(a) Using Prob. 3.15b (with b = a):
V(x, y)=\frac{4 V_{0}}{\pi} \sum_{n \text { odd }} \frac{\sinh (n \pi x / a) \sin (n \pi y / a)}{n \sinh (n \pi)} .
\sigma(y)=-\left.\epsilon_{0} \frac{\partial V}{\partial x}\right|_{x=0}=-\left.\epsilon_{0} \frac{4 V_{0}}{\pi} \sum_{n \text { odd }}\left(\frac{n \pi}{a}\right) \frac{\cosh (n \pi x / a) \sin (n \pi y / a)}{n \sinh (n \pi)}\right|_{x=0}
=-\frac{4 \epsilon_{0} V_{0}}{a} \sum_{n \text { odd }} \frac{\sin (n \pi y / a)}{\sinh (n \pi)} .
\lambda=\int_{0}^{a} \sigma(y) d y=-\frac{4 \epsilon_{0} V_{0}}{a} \sum_{n \text { odd }} \frac{1}{\sinh (n \pi)} \int_{0}^{a} \sin (n \pi y / a) d y .
But \int_{0}^{a} \sin (n \pi y / a) d y=-\left.\frac{a}{n \pi} \cos (n \pi y / a)\right|_{0} ^{a}=\frac{a}{n \pi}[1-\cos (n \pi)]=\frac{2 a}{n \pi}(\text { since } n \text { is odd }) .
=-\frac{8 \epsilon_{0} V_{0}}{\pi} \sum_{n \text { odd }} \frac{1}{n \sinh (n \pi)}=-\frac{\epsilon_{0} V_{0}}{\pi} \ln 2 .
[Summing the series numerically (using Mathematica) gives 0.0866434, which agrees precisely with ln 2/8. The series can be summed analytically, by manipulation of elliptic integrals—see “Integrals and Series, Vol. I: Elementary Functions,” by A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (Gordon and Breach, New York, 1986), p. 721. I thank Ram Valluri for calling this to my attention.]
Using Prob. 3.54 (with b = a/2):
V(x, y)=V_{0}\left[\frac{y}{a}+\frac{2}{\pi} \sum_{n} \frac{(-1)^{n} \cosh (n \pi x / a) \sin (n \pi y / a)}{n \cosh (n \pi / 2)}\right] .
\sigma(x)=-\left.\epsilon_{0} \frac{\partial V}{\partial y}\right|_{y=0}=-\left.\epsilon_{0} V_{0}\left[\frac{1}{a}+\frac{2}{\pi} \sum_{n}\left(\frac{n \pi}{a}\right) \frac{(-1)^{n} \cosh (n \pi x / a) \cos (n \pi y / a)}{n \cosh (n \pi / 2)}\right]\right|_{y=0}
=-\epsilon_{0} V_{0}\left[\frac{1}{a}+\frac{2}{a} \sum_{n} \frac{(-1)^{n} \cosh (n \pi x / a)}{\cosh (n \pi / 2)}\right]=-\frac{\epsilon_{0} V_{0}}{a}\left[1+2 \sum_{n} \frac{(-1)^{n} \cosh (n \pi x / a)}{\cosh (n \pi / 2)}\right] .
\lambda=\int_{-a / 2}^{a / 2} \sigma(x) d x=-\frac{\epsilon_{0} V_{0}}{a}\left[a+2 \sum_{n} \frac{(-1)^{n}}{\cosh (n \pi / 2)} \int_{-a / 2}^{a / 2} \cosh (n \pi x / a) d x\right] .
But \int_{-a / 2}^{a / 2} \cosh (n \pi x / a) d x=\left.\frac{a}{n \pi} \sinh (n \pi x / a)\right|_{-a / 2} ^{a / 2}=\frac{2 a}{n \pi} \sinh (n \pi / 2) .
=-\frac{\epsilon_{0} V_{0}}{a}\left[a+\frac{4 a}{\pi} \sum_{n} \frac{(-1)^{n} \tanh (n \pi / 2)}{n}\right]=-\epsilon_{0} V_{0}\left[1+\frac{4}{\pi} \sum_{n} \frac{(-1)^{n} \tanh (n \pi / 2)}{n}\right]
=-\frac{\epsilon_{0} V_{0}}{\pi} \ln 2
[The numerical value is -0.612111, which agrees with the expected value (\ln 2-\pi) / 4 .]
