ0 +q

4πa2Charged with charge q (2)

Where θ is measured from the positive z axis. Finally, recall that the force per unit area on a differential element is:

F⊥/A =σ2

2ε0(3)

To find the force on each hemisphere we must only integrate the above formula over each hemisphere. Note that the forceon the hemisperes will be equal and opposite, since the charge densities are equal and opposite. Furthermore, note that thecomponent of force on a surface element not in the z direction will cancel due to the azimuthal symmetry so we can integrateinstead only the z components of the force.

Applying these observations, the force required to keep the hemispheres from separating is given by

F = 2πa2

∫ π2

0

σ2

2εcos(θ) sin(θ)dθ (4)

5

With 4 and 2, we can calculate the force for the two scenarios given:

a.)

F0 = 9π(aE)2ε0

∫ π2

0

cos(θ)3 sin(θ)dθ

=9

4π(aE)2ε0

b.)

For a charged sphere we get

Fq = 2πa2

∫ π2

0

(−3ε0E cos(θ) + q4πa2 )2

2εcos(θ) sin(θ)dθ

= F0 + 2πa2

∫ π2

0

(−6ε0E cos(θ) q4πa2 + q2

16π2a4 )

2εcos(θ) sin(θ)dθ

= F0 +q

4ε0

∫ π2

0

(−6ε0E cos(θ) +q

4πa2) cos(θ) sin(θ)dθ

= F0 +q

4ε0(

q

8πa2− 2ε0E)

And the total force is given by

Fq = F0 +q

4ε0(

q

8πa2− 2ε0E) (5)

2.17

a.)

The free space greens function for three dimentions (apart from an arbitrary multiplicative constant) is

G(x, y, z;x′, y′, z′) =1

‖x− x′‖=

1√(x− x′)2 + (y − y′)2 + (z − z′)2

By defining a =√

(x− x′)2 + (y − x′)2 we can write the 2-dimentional free-space greens function as

G(x, y;x′, y′) = 2

∫ ∞0

1√a2 + x2

dx x = ‖z − z′‖

Which has the well-known solution (apart from an inessential, though infinite constant. This constant is hand-wavedaway by instead integrating to a large value Z instead of to infinity, but I think this will be good enough.)

G(x, y;x′, y′) = − log((x− x′)2 + (y − y′)2)

= − log(ρ2 + ρ′2 − 2ρρ′ cos(φ− φ′))

Where the last line follows from the ”law of cosines”.

b.)

The general solution to laplace’s equation is given by free-space solutions for ρ > ρ′ and ρ < ρ′ which are then patchedtogether at ρ = ρ′. To this end, we define two new functions, R and Φ, by the assumption G = R(ρ, ρ′)Φ(φ, φ′). Let uscompute the fourier coefficients of the Φ function for the laplace equation sourced by delta functions.

1

ρ

∂

∂ρ

(ρ∂R(ρ, ρ′)

∂ρ

)1√2π

∫eimφΦ(φ, φ′)dφ− R(ρ, ρ′)

ρ2

1√2π

∫m2eimφΦ(φ, φ′)dφ = − 4π√

2πδ(ρ− ρ′)eimφ

′

6

The surface terms in the second term of the l.h.s. canceled because of the cyclic boundary conditions on Φ. We can divideby eimφ

′and pull out an overall fourier integral to arrive at(

1

ρ

∂

∂ρρ∂R(ρ, ρ′)

∂ρ−m2R(ρ, ρ′)

ρ2

)∫ 2π

0

eim(φ−φ′)Φ(φ, φ′)dφ = −4πδ(ρ− ρ′)ρ

The fourier-transform factor is constant w.r.t. ρ and likewise the rest is a constant w.r.t. φ so w.l.o.g. we can set the fouriertransform factor equal to a constant, we choose 1, so that

∫ 2π

0

eim(φ−φ′)Φ(φ, φ′)dφ =1

2π→ Φ(φ, φ′) =

1

2πeim(φ′−φ)

∂

∂ρ

(ρ∂R(ρ, ρ′)

∂ρ

)− m2

ρ2R(ρ, ρ′) = −4π

ρδ(ρ− ρ′)

Thus we arrive at the general solution in the form

G =1

2π

∞∑−∞

eim(φ−φ′)Rm(ρ, ρ′)

where Rm satisfies the above equation.

c.)

