Question

A cubical box (sides of length a) consists of five metal plates, which are welded together and grounded (Fig. 3.23). The top is made of a separate sheet of metal, insulated from the others, and held at a constant potential${\mathit{V}}_{\mathbf{0}}$. Find the potential inside the box. [What should the potential at the center $\left(a/2,a/2,a/2\right)$be ? Check numerically that your formula is consistent with this value.]

Short answer

Answer

The potential inside the box is,

$V_{\left(x,y,\right)}=\frac{16V_0}{{\pi }^2}\sum _{n=1,3,5}^a\sum _{m=1,3,5}^a\frac{1}{nm}\sin \left(\frac{n\mathrm{\pi x}}{a}\right)\frac{\sin h\left(\pi \sqrt{n^2+m^2}{\displaystyle \raisebox{1ex}{$z$}\!\left/ \!\raisebox{-1ex}{$a$}\right.}\right)}{\sin h\left(\pi \sqrt{n^2+m^2}\right)}$

Step-by-step solution

Given data

The figure of cube is follow.

Here, a is the side of cube, V is the potential.

Determine boundary conditions

Write the Laplace equation in Cartesian co-ordinate system.

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

Let’s consider $V=X\left(x\right)Y\left(y\right)Z\left(z\right)$

Substitute $X\left(x\right)Y\left(y\right)Z\left(z\right)$for $V$in equation (1) and divide by $V$.

$\frac{1}{x}\frac{{\partial }^{2}X}{\partial x^2}+\frac{1}{Y}\frac{{\partial }^{2}X}{\partial x^2}+\frac{1}{Z}\frac{{\partial }^{2}Z}{\partial z^2}=0$

Then,

$$\begin{gathered} X\left(x\right)=A\sin kx+B\cos kx \\ Y\left(y\right)=C\sin \left(Iy\right)+D\cos \left(Iy\right) \\ Z\left(z\right)=E{e}^{\sqrt{k^2+I^{2}z}} \end{gathered}$$

By boundary condition (a) to the above equations,

Substitute 0 for x in the equation.

$$\begin{gathered} X\left(x\right)=A\sin kx+B\cos kx \\ X\left(0\right)=A\sin k\left(0\right)+B\cos \left(0\right) \\ 0=0+B\left(I\right) \\ 0=B \end{gathered}$$

Thus, $B=0$

The boundary condition (b) to the above equations,

Substitute a for x in the equation,

$$\begin{gathered} X\left(a\right)=A\sin k\left(a\right)+B\cos k\left(a\right) \\ 0=A\sin k\left(a\right)+B\cos k\left(a\right) \\ 0=A\sin k\left(a\right)+B\cos k\left(a\right) \end{gathered}$$

Therefore,

$k=\frac{n\mathrm{\pi }}{a}$

The boundary condition (c) to the above equations,

Substitute 0 for y in the equation,

localid="1655805955716" $$\begin{gathered} Y\left(y\right)=C\sin \left(Iy\right)+D\cos \left(Iy\right) \\ Y\left(0\right)=C\sin \left(0\right)+D\cos \left(0\right) \\ 0=C\left(0\right)+D\left(1\right) \\ 0=D \end{gathered}$$

Thus,

$D=0$

The boundary condition (d) to the above equations,

Substitute a for y in the equation

$$\begin{gathered} Y\left(y\right)=C\sin \left(Iy\right)+D\cos \left(Iy\right) \\ Y\left(0\right)=C\sin \left(Ia\right)+D\cos \left(Ia\right) \end{gathered}$$

Thus,

$I=\frac{m\mathrm{\pi }}{a}$

The boundary condition (e) to the above equations,

Substitute 0 for z in the equation,

$$\begin{gathered} Z\left(z\right)=E{e}^{\sqrt{k^2+I^{2}z}}+G{e}^{-\sqrt{k^2+I^{2}z}} \\ Z\left(0\right)=E{e}^{\sqrt{k^2+I^2}\left(0\right)}+G{e}^{-\sqrt{k^2+I^2}\left(0\right)} \\ 0=E+G \end{gathered}$$

Hence,

$$\begin{gathered} E+G=0 \\ E=-G \end{gathered}$$

Determine Potential

As,

$Z\left(z\right)=E{e}^{\sqrt{k^2+I^{2}z}}+G{e}^{-\sqrt{k^2+I^{2}z}}$

Substitute $-E$for $G$in above equation.

