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Write the expression for the potential inside the rectangular pipe.

$V\left(x,y\right)=\frac{4V_0}{\mathrm{π}}\underset{n=1,3,5}{∑}\frac{1}{n}\left(\frac{cosh\left(\frac{n\mathrm{π}x}{a}\right)}{cosh\left(\frac{n\mathrm{π}b}{a}\right)}\right)sin\left(\frac{n\mathrm{π}y}{a}\right)$…… (1)

Here, $V_0$is the constant, are the dimensions of the rectangular pipe on x and y axes.

The below figure shows, the rectangular pipe is placed on the coordinates.

Write the expression for potential when $V_0\left(y\right)=V_0$.

$V\left(x,y\right)=\frac{4V_0}{\mathrm{π}}\underset{n=1,3,5,...}{∑}\frac{1}{n}\left(\frac{\sinh \left(\frac{n\mathrm{π}x}{a}\right)}{\sinh \left(\frac{n\mathrm{π}b}{a}\right)}\right)\sin \left(\frac{n\mathrm{π}y}{a}\right)$

Substitute $V_1$for $V_0$, $y$for $x_l$and X for Y .

$V\left(x,y\right)=\frac{4V_1}{\mathrm{π}}\underset{N=1,3,5,...}{∑}\frac{1}{n}\left(\frac{\sinh \left(\frac{n\mathrm{π}y}{a}\right)}{\sinh \left(\frac{n\mathrm{π}b}{a}\right)}\right)\sin \left(\frac{n\mathrm{π}x}{a}\right)$ …… (2)

Here, $V_1$is the constant voltage through one side of the rectangular cube.

The above expression defines that the rectangle is placed in reference with $x=0$axis.

Substitute $\left(x+\frac{a}{2}\right)$for $x_1$, a for$b_1$and b for $\left(\frac{a}{2}\right)$in equation (2).

$\begin{array}{rcl}V\left(x,y\right)& =& \frac{4V_1}{\mathrm{π}}\underset{N=1,3,5,...}{∑}\frac{1}{n}\left(\frac{\sinh \left(\frac{n\mathrm{π}y}{2b}\right)}{\sinh \left(\frac{n\mathrm{π}a}{2b}\right)}\right)\sin \left(\frac{n\mathrm{π}\left(x+\frac{a}{2}\right)}{2b}\right)\\ & =& \frac{4V_1}{\mathrm{π}}\underset{N=1,3,5,...}{∑}\frac{1}{n}\left(\frac{\sinh \left(\frac{n\mathrm{π}y}{2b}\right)}{\sinh \left(\frac{n\mathrm{π}a}{2b}\right)}\right)\sin \left(\frac{n\mathrm{π}\left(x+b\right)}{2b}\right)\\ & & \end{array}$

From the above expression it is clear that the rectangle is mirrored by y axis.

Now, find the potential inside the rectangle pipe with two sides facing.

$\begin{array}{rcl}V& =& \frac{4V_0}{\mathrm{π}}\underset{n=1,3,5,...}{∑}\frac{1}{n}\left(\frac{\cosh \left(\frac{n\mathrm{π}x}{a}\right)}{\cosh \left(\frac{n\mathrm{π}b}{a}\right)}\right)\sin \left(\frac{n\mathrm{π}y}{a}\right)+\frac{4V_1}{\mathrm{π}}\underset{N=1,3,5,...}{∑}\frac{1}{n}\left(\frac{\sinh \left(\frac{n\mathrm{π}y}{2b}\right)}{\sinh \left(\frac{n\mathrm{π}a}{2b}\right)}\right)\sin \left(\frac{n\mathrm{π}\left(x+b\right)}{2b}\right)\\ & =& \frac{4}{\mathrm{π}}\underset{n=1,3,5,...}{∑}\frac{1}{n}\left(V_0\frac{\cosh \left(\frac{n\mathrm{π}x}{a}\right)}{\cosh \left(\frac{n\mathrm{π}b}{a}\right)}\sin \left(\frac{n\mathrm{π}y}{a}\right)+V_1\frac{\sinh \left(\frac{n\mathrm{π}y}{2b}\right)}{\sinh \left(\frac{n\mathrm{π}a}{2b}\right)}\sin \left(\frac{n\mathrm{π}\left(x+b\right)}{2b}\right)\right)\\ & & \end{array}$

Hence, total potential inside a rectangular pipe with two sides at potential $V_0$and at $V_1$is .

$\frac{4}{\mathrm{π}}\underset{n=1,3,5,...}{∑}\frac{1}{n}\left(V_0\frac{\cosh \left(\frac{n\mathrm{π}x}{a}\right)}{\cosh \left(\frac{n\mathrm{π}b}{a}\right)}\sin \left(\frac{n\mathrm{π}y}{a}\right)+V_1\frac{\sinh \left(\frac{n\mathrm{π}y}{2b}\right)}{\sinh \left(\frac{n\mathrm{π}a}{2b}\right)}\sin \left(\frac{n\mathrm{π}\left(x+b\right)}{2b}\right)\right)$