To complete the solution we solve the above equation in the homogenious case and then patch together the solution at ρ = ρ′.The general solution of the homogenious equation is given in Jackson as

R(ρ, ρ′) = am(ρ′)ρm m ∈ Z\{0}R(ρ, ρ′) = a0(ρ′) + b0(ρ′) log(ρ) m = 0

We need two solutions one for ρ > ρ′ and one for ρ < ρ′, we will distinguish them with a subscript > and < respectivelyon the coefficients. Then, the two full solutions are give by

R<(ρ, ρ′) = a0<(ρ′) + b0<(ρ′) log(ρ) +∑

m∈Z\{0}

am<(ρ′)ρm

R>(ρ, ρ′) = a0>(ρ′) + b0>(ρ′) log(ρ) +∑

m∈Z\{0}

am>(ρ′)ρm

There are a few conditions we can impose right off the bat. Namely, the solution needs to be well-behaved at zero, andbe logarithmic as ρ → ∞. With this we can determine that b0<(ρ′) = 0, and that R< has no negative powers of ρ and R>has no positive powers ofρ. This leaves us with

R<(ρ, ρ′) = a0<(ρ′) +∑m>0

am<(ρ′)ρm

R>(ρ, ρ′) = a0>(ρ′) + b0>(ρ′) log(ρ) +∑m<0

am>(ρ′)ρm

Next, the solution must be symmetric. That is to say that if ρ < ρ′ then R<(ρ, ρ′) = R>(ρ′, ρ) and likewise, for ρ′ > ρ,R>(ρ, ρ′) = R<(ρ′, ρ). These conditions state that

a0<(ρ′) +∑m>0

am<(ρ′)ρm = a0>(ρ) + b0>(ρ) log(ρ′) +∑m<0

am>(ρ)ρ′m

a0>(ρ′) + b0>(ρ′) log(ρ) +∑m<0

am>(ρ′)ρm = a0<(ρ) +∑m>0

am<(ρ)ρ′m

7

Note that the solutions for various m must independently obey this symmetry, since they are all linearly independent.Then this tells us that a<0 = A log(ρ′), a>0 = 0 and b>0 = A, with A a yet to be determined constant. In addition it must bethat am>(ρ′) = Am(ρ′)m and am<(ρ′) = Am(ρ′)−m with Am constants that will be determined by considering the derivativediscontinuity. So that we finally have the solutions, us to an overall constant in each term

R<(ρ, ρ′) = A log(ρ′) +∑m>0

Am

(ρ

ρ′

)mR>(ρ, ρ′) = A log(ρ) +

∑m>0

Am

(ρ′

ρ

)m

The last consideration to make is that the difference in the derivatives at ρ = ρ′ should be −4π/ρ′. The derivativeevaluated (with respect to ρ) near ρ = ρ′ are

R<,ρ(ρ→ ρ′, ρ′) =∑m>0

Amm

ρ′

R>,ρ(ρ→ ρ′, ρ′) =A

ρ′−∑m>0

Amm

ρ′

Each term satisfies the discontinuity condition individually so we get that A = −4π and that Am = 2πm . So that the two

solutions can be written in the form

R<(ρ, ρ′) = −2π log(ρ′2) + 4π∑m>0

1

m

(ρ

ρ′

)mR>(ρ, ρ′) = −2π log(ρ2) + 4π

∑m>0

1

m

(ρ′

ρ

)mWhich can be conveniently expressed as

R<(ρ, ρ′) = −2π log(ρ2>) + 4π

∑m>0

1

m

(ρ<ρ>

)m

With ρ<(ρ>) the lesser (greater) of the two radii. Lastly we can plug this back into our original solution. Note that thefull solution (for each m) of the radial equation has a positive and a negative power of m. We threw out the poorly-behavedpowers of m, but for the well behaved ones we get 2 contributions, one from the positive m in the series, and one from thenegative m in the above series. Thus this ammounts to adding the positive-m term to the negative m term, causing theimaginary parts of the exponential to cancel (as needs to be the case to itnerprit this as a potential), leaving a factor of 2 cos,so in the end we get

G = − log(ρ2>) + 2

∞∑1

1

m

(ρ<ρ>

)mcos(m(φ− φ′))

2.20

a.)

