91Ó°ÊÓ

The boundary conditions are:

$$\begin{gathered} \mathit{V}\mathbf{(}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \\ \mathit{V}\mathbf{(}\mathit{y}\mathbf{=}\mathit{a}\mathbf{)}\mathbf{=}{\mathit{V}}_{\mathbf{0}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)} \\ \mathit{V}\mathbf{(}\mathit{x}\mathbf{=}\mathit{b}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{3}\mathbf{\right)} \\ \mathit{V}\mathbf{(}\mathit{x}\mathbf{=}\mathbf{-}\mathit{b}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{4}\mathbf{\right)} \end{gathered}$$

The Laplace's equation in two dimensions separated into two variables is written as,

$$\begin{gathered} \frac{d^{2}X}{d{t}^2}=k^{2}X \\ \frac{d^{2}Y}{d{t}^2}=k^{2}Y......\left(5\right) \end{gathered}$$

The potential is separated into two parts, one for k = 0, and the other for $k≠0$.

For k = 0, the solution to equations (5) is,

$$\begin{gathered} X=Ax+B \\ Y=Cy+D \end{gathered}$$

Since the situation defined in the problem is symmetric in x .

$A=0$

B is be absorbed into C and D.

Thus,

$V(x,y)=Cy+D$

Substitution of the first boundary condition mentioned in equation (1).

D = 0

Substitution of the second boundary condition mentioned in equation (2).

$C=\frac{V_0}{a}$

Thus, the k= 0 part of the potential is,

$V=\frac{V_{0}y}{a}$

This takes care of the first boundary condition.

The boundary condition for the rest of the potential $\overline{V}(x,y)$ at y = a is 0 and at $\overline{V}(x,y)$at $x=Â\pm b$is$-\frac{V_{0}y}{a}$, the general solution to which is

role="math" localid="1657621230345" $\stackrel{¯}{V}(x,y)=\underset{}{∑C_n\cos h\left(\frac{n\mathrm{Ï„}\mathrm{Ï„}x}{a}\right)}\sin \left(\frac{n\mathrm{Ï€²â}}{a}\right)$

Substitution of the boundary condition gives

$$\begin{gathered} \stackrel{¯}{V}(x,y)=\underset{}{∑C_n\cos h\left(\frac{n\mathrm{Ï„}\mathrm{Ï„}x}{a}\right)}\sin \left(\frac{n\mathrm{Ï€²â}}{a}\right) \\ =-\frac{V_{0}y}{a} \end{gathered}$$

Application of Fourier series gives

$$\begin{gathered} \underset{}{∑}{C}_n\cos h\left(\frac{n\mathrm{Ï€²ú}}{a}\right){∫}_0^a\sin \left(\frac{n\mathrm{Ï€}y}{a}\right)\sin \left(\frac{n^{\text{'}}\mathrm{Ï€²ú}}{a}\right)dy=\frac{V_0}{a}{∫}_0^{a}y\sin \left(\frac{n^{\text{'}}\mathrm{Ï€²ú}}{a}\right)dy \\ \frac{a}{2}{C}_n\cos h\left(\frac{n\mathrm{Ï€²ú}}{a}\right)=\frac{V_0}{a}{\left[{\left(\frac{a}{n\mathrm{Ï€}}\right)}^2\sin \left(\frac{n\mathrm{Ï€}y}{a}\right)-\left(\frac{ay}{n\mathrm{Ï€}}\right)\cos \left(\frac{n\mathrm{Ï€}y}{a}\right)\right]}_0^a \\ {C}_n=\frac{2V_0}{a^2\cos h\left(\frac{n\mathrm{Ï€²ú}}{a}\right)}\left(\frac{a^2}{n\mathrm{Ï€}}\right)\cos \left(n\mathrm{Ï€}\right) \\ {\mathrm{C}}_{\mathrm{n}}=\frac{2{\mathrm{V}}_0}{\mathrm{²ÔÏ€}}\frac{{(-1)}^{\mathrm{n}}}{\cosh \left(\frac{\mathrm{²ÔÏ€²ú}}{\mathrm{a}}\right)} \end{gathered}$$

Thus, the net potential is