Phys 742 – Assignment 3 – Due 13 Feb 96

1. (Similar to Jackson’s 2.6; see also 2.14d)

A parallel plate capacitor has a small hemispherical boss of radius a on one of its plates. Assume that A is the area of each plate, d the separation between them, with \sqrt{A} >> d >> a. To be definite, assume that the plates are flat circular disks perpendicular to the z axis and put the boss at the center of the lower plate. However, the result should hold more generally, as long as a is much smaller than the distance of the boss from the closest edge.

(a) Where on the lower plate does the maximum electric field occur, and how much larger than average is it? Comment on the chances of being struck by lightning in hilly country.

(b) Show that the presence of the boss does not change the capacitance, to order a^{2}/A. Do you expect higher order corrections to increase or decrease C?

2. (Similar to Jackson’s 2.5)

(a) Solve the problem of an uncharged conducting cylindrical shell of radius a in a uniform field E_{0} perpendicular to its axis. You can use the method of images if you wish. More simply, you can superimpose the external field and that of an induced line dipole, so that the shell is an equipotential.

(b) If the cylinder is cut into half by a plane perpendicular to the field, what force per unit length is required to keep the half-cylinders from separating?

(c) Same question as in (b), when the charge on the shell is λ per unit length.

3.

Jackson’s 2.4, but you can take parts (b) and (d) together and look at Estevez and Suen, page 42, for guidance. Be sure to correct the misprints, however.

4.

(a) For the problem of Jackson’s Fig. 2.10, what is the behavior of Φ(x, y) near the origin, i.e., for ρ << a? Sketch equipotentials and field lines in this (two-dimensional) corner.

(b) For the same geometry as in (a), put Φ = 0 on the planes x = 0 and y = 0, Φ = V on the plane x = a. What is now the behavior of Φ(x, y) near the origin? Sketch equipotentials and field lines in this (two-dimensional) corner.

(c) For the problem of Jackson’s Fig. 2.9, what is the behavior of Φ(x, y, z) near the origin? What are the charge densities on the three faces as they meet in this (threedimensional) corner?

Answer 1

(a) This problem is equivalent to that of a sphere in a uniform field. Putting the origin at the center of the boss and using Jackson’s (2.13), the potential is

Φ = −E_{0} (r − \frac{a^{3}}{r^{2}} ) cos θ = −E_{0} (1 − \frac{a^{3}}{r^{3}} ) z  (1)

where E_{0} is the field on the lower plate, far from the boss. It is also the average field on the plate, to order a^{2}/A. The field on the boss is 3E_{0} cos θ. It has a maximum value 3E_{0} at θ = 0. Hilltops are to be avoided during thunderstorms; it is best to be at the foot of the hill (if the top is a safe distance away).

(b) The charge density on the boss is

σ_{boss} = \frac{3}{4π} E_{0} cos θ

as in Jackson’s (2.15). The charge on the boss is

Q_{boss} = \frac{3}{2} E_{0}a^{2} ∫_{0}^{π/2} cos θ sin θ dθ = \frac{3}{4} E_{0}a^{2}

The charge density on the flat part of the lower plate is

σ_{flat} = −\frac{1}{4π} \frac{∂Φ}{∂z}|_{z=0} = \frac{1}{4π} E_{0} (1 − \frac{a^{3}}{ρ^{3}} )

and to order a^{2}/A the charge is

Q_{flat} ≃ E_{0} (\frac{1}{4π} (A − πa^{2}) − \frac{1}{2} ∫_{a}^{∞} \frac{a^{3}}{ρ^{3}} ρdρ) = E_{0} (\frac{A}{4π} − \frac{3}{4} a^{2})

(The approximation of extending the integral to infinity neglects a term of order E_{0}a^{3}/\sqrt{A}). The total charge on the lower plate is then Q_{lower} = Q_{boss} + Q_{flat} = E_{0}A/4π, unchanged by the boss. This is enough to conclude that C is unchanged to this order.

