Question

For the infinite slot (Ex. 3.3), determine the charge density $\mathbf{\sigma }\mathbf{\left(}\mathbf{y}\mathbf{\right)}$ on the strip at $\mathrm{x}=0$, assuming it is a conductor at constant potential ${\mathbf{v}}_{\mathbf{0}}$.

Short answer

The expression for the charge density on the strip at$\mathrm{x}=0$ is $\mathrm{\sigma }\left(\mathrm{y}\right)=\frac{4{\mathrm{\epsilon }}_0{\mathrm{V}}_0}{\mathrm{a}}\sum _{\mathrm{n}=1,3.5\dots } \sin \left(\frac{\mathrm{n\pi y}}{\mathrm{a}}\right)$.

Step-by-step solution

Define functions

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

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

Here, ${\mathrm{v}}_0$is the constant potential along the conductor, $\mathrm{x}$is the x-coordinate, $\mathrm{y}$is the y-coordinate and is the positive integer.

Determine charge density

Derive the charge density in the terms of electric potential.

$\mathrm{\sigma }=-{\mathrm{\epsilon }}_0\frac{\mathrm{\partial }\mathrm{V}}{\mathrm{\partial }\mathrm{n}}$

$\mathrm{\sigma }\left(\mathrm{y}\right)=-{\mathrm{\epsilon }}_0{\left[\frac{\mathrm{\partial }\mathrm{V}}{\mathrm{\partial }\mathrm{x}}\right]}_{\mathrm{x}-0}$…… (2)

Substitute $\frac{4{\mathrm{V}}_0}{\mathrm{\pi }}\sum _{\mathrm{n}} \frac{1}{\mathrm{n}}{\mathrm{e}}^{-{\mathrm{n}}_{\mathrm{i},5}}\sin \frac{\mathrm{n\pi y}}{\mathrm{a}}$for $\mathrm{V}(\mathrm{x},\mathrm{y})$in equation (2).

$\mathrm{\sigma }\left(\mathrm{y}\right)=-{\left.{\mathrm{\epsilon }}_0\frac{\mathrm{\partial }}{\mathrm{\partial }\mathrm{x}}\left\{\frac{4{\mathrm{V}}_0}{\mathrm{\pi }}\sum \frac{1}{\mathrm{n}}{\mathrm{e}}^{\frac{\mathrm{n\pi x}}{\mathrm{a}}}\sin \left(\frac{\mathrm{n\pi y}}{\mathrm{a}}\right)\right\}\right|}_{\mathrm{x}-0}$

$=-{\left.{\mathrm{\epsilon }}_0\frac{4{\mathrm{V}}_0}{\mathrm{\pi }}\frac{\mathrm{\partial }}{\mathrm{\partial }\mathrm{x}}\left\{\sum \frac{1}{\mathrm{n}}{\mathrm{e}}^{\mathrm{n\pi x}}\sin \left(\frac{\mathrm{n\pi y}}{\mathrm{a}}\right)\right\}\right|}_{\mathrm{x}=0}$

$={\left.{\mathrm{\epsilon }}_0\frac{4{\mathrm{V}}_0}{\mathrm{\pi }}\frac{1}{\mathrm{n}}\left(\frac{\mathrm{n\pi }}{\mathrm{a}}\right)\sum {\mathrm{e}}^{\mathrm{n\pi x}}\sin \left(\frac{\mathrm{n\pi y}}{\mathrm{a}}\right)\right|}_{\mathrm{x}=0}$

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

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