Using the greens function of problem 2.17 the potential is given as

Φ(ρ, φ) =

∫σ(ρ′, φ′)

4πε0G(ρ, φ; ρ′, φ′)ρdρdφ

=λ

4πε0

3∑n=0

(−1)n

(− log(ρ2

>) + 2

∞∑m=1

1

m

(ρ<ρ>

)mcos(m(φ− nπ/2))

)

8

Where ρ>(ρ<) is the greater (lesser) of ρ and a, but since the log term has no φ dependence it cancels when the sum overn is performed, thus we can write the solution as

Φ(ρ, φ) =λ

2πε0

∞∑m=1

1

m

(ρ<ρ>

)m 3∑n=0

(−1)n cos(m(φ− nπ/2))

for a moment lets focus on the sum of cosines. Writing this out in full and manipulating we find that

cos(mφ)− cos(mφ−mπ/2) + cos(mφ− 2mπ/2)− cos(mφ− 3mπ/2) = (cos(mφ)− cos(mφ−mπ/2))(1 + (−1)m)

= { 2 cos(mφ) ifm = 4k + 2, k ∈ Z0 else

So that we can write the solution as

Φ(ρ, φ) =λ

πε0

∞∑k=0

1

2k + 1

(ρ<ρ>

)4k+2

cos((4k + 2)φ)

b.)

By manipulating our solution for Φ we quickly arrive at

Φ(ρ, φ) =λ

πε0Re

( ∞∑k=0

1

2k + 1

(ρ>ρ<

eiφ)4k+2

)Note the identity (page 75 of jackson) ∑

nodd

Zn

n=

1

2log

(1 + Z

1− Z

)

Where for our case we have Z =(ρ>e

iφ

ρ<

)2

so that

Φ =λ

2πε0Re

(log

(1 + (ρ>ρ< e

iφ)2

1− (ρ>ρ< eiφ)2

))

=λ

2πε0Re

(log

((1− iρ>ρ< e

iφ)(1 + iρ>ρ< eiφ)

(1− ρ>ρ<eiφ)(1 + ρ>

ρ<eiφ)

))

In the interior we have that ρ< = ρ and ρ> = a and the solution takes the form

Φ =λ

2πε0Re

(log

((z − ia)(z + ia)

(z − a)(z + a)

))= Re(w(z))

In the exterior ρ< = a and ρ> = ρ and we have

Φ =λ

2πε0Re

(log

((a− iz)(a+ iz)

(a− z)(a+ z)

))

But we can multiply the interior by i2 = −1 without changing the real part (since the real part of log depends only onthe magnitude of z).

9

Φ =λ

2πε0Re

(log

((ia+ z)(ia− z)(a− z)(a+ z)

))=

λ

2πε0Re

(log

((z − ia)(ia+ z)

(z − a)(z + a)

))= Re(w(z))

Thus, we can identify

Φ = Re(w(z))

So that the potential is exactly the real part of the funtion w(z) defined in Jackson. Now, as is stated in problem 2.3 ofJackson, the potential of an infinite line charge is the logarithm of the distance. This solution reflects that beautifully byconverting the cartesian components of position into a complex variable and then taking the real part. The four factors inthe argument of teh logarithm correspond exactly to two positive line charges at ±ia, or ±ay, and two negative line chargesat ±a, or ±ax, where x and y have been identified with the real and imaginary axes respectively.

c.)

The cartesian components of the field are given by

Ex = −(cos(θ)∂Φ

∂ρ− sin θ

ρ

∂Φ

∂φ)

Ey = −(sin(θ)∂Φ

∂ρ+

cos θ

ρ

∂Φ

∂φ)

Note that for ρ < a

∂Φ

∂ρ=

2λ

aπε0

∞∑k=0

ρ

a

4k+1cos((4k + 2)φ)

1

ρ

∂Φ

∂ρ= − 2λ

aπε0

∞∑k=0

ρ

a

4k+1sin((4k + 2)φ)

So the full expressions are

Ex = − 2λ

aπε0

( ∞∑k=0

ρ

a

4k+1(cos((4k + 2)φ) cos(θ) + sin(θ) sin((4k + 2)φ))

)

=2λ

aπε0

( ∞∑k=0

ρ

a

4k+1cos((4k + 1)φ)

)

Ey = − 2λ

aπε0

( ∞∑k=0

ρ

a

4k+1(cos((4k + 2)φ) sin(θ)− cos(θ) sin((4k + 2)φ))

)

=2λ

aπε0

( ∞∑k=0

ρ

a

4k+1sin((4k + 3)φ)

)

Keeping only the first two terms, we get

Ex = − 2λ

aπε0

(ρ

acos(φ) +

ρ

a

5cos(5φ)

)Ey = − 2λ

aπε0

(ρ

asin(3φ) +

ρ

a

5sin(7φ)

)

10

When y = 0, φ = 0 orπ and the x component of the electric field is given by

Ex = ± 2λ

aπε0

(ρ

a± ρ

a

5)

with the ± corresponding to the negative or positive axis respectively. The relative strength of the k = 1 contributionand k = 2 contribution is

k = 1

k = 2= ± 1

ρ4

2.26

Consider the 2-dimentional region ρ ≥ α, 0 ≤ φ ≤ β bounded by conducting surfaces held at zero potential.The large ρvalue of the potential Φ is determined by some unknown charge distribution at ρ >> α and thus this may be considered asa laplace problem with dirichlet boundary conditions.

a.)