To make sure, we should also compute the charge on the upper plate. We must correct the potential in order to make it vanish on the upper plate. This is achieved by adding an image of the potential of the boss, so that

Φ ≃ −E_{0} (1 − \frac{a^{3}}{r^{3}} ) z + E_{0} \frac{a^{3}}{r^{′ 3}} (z − 2d)

where r′ = \sqrt{ρ^{2} + (z − 2d)^{2}} (This potential does not vanish on the lower plate, hence we should add another correction of higher order in a/d, and so on ad infinitum). The charge density is

σ_{upper} = \frac{1}{4π} \frac{∂Φ}{∂z}|_{z=d}

The term E_{0}a^{3}z/r^{3} gives a charge

\frac{1}{2} E_{0}a^{3} ∫_{0}^{∞} \frac{ρ^{2} − 2d^{2}}{(ρ^{2} + d^{2})^{5/2}} ρ dρ = 0

(Without computing the integral, we can note the total Gaussian flux of a dipole through a plane is zero). The correction term E_{0}a^{3}(z − 2d)/r^{′ 3} gives the same result, zero charge. (It is always true that adding an image term doubles E_{z}, and thus doubles the charge). The remaining term −E_{0}z gives Q_{upper} = −E_{0}A/4π, as anticipated.

We confirm then that, to order a^{2}/A, C is unchanged:

C = \frac{Q_{lower}}{E_{0}d} = −\frac{Q_{upper}}{E_{0}d} = \frac{A}{4πd}

We expect that the boss will increase C, because it increases the area of one plate and decreases the average distance between the plates. However, the change is not of order a^{2}/A, it is smaller.

Answer 2

(a) Φ = −E_{0}x + px/ρ^{2} with p = E_{0}a^{2}. Put it another way, Φ = −E_{0}a(1 − a^{2}/ρ^{2})ρ cos ϕ (b) At ρ = a, the field is radial and E_{ρ} = 2E_{0} cos ϕ. The force per unit area is

E_{ρ}^{2}/8π = (E_{0}^{2}/2π) cos^{2} ϕ

The force per unit length on the right half-cylinder is:

F_{x} = (aE_{0}^{2}/2π) ∫_{−π/2}^{π/2} cos^{3} ϕ dϕ = 2aE_{0}^{2}/3π

F_{y} = (aE_{0}^{2}/2π) ∫_{−π/2}^{π/2} cos^{2} ϕ sin ϕ dϕ = 0

Answer 3

Center one cylinder at the origin, the other at x = x_{0}. The potential is

Φ = − ln (x^{2} + y^{2}) + ln ((x − x_{0})^{2} + y^{2})

The equipotential Φ = V_{1} has equation

e^{V_{1}}(x^{2} + y^{2}) = (x − x_{0})^{2} + y^{2}
(e^{V_{1}} − 1) (x^{2} + y^{2}) + 2xx_{0} = x_{0}^{2}

Comparing this with the standard equation for a circle of radius a, centered at x = x_{c} :

x^{2} + y^{2} − 2xx_{c} = a^{2} − x_{c}^{2}

we see that the equipotential is a circle centered at

x_{1} = \frac{−x_{0}}{e^{V_{1}} − 1}

and that the radius of thic circle, a_{1}, is given by

a_{1}^{2} = x_{1}^{2} + \frac{x_{0}^{2}}{e^{V_{1}} − 1}

Using the shorthand notation

A_{1} = \frac{1}{e^{V_{1}} − 1}

we have

a_{1}^{2} = x_{0}^{2} (A_{1}^{2} + A_{1}) = x_{0}^{2}A_{1}^{2}e^{V_{1}}  (2)

