91Ó°ÊÓ

Consider that, the configuration is independent of z,

$\frac{{\mathbf{∂}}^{\mathbf{2}}\mathbf{V}}{\mathbf{∂}{\mathbf{x}}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{∂}}^{\mathbf{2}}\mathbf{V}}{\mathbf{∂}{\mathbf{y}}^{\mathbf{2}}}\mathbf{=}\mathbf{0}$ …… (1)

Here, V is the potential, x,y are the Cartesian co-ordinate constant.

The two infinite grounded metal plates are running parallel to XZ plane.

Take, $\mathit{V}\left(X,Y\right)\mathbf{=}\mathit{X}\left(x\right)\mathbf{,}\mathit{Y}\left(y\right)$

Apply the boundary conditions,

$\mathit{V}\left(x,y\right)\mathbf{=}\underset{\mathbf{a}\mathbf{=}\mathbf{1}}{\overset{\mathbf{∞}}{\mathbf{∑}}}{\mathit{C}}_{\mathbf{n}}{\mathit{e}}^{\mathbf{-}\mathbf{n}\mathbf{Ï€³\ae }\mathbf{/}\mathbf{a}}\mathit{s}\mathit{i}\mathit{n}\frac{\mathbf{n}\mathbf{Ï€²â}}{\mathbf{a}}$ …… (2)

Here, ${\mathit{C}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{2}}{\mathbf{a}}\underset{\mathbf{0}}{\overset{\mathbf{a}}{\mathbf{∫}}}{\mathit{V}}_{\mathbf{0}}\left(y\right)\mathit{s}\mathit{i}\mathit{n}\frac{\mathbf{n}\mathbf{π}\mathbf{y}}{\mathbf{a}}$

The boundary conditions are $\mathbf{0}\mathbf{<}\mathit{y}\mathbf{<}\mathit{a}$is split in to $\mathbf{0}\mathbf{<}\mathit{y}\mathbf{<}\mathit{a}\mathbf{/}\mathbf{2}$and $\mathit{a}\mathbf{/}\mathbf{2}\mathbf{<}\mathit{y}\mathbf{<}\mathit{a}$.

Write the limits for the potential $\mathit{V}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{)}$

$\mathit{V}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{\{}\begin{array}{l}\mathbf{+}{\mathbf{V}}_{\mathbf{0}}\\ \mathbf{-}{\mathbf{V}}_{\mathbf{0}}\end{array}\begin{array}{r}\mathbf{0}\mathbf{<}\mathbf{y}\mathbf{<}\mathbf{a}\mathbf{/}\mathbf{2}\\ \mathbf{a}\mathbf{/}\mathbf{2}\mathbf{<}\mathbf{y}\mathbf{<}\mathbf{a}\end{array}\mathbf{\}}$

Write the expression for ${\mathit{C}}_{\mathbf{n}}$.

$$\begin{gathered} {\mathit{C}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{2}}{\mathbf{a}}\left[\underset{0}{\overset{a/2}{∫}}{V}_{0}sin\frac{n\mathrm{Ï€²â}}{a}dy-\underset{a/2}{\overset{a}{∫}}{V}_{0}sin\frac{n\mathrm{Ï€}y}{a}dy\right] \\ \mathbf{}\mathbf{}\mathbf{}{\mathbf{}}_{\mathbf{}}\mathbf{=}\frac{\mathbf{2}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{a}}\mathbf{×}\frac{\mathbf{a}}{\mathbf{n}\mathbf{Ï€}}\left\{{\left[-cos\frac{n\mathrm{Ï€}y}{a}\right]}_0^{a/2}+{\left[cos\frac{n\mathrm{Ï€}y}{a}\right]}_0^{a/2}\right\} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{2}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{n}\mathbf{Ï€}}\left\{-cos\frac{n\mathrm{Ï€}}{2}+cos0+cosn\mathrm{Ï€}-\cos \frac{\mathrm{²ÔÏ€}}{2}\right\} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{2}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{n}\mathbf{Ï€}}\left\{1+{\left(-1\right)}^n-2cos\frac{n\mathrm{Ï€}}{2}\right\} \end{gathered}$$

The value of ${\mathit{C}}_{\mathbf{n}}$is zero for all odd values of n,

If n = 1,3,5,7,.....then ${\mathit{C}}_{\mathbf{n}}\mathbf{=}\mathbf{0}$.

If n = 4then the value of ${\mathit{C}}_{\mathbf{n}}$is,

$$\begin{gathered} {\mathit{C}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{2}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{n}\mathbf{Ï€}}\left\{1+{\left(-1\right)}^4-2cos\frac{4\mathrm{Ï€}}{2}\right\} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{2}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{n}\mathbf{Ï€}}\left\{1+1-2\right\} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0} \end{gathered}$$

Thus, if n = 4,8,12,...... then ${\mathit{C}}_{\mathbf{n}}\mathbf{=}\mathbf{0}$.

For all values of n = 2,6,10,......then the value of ${\mathit{C}}_{\mathbf{n}}$is

${\mathit{C}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{8}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{n}\mathbf{Ï€}}$.

Substitute the all n values in equation (2).

Now, write the solution using the boundary conditions.

$\mathit{V}\left(x.y\right)\mathbf{=}\underset{\mathbf{n}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{6}\mathbf{,}\mathbf{10}}{\mathbf{∑}}\mathbf{}{\mathit{C}}_{\mathbf{n}}{\mathit{e}}^{\mathbf{-}\mathbf{n}\mathbf{Ï€³\ae }\mathbf{/}\mathbf{a}}\mathit{s}\mathit{i}\mathit{n}\frac{\mathbf{n}\mathbf{Ï€²â}}{\mathbf{a}}$

Substitute $\frac{\mathbf{8}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{Ï€}}$for ${\mathit{C}}_{\mathbf{n}}$in above equation.

$\mathit{V}\mathbf{(}\mathbf{x}\mathbf{.}\mathbf{y}\mathbf{)}\mathbf{=}\frac{\mathbf{8}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{Ï€}}\underset{\mathbf{n}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{6}\mathbf{,}\mathbf{10}}{\mathbf{∑}}\mathbf{}\frac{{\mathbf{e}}^{\mathbf{-}\mathbf{²ÔÏ€³\ae }\mathbf{/}\mathbf{a}}\mathbf{s}\mathbf{i}\mathbf{n}\frac{\mathbf{²ÔÏ€²â}}{\mathbf{a}}}{\mathbf{n}}$

Thus, the potential in the infinite slot of boundary is $\mathit{V}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\frac{\mathbf{8}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{Ï€}}\underset{\mathbf{n}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{6}\mathbf{,}\mathbf{10}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}}{\mathbf{∑}}\frac{{\mathbf{e}}^{\mathbf{-}\mathbf{²ÔÏ€³\ae }\mathbf{/}\mathbf{a}}\mathbf{sin}{\displaystyle \frac{\mathbf{²ÔÏ€²â}}{\mathbf{a}}}}{\mathbf{n}}$.