$Z\left(z\right)=E{e}^{\sqrt{k^2+l^2}z}-{E^e}^{-\sqrt{k^2+l^2}z}$

Now, Substitute $\frac{n\mathrm{\pi }}{a}$for k and $\frac{\mathrm{m\pi }}{a}$for I

$$\begin{gathered} Z\left(z\right)=E\left\{exp\left[\sqrt{{\left(\frac{n\mathrm{\pi }}{a}\right)}^2+{\left(\frac{m\mathrm{\pi }}{a}\right)}^{2}z}\right]-exp\left[-\sqrt{{\left(\frac{n\mathrm{\pi }}{a}\right)}^2+{\left(\frac{m\mathrm{\pi }}{a}\right)}^{2}z}\right]\right\} \\ =E\left\{exp\left[\frac{\mathrm{\pi }}{a}\sqrt{n^2+m^2}z\right]-exp\left[-\frac{\pi }{a}\sqrt{n^2+m^2}z\right]\right\} \\ =E\left\{exp\left[\pi \sqrt{n^2+m^2}\frac{z}{a}\right]-exp\left[-\mathrm{\pi }\sqrt{n^2+m^2}\frac{\mathrm{z}}{\mathrm{a}}\right]\right\} \\ =2E\left\{\frac{exp\left[\pi \sqrt{n^2+m^2}\frac{z}{a}\right]-exp\left[-\mathrm{\pi }\sqrt{n^2+m^2}\frac{\mathrm{z}}{\mathrm{a}}\right]}{2}\right\} \end{gathered}$$

Using the trigonometry formula of $\sin h$the equation becomes,

$Z\left(z\right)=2E\sinh \left[\left(\pi \sqrt{n^2+m^2}\right)\frac{z}{a}\right]$

$Z\left(z\right)=2E\sin \left[\left(\mathrm{\pi }\sqrt{{\mathrm{n}}^2+{\mathrm{m}}^2}\right)\frac{z}{a}\right]$

Therefore,

$Z\left(z\right)=2E\sin h\left[\left(\mathrm{\pi }\sqrt{{\mathrm{n}}^2+{\mathrm{m}}^2}\right)\frac{z}{a}\right]$

Then,

$V_{\left(x,y,z\right)}=\sum _{n=1}^a\sum _{m=1}^a{C}_{n,m}\sin \left(\frac{n\mathrm{\pi x}}{a}\right)\sin \left(\frac{m\mathrm{\pi y}}{a}\right)\sin h\left(\mathrm{\pi }\sqrt{{\mathrm{n}}^2+{\mathrm{m}}^2}\frac{\mathrm{z}}{\mathrm{a}}\right)$ …… (2)

Apply $z=a$to equation (2)

$$\begin{gathered} V_0\sum _{n=1}^a\sum _{m=1}^a{C}_{n,m}\sin h\left(\mathrm{\pi }\sqrt{{\mathrm{n}}^2+{\mathrm{m}}^2}\frac{\mathrm{a}}{\mathrm{a}}\right)\sin \left(\frac{n\mathrm{\pi x}}{a}\right)\sin \left(\frac{m\mathrm{\pi y}}{a}\right) \\ {V}_0\sum _{n=1}^a\sum _{m=1}^a{C}_{n,m}\sin h\left(\mathrm{\pi }\sqrt{{\mathrm{n}}^2+{\mathrm{m}}^2}\right)\sin \left(\frac{n\mathrm{\pi x}}{a}\right)\sin \left(\frac{m\mathrm{\pi y}}{a}\right) \\ \left(C_{n,m}\right)\sin h\left(\mathrm{\pi }\sqrt{{\mathrm{n}}^2+{\mathrm{m}}^2}\right)={\left(\frac{2}{a}\right)}^2{V}_0\underset{0}{\overset{a}{\int }}\underset{0}{\overset{a}{\int }}\sin \left(\frac{n\mathrm{\pi x}}{a}\right)\sin \left(\frac{m\mathrm{\pi y}}{a}\right)dxdy \\ {C}_{n,m}\left(\sin h\left(\mathrm{\pi }\sqrt{{\mathrm{n}}^2+{\mathrm{m}}^2}\right)\right)=\left\{\begin{array}{ll}0& normiseven\\ \frac{16V}{{\mathrm{\pi }}^2\mathrm{nm}}& ifbothareodd\end{array}\right. \end{gathered}$$

Thus, the potential is,

$V_{\left(x,y,z\right)}=\frac{16V_0}{{\mathrm{\pi }}^2}\sum _{n=1,3,5,...}^a\sum _{m=1,3,5,...}^a\frac{1}{nm}\sin \left(\frac{n\mathrm{\pi x}}{a}\right)\sin \left(\frac{m\mathrm{\pi y}}{a}\right)\frac{\sin h\left(\pi \sqrt{n^2+m^2}z/a\right)}{\sin h\left(\mathrm{\pi }\sqrt{{\mathrm{n}}^2+{\mathrm{m}}^2}\right)}$