PHYSICS 505: CLASSICAL ELECTRODYNAMICS HOMEWORK 2 5

2.8 Let us consider two parallel, straight line charges separated by a distance R and with equal andopposite linear charge densities ±λ.

a) Let us find the surfaces of constant potential. We will show that these are circular cylinders.With the anticipation only available in hindsight, let us work with polar coordinates—

suppressing the longitudinal direction—defined such that the origin is located at thecenter of one of the circles of constant potential. Specifically, let us say that the firstline charge is located at (η, 0) and the second at (R + η, 0). Although we do not yetknow the displacement η from the first line charge, we know that if η is properlyspecified, the potential should be independent of the angular coordinate.

In general, the potential will be the linear superposition the potentials of each of thetwo line-charges. Having calculated this in homework one, we see

ϕ(ρ, θ) =λ

4πε0

[log(η2 − 2ηρ cos θ + ρ2)− log((R + η)2 − 2(R + η)ρ cos θ + ρ2)

].

The requirement that ϕ be constant on a cylinder centered at the origin is equivalent tothe condition that ∂ϕ

∂θ = 0. Using this identity, we compute

∂ϕ

∂θ= 0 =

λ

2πε0

(ηρ sin θ

η2 − 2ηρ cos θ + ρ2− (R + η)ρ sin θ

(R + η)2 − 2(R + η)ρ cos θ + ρ2

),

=λρ sin θ

2πε0

(η

η2 − 2ηρ cos θ + ρ2− (R + η)

(R + η)2 − 2(R + η)ρ cos θ + ρ2

),

∴ η

η2 − 2ηρ cos θ + ρ2=

(R + η)(R + η)2 − 2(R + η)ρ cos θ + ρ2

.

Expanding and collecting terms, we arrive at the constraint

ρ2 = η(η + R).

We must now determine the radius ρ such that ϕ(ρ, θ) = V ∀ θ. To do this, we willinsert the expression derived above for ρ2 into our expression for the potential withthe condition that ϕ = V . Because this will be independent of the angle, let uschoose θ = π/2 to simplify our expressions. We will need to remember this choicelater.

Computing directly, we see that

ϕ(ρ, θ) = V =λ

4πε0log

(η2 − 2ηρ cos θ + ρ2

(R + η)2 − 2(R + η)ρ cos θ + ρ2

),

=λ

4πε0log

(η2 + η2 + ηR

(R + η)2 + η2 + ηR

),

=λ

4πε0log

(2η2 + ηR

2η2 + 3ηR + R2

).

By rearranging terms, exponentiating, and simplifying, we obtain the quadratic ex-pression

η22(1− e4πε0V/λ

)+ ηR

(1− 3e4πε0V/λ

)−R2e4πε0V/λ = 0.

Simply using the quadratic formula and a bit of algebra, we see that

η = − R

2(1− e4πε0V/λ)

(1− 3e4πε0V/λ ± 1± e4πε0V/λ

),

∴ η = Re4πε0V/λ

1− e4πε0V/λ=

R

e−4πε0V/λ − 1= − Re2πε0V/λ

2 sinh(2πε0V/λ)or η = −R/2.

Notice that the − solution of the quadratic, η = −R/2, simply demonstrates that be-tween the two wires the potential is constant along the line (ρ, π/2). We could haveanticipated this solution because we simplified our work to determine η using thecondition that at ρ, ϕ(ρ, π/2) = ϕ(ρ,−π/2) = V . The fact that this cannot be a so-lution is evident from the last line of our expression for ϕ(ρ, θ) above: the numeratorin the logarithm vanishes if η = −R/2 and hence this is not a physical solution.

6 JACOB LEWIS BOURJAILY

Using our expression for ρ, we have

ρ2 = R2 e8πε0V/λ

(1− e4πε0V/λ)2+ R2 e4πε0V/λ

(1− e4πε0V/λ)= R2 e4πε0V/λ

(1− e4πε0V/λ)2,

∴ ρ = Re2πε0V/λ

1− e4πε0V/λ=

Re−2πε0V/λ

e−4πε0V/λ − 1= − R

2 sinh(2πε0V/λ).

Therefore, we have shown that if

ρ =R

2 sinh(2πε0V/λ)and η = − Re2πε0V/λ

2 sinh(2πε0V/λ),

the potential along a right circular cylinder in the longitudinal direction of radius ρpositioned precisely η to the left of the first line-charge is constant and equal to V .

