Some Answers

Griffiths (3.12)

$$\Phi(x,y) = \frac{8V_0}{\pi} \sum_{j=0}^\infty \frac{1}{4j+... ...4j+2)\pi x}{a}\right) \sin\left(\frac{(4j+2)\pi y}{a}\right)$$ (1)

Griffiths (3.14)

$$\Phi(x,y) = \frac{4V_0}{\pi} \sum_{\textrm{odd} n}^\infty ... ...ht) \sin\left(n\pi y/a\right)} {\sinh\left(n\pi b/a\right)}$$ (2)

Griffiths (3.15)

$$\Phi(x,y) = \frac{16V_0}{\pi^2} \sum_{\textrm{odd } m}\sum_... ...\left(n\pi y/a\right)} {\sinh\left(\sqrt{m^2+n^2}\pi\right)}$$ (3)

Jackson (2.13)

(Unfortunately I chose wrong geometry. Change $\phi\rightarrow\phi-\pi/2$ to get Jackson's answer). For $\rho < b$, the general solution is

$$\Phi(\rho,\phi) = a_0 + \sum_{n=1}^\infty \rho^n(a_n\sin(n\phi)+b_n\cos(n\phi))$$ (4)

Boundary conditions:

$$\Phi(b,\phi) = V_1 \quad\quad 0\le\phi\le\pi/2$$ (5)

$$= V_2 \quad\quad \pi/2\le\phi\le\pi$$ (6)

Then, $a_0 = (V_1+V_2)/2$, $a_n=0$ for all even $n > 0$, $a_n=2(V_1-V_2)/(\pi nb^n)$ for all odd $n > 0$. $b_n = 0$ for all $n$. Thus,

$$\Phi(\rho,\phi) = (V_1+V_2)/2 + \frac{2(V_1-V_2)}{\pi} \sum_{n=0}^\infty \frac{1}{2n+1} \left(\frac{\rho}{b}\right)^{2n+1} \sin((2n+1)\phi)$$ (7)

$$= (V_1+V_2)/2 + \frac{2(V_1-V_2)}{\pi}\,\textrm{Im } \sum_{n=0}^\infty \frac{1}{2n+1} \left(\frac{\rho}{b}e^{i\phi}\right)^{2n+1}$$ (8)

$$= (V_1+V_2)/2 + \frac{(V_1-V_2)}{\pi}\,\textrm{Im } \ln \left( \frac{1+\frac{\rho}{b}e^{i\phi}} {1-\frac{\rho}{b}e^{i\phi}}\right)$$ (9)

$$= (V_1+V_2)/2 + \frac{(V_1-V_2)}{\pi} \tan^{-1} \left( \frac{2\rho b\sin\phi}{b^2-\rho^2}\right)$$ (10)

And the charge density is

$$\sigma(\phi) = -\epsilon_0 \frac{(V_1-V_2)}{\pi b\cos\phi}$$ (11)

Jackson (2.14)

Jackson (2.17)

Use following identities to prove the result:

$$\delta(\v{r}-\v{r'}) =\frac{1}{\rho}\delta(r-r')\delta(\phi-\phi')$$ (12)

$$\delta(\phi-\phi') = \frac{1}{2\pi}\sum_{-\infty}^{\infty} e^{i m(\phi-\phi')}$$ (13)

To find $g_m$ when $m=0$, choose

$$g_0(\rho,\rho') = A \quad\quad \rho < \rho'$$ (14)

$$= B\ln\rho \quad\quad \rho' < \rho$$ (15)

But then $g_0$ does not vanish as $\rho\rightarrow\infty$. When can you use this Green's Function?

Jackson (2.20)

Use the Green's Function derived above.