Question

For the infinite slot (Ex. 3.3), determine the charge density $\mathit{\sigma }\left(y\right)$on

the strip at $\mathit{x}\mathbf{=}\mathbf{0}$, assuming it is a conductor at constant potential ${\mathit{V}}_{\mathbf{0}}$.

Short answer

Answer

The equation for the charge density on the strip at $x=0$is $\sigma \left(y\right)=\frac{4{\epsilon }_0{V}_0}{a}\sum _{n=1,3,5...}\sin \left(\frac{n\mathrm{\pi y}}{a}\right)$.

Step-by-step solution

Define functions

Write the expression for the potential $\mathit{V}\left(x,y\right)$in the infinite slot.

$\mathit{V}\left(x,y\right)\mathbf{=}\frac{\mathbf{4}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{\pi }}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{5}\mathbf{.}\mathbf{.}\mathbf{.}}\frac{\mathbf{1}}{\mathbf{n}}{\mathit{e}}^{\frac{\mathbf{-}\mathbf{n}\mathbf{\pi x}}{\mathbf{a}}}\mathit{s}\mathit{i}\mathit{n}\left(\frac{n\pi y}{a}\right)$ …… (1)

Here, $V_0$is the constant potential along the conductor, $x$is the x-coordinate, $y$is the y-coordinate, and $n$is the positive integer.

Determine the charge density

Derive the charge density in terms of electric potential.

$$\begin{gathered} \sigma =-e_0\frac{\partial V}{\partial n} \\ \sigma \left(y\right)=-e_0{\left[\frac{\partial V}{\partial x}\right]}_{x=0} \end{gathered}$$ …… (2)

Substitute $\frac{4V_0}{\mathrm{\pi }}\sum _{n=1,3,5...}\frac{1}{n}{e}^{\frac{-n\mathrm{\pi x}}{a}}\sin \frac{n\mathrm{\pi y}}{a}$for $V\left(x,y\right)$in equation (2).

$$\begin{gathered} {\overline{)\sigma \left(y\right)={\epsilon }_0\frac{\partial }{\partial x}\left\{\frac{4V_0}{\mathrm{\pi }}\underset{}{\sum \frac{1}{n}{e}^{\frac{n\mathrm{\pi x}}{a}}}\sin \left(\frac{n\mathrm{\pi y}}{a}\right)\right\}}}_{x=0} \\ {\overline{)={\epsilon }_0\frac{4V_0}{\mathrm{\pi }}\frac{\partial }{\partial x}\left\{\underset{}{\sum \frac{1}{n}{e}^{\frac{n\mathrm{\pi x}}{a}}}\sin \left(\frac{n\mathrm{\pi y}}{a}\right)\right\}}}_{x=0} \\ {\overline{)={\epsilon }_0\frac{4V_0}{\mathrm{\pi }}\sum \frac{1}{n}\left\{\left(-\frac{nx}{a}\right)\underset{}{e^{\frac{n\mathrm{\pi x}}{a}}}\sin \left(\frac{n\mathrm{\pi y}}{a}\right)\right\}}}_{x=0} \\ {\overline{)={\epsilon }_0\frac{4V_0}{\mathrm{\pi }}\frac{1}{n}\left(\frac{nx}{a}\right)\underset{}{\sum \frac{1}{n}{e}^{\frac{n\mathrm{\pi x}}{a}}}\sin \left(\frac{n\mathrm{\pi y}}{a}\right)}}_{x=0} \end{gathered}$$

$\sigma \left(y\right)=\frac{4{\epsilon }_0{V}_0}{a}\sum _{n=1,3,5...}\sin \left(\frac{n\mathrm{\pi y}}{a}\right)$

Hence, the equation for the charge density on the strip at $x=0$is $\sigma \left(y\right)=\frac{4{\epsilon }_0{V}_0}{a}\sum _{n=1,3,5...}\sin \left(\frac{n\mathrm{\pi y}}{a}\right)$.