It is important to notice that the sign of η is ‘somewhat’ important. In particular, ifV/λ < 0 then η is a positive displacement to the left of and encloses the first line-charge. However, if V/λ > 0, then η is a negative displacement to the left of the firstline-charge; in particular, −η > R and so the cylinder is displaced to the right of andencloses the second line-charge.

b) We are to demonstrate that the capacitance per unit length of two cylindrical conductors withradii a and b separated by a distance d is given by

C =2πε0

arccosh(

d2−a2−b2

2ab

) .

proof: We know from the definition of capacitance that it is given by C/L = λV+−V−

. Wemust show that the above equation is consistent with this fact.

First, we know that the required problem is equivalent to one in which there are onlytwo line charges, separated by a distance R = η− + η+ with radii such that thetwo cylinders are at constant voltages V+ and V−, respectively. Specifically, we see,reorganizing the expressions derived above,

d = η− − η+ =R

(e−4πε0V−/λ − 1)− R

(e−4πε0V+/λ − 1),

and,

a =Re−2πε0V+/λ

(e−4πε0V+/λ − 1)and b =

Re−2πε0V−/λ

(e−4πε0V−/λ − 1).

Looking at the equation we must verify, it may be helpful to compute d2−a2−b2

2ab and seeif it points us toward the solution. Let us first compute the numerator. We find that

d2 − a2 − b2 =

R2

(e−4πε0V−

λ − 1)2 −

2R2

(e−4πε0V−

λ − 1) (

e−4πε0V+

λ − 1) +

R2

(e−4πε0V+

λ − 1)2 −

R2e−4πε0V+

λ

(e−4πε0V+

λ − 1))2 −

R2e−4πε0V−

λ

(e−4πε0V−

λ − 1)2

,

= R2

1− e

−4πε0V+λ

(e−4πε0V+

λ − 1)2 +

1− e−4πε0V−

λ

(e−4πε0V−

λ − 1)2 +

2(e−4πε0V−

λ − 1)(

e−4πε0V+

λ − 1)

,

=R2

(e−4πε0V−

λ − 1)(

e−4πε0V+

λ − 1)

(−e

−4πε0V+λ − e

−4πε0V−λ

).

Now, because

2ab = − 2R2e−2πε0(V+−V−)/λ

(e−4πε0V−

λ − 1)(

e−4πε0V+

λ − 1) ,

we have thatd2 − a2 − b2

2ab=

12e−2πε0(V+−V−)/λ

(e−4πε0V+/λ + e−4πε0V−/λ

),

=12

(e2πε0(V+−V−)/λ + e−2πε0(V+−V−)/λ

),

PHYSICS 505: CLASSICAL ELECTRODYNAMICS HOMEWORK 2 7

∴ d2 − a2 − b2

2ab= cosh

(2πε0 (V+ − V−)

λ

).

Therefore, we see that the capacitance per unit length agrees with the desired formula:

C =2πε0

arccosh(

d2−a2−b2

2ab

) =2πε0

2πε0(V+−V−)λ

=λ

V+ − V−.

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c) Let us verify that the expression above for the capacitance agrees with that derived in homeworkone for a similar problem in the limit where d2 >> a2 + b2.

proof: Similar to our previous results, we can make extraordinary progress by simply guess-ing the form of the answer. First, it should be true that

arccosh(

d2 − a2 − b2

2ab

)= log ξ,

for some ξ. In fact, because we are simply verifying a result, we could presume it tobe true and then it would be obvious what ξ must be. However, let us simply see ifsuch an equation makes sense.

By taking the hyperbolic cosine of each side, we effectively exponentiate the logarithm.Therefore, ξ must satisfy

d2 − a2 − b2

2ab=

12

(elog ξ + e− log ξ

)=

12

(ξ +

1ξ

)=

ξ2 + 12ξ

.

This reduces to the following quadratic equation,

ξ2 − ξ

(d2 − a2 − b2

ab

)+ 1 = 0.

It is clear that in the limit where d2−a2−b2

ab >> 1, the solutions approach the trivialsolution and that given by

ξ ≈ d2 − a2 − b2

ab.

Furthermore, in the limit of d2 >> a2 + b2, this approaches ξ ∼ d2

ab .Therefore, we see that

C =πε0

arccosh(

d2−a2−b2

2ab

) ≈ πε0

log(

d2−a2−b2

ab

) ∼ πε0

log(

d2

ab

) ∼ πε0

2 log(

d√ab

) .

∴ C ∼ 2πε0

log(

d√ab

) .

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