(b) From Prob. 3.24:
V(s, \phi)=a_{0}+b_{0} \ln s+\sum_{k=1}^{\infty}\left(a_{k} s^{k}+b_{k} \frac{1}{s^{k}}\right)\left[c_{k} \cos (k \phi)+d_{k} \sin (k \phi)\right].
In the interior (s < R) b0 and bk must be zero (ln s and 1/s blow up at the origin). Symmetry \Rightarrow d_{k}=0 . So
V(s, \phi)=a_{0}+\sum_{k=1}^{\infty} a_{k} s^{k} \cos (k \phi) .
At the surface:
V(R, \phi)=\sum_{k=0} a_{k} R^{k} \cos (k \phi)=\left\{\begin{array}{l}V_{0}, \text { if }-\pi / 4<\phi<\pi / 4 \\0, \text{otherwise }\end{array}\right.
Fourier’s trick: multiply by \cos \left(k^{\prime} \phi\right) \text { and integrate from }-\pi \text { to } \pi :
\sum_{k=0}^{\infty} a_{k} R^{k} \int_{-\pi}^{\pi} \cos (k \phi) \cos \left(k^{\prime} \phi\right) d \phi=V_{0} \int_{-\pi / 4}^{\pi / 4} \cos \left(k^{\prime} \phi\right) d \phi=\left\{\begin{array}{l}V_{0} \sin \left(k^{\prime} \phi\right) /\left.k^{\prime}\right|_{-\pi / 4} ^{\pi / 4}=\left(V_{0} / k^{\prime}\right) \sin \left(k^{\prime} \pi / 4\right), \text { if } k^{\prime} \neq 0 ,\\V_{0} \pi / 2, \text { if } k^{\prime}=0 .\end{array}\right.
But
\int_{-\pi}^{\pi} \cos (k \phi) \cos \left(k^{\prime} \phi\right) d \phi= \begin{cases}0, & \text { if } k \neq k^{\prime} \\ 2 \pi, & \text { if } k=k^{\prime}=0 , \\ \pi, & \text { if } k=k^{\prime} \neq 0.\end{cases}
So 2 \pi a_{0}=V_{0} \pi / 2 \Rightarrow a_{0}=V_{0} / 4 ; \pi a_{k} R^{k}=\left(2 V_{0} / k\right) \sin (k \pi / 4) \Rightarrow a_{k}=\left(2 V_{0} / \pi k R^{k}\right) \sin (k \pi / 4)(k \neq 0) ; hence
V(s, \phi)=V_{0}\left[\frac{1}{4}+\frac{2}{\pi} \sum_{k=1}^{\infty} \frac{\sin (k \pi / 4)}{k}\left(\frac{s}{R}\right)^{k} \cos (k \phi)\right] .
Using Eq. 2.49, and noting that in this case \hat{ n }=-\hat{ s }
\sigma=-\epsilon_{0} \frac{\partial V}{\partial n} (2.49)
\sigma(\phi)=\left.\epsilon_{0} \frac{\partial V}{\partial s}\right|_{s=R}=\left.\epsilon_{0} V_{0} \frac{2}{\pi} \sum_{k=1}^{\infty} \frac{\sin (k \pi / 4)}{k R^{k}} k s^{k-1} \cos (k \phi)\right|_{s=R}=\frac{2 \epsilon_{0} V_{0}}{\pi R} \sum_{k=1}^{\infty} \sin (k \pi / 4) \cos (k \phi) .
