A rectangular pipe, running parallel to the z-axis (from $\mathbf{-}\mathbf{\infty }$to $\mathbf{+}\mathbf{\infty }$), has three grounded metal sides, at $\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathit{y}\mathbf{=}\mathit{a}$and $\mathit{x}\mathbf{=}\mathbf{0}$The fourth side, at $\mathit{x}\mathbf{=}\mathit{b}$, is maintained at a specified potential ${\mathit{V}}_{\mathbf{0}}\left(y\right)$.

(a) Develop a general formula for the potential inside the pipe.

(b) Find the potential explicitly, for the case ${\mathit{V}}_{\mathbf{0}}\left(y\right)\mathbf{=}{\mathit{V}}_{\mathbf{0}}$(a constant).

Rectangular pipe is extending from $-a$to $+a$parallel to Z-axis. Three meal plate are grounded at $y=0,y=a,x=0$.

At $x=b$the plate is maintained at constant potential $V_0\left(y\right)$. Here, Laplacian is independent of Z. Then Laplace equation is,

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

Here, x and y are the coordinates.

Now, write the boundary conditions.

$$\begin{gathered} i)V\left(x,0\right)=0 \\ ii)V\left(x,a\right)=0 \\ iii)V\left(0,y\right)=0 \\ iv)V\left(b,y\right)=V_0\left(y\right) \end{gathered}$$

Here, V is the potential at different points.

By separation of variable solving equation (1),

$V\left(x,y\right)=\left(A{e}^{kx}+B{e}^{-kx}\right)\left(C\sin ky+D\cos ky\right)$ …… (2)

By applying boundary condition (1) into the equation (2),

$0=\left(A+B\right)C\sin ky$

Then, $D=0$

Apply the boundary condition (iii) to equation (2).

$0=\left(A+B\right)C\sin ky$

Then, $A=-B$.

By applying boundary condition (ii) in (2),

Then, $$\begin{gathered} V\left(x,y\right)=AC\left(e^{\frac{n\mathrm{\pi x}}{e}}-e^{\frac{-n\mathrm{\pi x}}{e}}\right)\sin \frac{n\mathrm{\pi y}}{a} \\ =2AC\sin h\frac{n\mathrm{\pi x}}{a}\sin \frac{n\mathrm{\pi y}}{a} \end{gathered}$$

a)

Write the general solution.

$V\left(x,y\right)=\sum _{n=1}^{\infty }{C}_n\sin h\left(\frac{n\mathrm{\pi x}}{a}\right)\sin \left(\frac{n\mathrm{\pi y}}{a}\right)$ …… (3)

Apply boundary condition (iv) in (3),

$V_0\left(y\right)=\sum _{n=1}^{\infty }{C}_n\sin h\left(\frac{n\mathrm{\pi x}}{a}\right)\sin \left(\frac{n\mathrm{\pi y}}{a}\right)$

By applying Fourier’s theorem,

Then,

$C_n\sin h\left(\frac{n\mathrm{\pi x}}{a}\right)=\frac{2}{a}\underset{0}{\overset{a}{\int }}{V}_0\left(y\right)\sin \left(\frac{n\mathrm{\pi y}}{a}\right)$

Solve for $C_n$.

$C_n=\frac{2}{a\sin h\left(\frac{n\mathrm{\pi x}}{a}\right)}\underset{0}{\overset{a}{\int }}{V}_0\left(y\right)\sin \left(\frac{n\mathrm{\pi y}}{a}\right)dy$ …… (4)

b)

Now,

$C_n=\frac{2}{a\sin h\left(\frac{n\mathrm{\pi x}}{a}\right)}\underset{0}{\overset{a}{\int }}{V}_0\left(y\right)\sin \left(\frac{n\mathrm{\pi y}}{a}\right)dy$

Given,

$V_0\left(y\right)=V_0$

Therefore,

$$\begin{gathered} V_0\left(y\right)=\frac{2V_0}{a\sin h\left({\displaystyle \frac{n\mathrm{\pi b}}{a}}\right)}\frac{a}{n\mathrm{\pi }}{\left[-\cos \frac{n\mathrm{\pi y}}{a}\right]}_0^a \\ =\frac{2V_0}{a\sin h\left({\displaystyle \frac{n\mathrm{\pi b}}{a}}\right)}\frac{a}{n\mathrm{\pi }}\left[-\cos n\mathrm{\pi }+1\right] \end{gathered}$$

Write the boundary conditions for cosine and sine.

role="math" localid="1655810358911" $1-\cos n\mathrm{\pi }=0$if $n$

$1-\cos n\mathrm{\pi }=2$if $n$is odd.

Then,

$C_n=\frac{4V_0}{n\mathrm{\pi sinh}\left({\displaystyle \frac{\mathrm{n\pi b}}{\mathrm{a}}}\right)}$

Then, the potential value is,

$V\left(x,y\right)=\frac{4V_0}{\mathrm{\pi }}\sum _{k=1,3,5}\frac{\sin h\left({\displaystyle \frac{n\mathrm{\pi x}}{a}}\right)\sin h\left({\displaystyle \frac{n\mathrm{\pi y}}{a}}\right)}{n\sin h\left({\displaystyle \frac{n\mathrm{\pi b}}{a}}\right)}$.