Similarly, the equipotential Φ = V_{2} is a circle of radius a_{2} given by

a_{2}^{2} = x_{0}^{2} (A_{2}^{2} + A_{2}) = x_{0}^{2}A_{2}^{2}e^{V_{2}}  (3)

where

A_{2} = \frac{1}{e^{V_{2}} − 1}

and this circle is centered at x_{2} = −x_{0}A_{2}. The separation of the centers of the two circles is then

x_{2} − x_{1} = x_{0}(A_{1} − A_{2})  (4)

Since a_{1}, a_{2} and x_{2} − x_{1} are given, we can solve eqs. (2), (3) and (4) for x_{0}, A_{1} and A_{2} and then compute V_{1} − V_{2}. This would be long and tedious, but we know what we must get and with perfect foresight we compute

a_{1}^{2} + a_{2}^{2} − (x_{2} − x_{1})^{2} = x_{0}^{2} (A_{1} + A_{2} + 2A_{1}A_{2})
= x_{0}^{2}A_{1}A_{2} (e^{V_{1}} + e^{V_{2}})

and

a_{1}a_{2} = x_{0}^{2}\sqrt{A_{1}^{2}e^{V_{1}}A_{2}^{2}e^{V_{2}}} = x_{0}^{2} |A_{1}A_{2}| e^{(V_{1}+V_{2})/2}

to obtain

\frac{|a_{1}^{2} + a_{2}^{2} − (x_{2} − x_{1})^{2}|}{2a_{1}a_{2}} = \frac{e^{V_{1}} + e^{V_{2}}}{2e^{(V_{1}+V_{2})/2}} = cosh \frac{V_{1} − V_{2}}{2}

(b) One method of solution is to write

Φ = \frac{V x}{a} + ϕ  (5)

where ϕ satisfies the boundary conditions: ϕ = 0 on the planes x = 0 and x = a for y > 0, ϕ = −V x/a on the plane y = 0 for 0 < x < a. Similarly to part (a), we obtain

ϕ = ∑_{n=1}^{∞} A_{n} sin(nπx/a) exp(−ny/a)  (6)

with

A_{n} = −\frac{2}{a} ∫_{0}^{a} \frac{V x}{a} sin \frac{nπx}{a} dx = \frac{2V}{nπ} (−1)^{n}

The series can be summed explicitly, giving

ϕ = −\frac{2V}{π} Im ln(1 + e^{iπz/a})

It is convenient to rewrite eq. (5) as Φ = (2V/π) Im ln (e^{iπz/2a}) + ϕ, which leads to

Φ = −\frac{2V}{π} Im ln (cos \frac{πz}{2a} )  (7)

The explicit form of Φ(x, y) is obtained by using cos(α + iβ) = cos α cosh β − i sin α sinh β. It is

Φ = \frac{2V}{π} arctan (tan \frac{πx}{2a} tanh \frac{πy}{2a} )  (8)

For small x and y (in the corner)

Φ ≃ \frac{πV}{2a^{2}} xy  (9)

The equipotentials are hyperbolae with asymptotes on the x and y axes. The flux lines are hyperbolae with asymptotes at ±45^{∘}, as can be seen by introducing the stream function Ψ through the complex potential

Φ + iΨ = \frac{2iV}{π} ln (cos \frac{πz}{2a} )

For small x and y

Φ + iΨ ≃ \frac{2iV}{π} ln (1 − \frac{1}{2} (\frac{πz}{2a} )^{2}) ≃ −\frac{iπV z^{2}}{4a^{2}}

This gives back eq. (9) and also gives

Ψ ≃ −\frac{πV}{4a^{2}} (x^{2} − y^{2})

The graph shows lines of constant Φ (solid) and constant Ψ (dashed).