The general solution to laplace’s equation in polar coordinaes is given, after separation of variables, by

R(p) = aρν + bρ−ν , Ψ(φ) = A cos(νφ) +B sin(νφ) | ν > 0

R(p) = a0 + b0 log(ρ), Ψ(φ) = A0 +B0φ | ν = 0

Given the boundary conditions we can conclude that A0 = B0 = A = 0 and that ν = nπβ . So, only the sin term of the

ν > 0 solution will appear. Lastly we have the condition that Φ(α, φ) = 0 which gives us the condition

a(α)ν + b(α)−ν = 0→ b = −a(α)2ν

Which gives us the general form of the solution

Φ(ρ, φ) =∑n∈Z

Cn(α)nπβ

(( ρα

)nπβ −

(α

ρ

)nπβ

)sin(

nπ

βφ)

b.)

Keeping only the lowest nonvanishing term we get

Φ0(ρ, φ) = C1(α)πβ

(( ρα

)πβ −

(α

ρ

)πβ

)sin(

π

βφ)

By which we can calculate the electric fields at the surfaces via the equation En = −∂Φ∂n to be

Eρ(ρ, φ) = C1π

βαπβ−1

(( ρα

)πβ−1

+

(α

ρ

)πβ+1

)sin(

π

βφ)

Eφ(ρ, φ) = −∂Φ

∂n= −1

ρ

∂Φ

∂φ= C1

π

βραπβ

(( ρα

)πβ −

(α

ρ

)πβ

)cos(

π

βφ)

11

With Eρ zero at φ = 0,β and Eφ = 0 at ρ = α. Then, the surface charge density is given by σ = ε0E⊥, or

σ |ρ=α = C1πε0βαπβ−1 sin(

π

βφ)

σ |φ=0 = C1πε0βρ

απβ

(( ρα

)πβ −

(α

ρ

)πβ

)= σ |φ=β

c.)

Consider β = π. The linear charge density looks like this

Figure 4: This plot shows the linear charge density. Note that the charge density on the cylinder goes as sin(φ) but since itis a half-circle, it works out that the linear charge density is constant! Of course, this is a plot of the non-dimentionalizedfunction with α = 1

Now lets convert the polar vectors into cartesian vectors! I will use x parallel to the plane and y perpendicular to theplane. Then the electric field in x and y coordinates, for large ρ as is required to be far away from the cylinder, is given by

(ExEy

)=

(Eρ cos(φ)− Eφ sin(φ)Eρ sin(φ) + Eφ cos(φ)

)

= C1

(sin(φ) cos(φ)− cos(φ) sin(φ)

cos(φ)2 + sin(φ)2

)

= E0

(01

)So, indeed the field is constant and perpendicular to the plane far from the cylinder.

Now, we have the charge density on the cylinder, we just need to integrate over φ. This works out to be

∫φ

σρ=α = 2αE0ε0

∫sin(φ) = 4E0ε0α.

Now, the surface charge density on an infinite conducting plane is σ = E0ε0 so indeed the surface charge on the cylinderis twice that contained on a strip on width 2α on an infinite conducting plane. Of course, all the surface charge densities areactually charge density per unit length in the z direction. Next we need to show that the extra charge density is taken fromthe surrounding area. To show this we will take the difference between the surface charge density of this system and that ofan infinite conducting plane, and integrate this from ρ = α to ∞, and multiply by 2, of course. This works out to be:

12

2

∫ ∞α

(σφ − σ0)dρ = 2E0ε0

∫ ∞α

(1− α2

ρ2− 1

)dρ

= 2E0ε0α

Exactly what we needed! Now, just for completeness’ sake, note that the difference between these two surface chargesgoes to zero for large ρ so indeed the statment that the charge is taken from nearby points on the plane is accurate. Thisalso answers the last question regarding the surface charge on a wide strip being the same wether or not the bump is present.Since the difference in surface charge densities between the two cases goes to zero as ρ becomes large, the total surface chargewithin that large ρ strip will be the same between the two cases, since the total charge density on the two surfaces is thesame, and the surface charge density at large ρ is the same.

13