We want the net (line) charge on the segment opposite to V_{0}(-\pi<\phi<-3 \pi / 4 \text { and } 3 \pi / 4<\phi<\pi) :
\lambda=\int \sigma(\phi) R d \phi=2 R \int_{3 \pi / 4}^{\pi} \sigma(\phi) d \phi=\frac{4 \epsilon_{0} V_{0}}{\pi}\sum_{k=1}^{\infty} \sin (k \pi / 4) \int_{3 \pi / 4}^{\pi} \cos (k \phi) d \phi
=\frac{4 \epsilon_{0} V_{0}}{\pi} \sum_{k=1}^{\infty} \sin (k \pi / 4)\left[\left.\frac{\sin (k \phi)}{k}\right|_{3 \pi / 4} ^{\pi}\right]=-\frac{4 \epsilon_{0} V_{0}}{\pi} \sum_{k=1}^{\infty} \frac{\sin (k \pi / 4) \sin (3 k \pi / 4)}{k} .
| k |
\sin (k \pi / 4) |
\sin (3 k \pi / 4) |
product |
| 1 |
1 / \sqrt{2} |
1 / \sqrt{2} |
1/2 |
| 2 |
1 |
-1 |
-1 |
| 3 |
1 / \sqrt{2} |
1 / \sqrt{2} |
1/2 |
| 4 |
0 |
0 |
0 |
| 5 |
-1 / \sqrt{2} |
-1 / \sqrt{2} |
1/2 |
| 6 |
-1 |
1 |
-1 |
| 7 |
-1 / \sqrt{2} |
-1 / \sqrt{2} |
1/2 |
| 8 |
0 |
0 |
0 |
\lambda=-\frac{4 \epsilon_{0} V_{0}}{\pi}\left[\frac{1}{2} \sum_{1,3,5 \ldots} \frac{1}{k}-\sum_{2,6,10, \ldots} \frac{1}{k}\right]=-\frac{4 \epsilon_{0} V_{0}}{\pi}\left[\frac{1}{2} \sum_{1,3,5 \ldots} \frac{1}{k}-\frac{1}{2} \sum_{1,3,5, \ldots} \frac{1}{k}\right]=0 .
Ouch! What went wrong? The problem is that the series \sum(1 / k) is divergent, so the “subtraction” \infty-\infty is suspect. One way to avoid this is to go back to V(s, \phi) , calculate \epsilon_{0}(\partial V / \partial s) \text { at } s \neq R , and save the limit s → R until the end:
\sigma(\phi, s) \equiv \epsilon_{0} \frac{\partial V}{\partial s}=\frac{2 \epsilon_{0} V_{0}}{\pi} \sum_{k=1}^{\infty} \frac{\sin (k \pi / 4)}{k} \frac{k s^{k-1}}{R^{k}} \cos (k \phi)
=\frac{2 \epsilon_{0} V_{0}}{\pi R} \sum_{k=1}^{\infty} x^{k-1} \sin (k \pi / 4) \cos (k \phi) \quad \text { (where } x \equiv s / R \rightarrow 1 at the end).
\lambda(x) \equiv \sigma(\phi, s) R d \phi=-\frac{4 \epsilon_{0} V_{0}}{\pi} \sum_{k=1}^{\infty} \frac{1}{k} x^{k-1} \sin (k \pi / 4) \sin (3 k \pi / 4)
=-\frac{4 \epsilon_{0} V_{0}}{\pi}\left[\frac{1}{2 x}\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\cdots\right)-\frac{1}{x}\left(\frac{x^{2}}{2}+\frac{x^{6}}{6}+\frac{x^{10}}{10}+\cdots\right)\right]
=-\frac{2 \epsilon_{0} V_{0}}{\pi x}\left[\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\cdots\right)-\left(x^{2}+\frac{x^{6}}{3}+\frac{x^{10}}{5}+\cdots\right)\right] .
But (see math tables) : \ln \left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\cdots\right) .
=-\frac{2 \epsilon_{0} V_{0}}{\pi x}\left[\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right)-\frac{1}{2} \ln \left(\frac{1+x^{2}}{1-x^{2}}\right)\right]=-\frac{\epsilon_{0} V_{0}}{\pi x} \ln \left[\left(\frac{1+x}{1-x}\right)\left(\frac{1+x^{2}}{1-x^{2}}\right)\right]
=-\frac{\epsilon_{0} V_{0}}{\pi x} \ln \left[\frac{(1+x)^{2}}{1+x^{2}}\right] ; \quad \lambda=\lim _{x \rightarrow 1} \lambda(x)=\frac{-\epsilon_{0} V_{0}}{\pi} \ln 2 .