/x

Alternative method

Use the Fourier integral representation

Φ(x, y) = ∫_{0}^{∞} \frac{dk}{2π} A(k) sin ky sinh kx  (10)

and impose the boundary condition (for 0 < y < ∞)

V = ∫_{0}^{∞} \frac{dk}{2π} A(k) sin ky sinh ka  (11)

(If you are worried by the behavior for y → ∞, use V e^{−αy} with α → 0^{+}). The functions sin ky have the orthonormality property

∫_{0}^{∞} sin ky sin qy dy = \frac{π}{2} δ(k − q)  (12)

for k̇ > 0 and q > 0. This is well known, but if proof is needed start from

∫_{−∞}^{∞} e^{iqx} e^{ikx}dx = 2πδ(k + q)
∫_{−∞}^{∞} e^{−iqx} e^{ikx}dx = 2πδ(k − q)

and subtract side by side. Noting that k + q cannot vanish, obtain 2i ∫_{−∞}^{∞} e^{iqx} sin kx dx = −2πδ(k − q) and then fold the integral back into the interval from 0 to ∞.

From (11) and (12) we obtain

A(k) sinh ka = 4V/k

(or 4V k/(α^{2} + k^{2}) for the rigor-minded). Thus (putting α = 0 with impunity):

Φ(x, y) = \frac{2V}{π} ∫_{0}^{∞} \frac{sin ky sinh kx}{k sinh ka} dk  (13)

Amazingly, the integral is found in Gradshteyn-Ryzhik, page 513, and we obtain

Φ(x, y) = \frac{2V}{π} arctan (tan \frac{πx}{2a} tanh \frac{πy}{2a} )

in agreement with the other method. For small x and y this reduces to

Φ(x, y) ≃ \frac{πV}{2a^{2}} xy

as before. We can also get this result directly from eq. (13). For small x and y we get

Φ(x, y) ≃ \frac{2V}{π} ∫_{0}^{∞} \frac{xy k dk}{sinh ka}

and we can use the standard result (Gradshteyn-Ryzhik, page 347)

∫_{0}^{∞} \frac{q dq}{sinh q} = \frac{1}{4}π^{2}

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(c) Following Jackson, page 70,

Φ = ∑_{n,m=1}^{∞} A_{nm} sin \frac{nπx}{a} sin \frac{mπy}{b} sinh(γ_{mn}z)

with

γ_{mn} = π\sqrt{\frac{n^{2}}{a^{2}} + \frac{m^{2}}{b^{2}}}

A_{nm} = \frac{4V}{π^{2}nm sinh(γ_{mn}c)}

For small x, y, z

Φ ≃ \frac{4V xyz}{abc} ∑_{n,m=1}^{∞} \frac{γ_{mn}c}{sinh(γ_{mn}c)}

The sum must be evaluated numerically (see below for typical values). In any case, the charge density on the xy face, for instance, is proportional to −V xy.

For a = b = c we get numerically

∑_{n=1}^{∞} ∑_{m=1}^{∞} \frac{π\sqrt{m^{2} + n^{2}}}{sinh (π\sqrt{m^{2} + n^{2}})} = 0.1346866

For a = b >> c we get

\frac{1}{abc} ∑_{n,m=1}^{∞} \frac{γ_{mn}c}{sinh(γ_{mn}c)} → \frac{1}{π^{2}} ∫_{0}^{∞} dk_{x} ∫_{0}^{∞} dk_{y} \frac{k}{sinh kc} = \frac{1}{2πc^{3}} ∫_{0}^{∞} \frac{q^{2}dq}{sinh q}

Using a formula in Gradshteyn-Ryzhik, page 348:

∫_{0}^{∞} \frac{q^{2}dq}{sinh q} = \frac{2^{3} − 1}{4} Γ(3)ζ(3) = \frac{7}{2} ζ(3) = 4.207199

as can also be checked numerically. Thus in 3-d we have

Φ ≃ \frac{2V xyz}{πc^{3}} \frac{7}{2}ζ(3) = 2.678386 \frac{V xyz}{c^{3}}

The corresponding result in 2-d, in the present notation, is

Φ ≃ \frac{2V xy}{πc^{2}} \frac{3}{2}ζ(2) = 1.570796 \frac{V xy}{c^{2}}

This is the same as found in part (b), since ζ(2) = \frac{1}{6}π^{2}.