91Ó°ÊÓ

Consider the shadow of hollow of spherical shell with inner radius is aand outer radius is b.

Write the expression for the charge density of the sphere.

$\mathbf{P}\mathbf{=}\frac{\mathbf{k}}{{\mathbf{r}}^{\mathbf{2}}}$

Here, kis constant, ris the radius of the Gaussian surface and pVolume charge density

Write the electric field configuration.

$E\left(r\right)=\left\{\begin{array}{cc}0,& r<a\\ \frac{k(r-a)}{{\mathrm{ξ}}_0{r}^2}& a<r<b\\ \frac{k(b-a)}{{\mathrm{ξ}}_0{r}^2}& r>b\end{array}\right.$

Write the expression for potential at the center.

$$\begin{gathered} V_{center}={∫}_{∞}^{center}E\left(r\right)dr \\ =-{∫}_{∞}^{b}E\left(r\right)dr-{∫}_b^{a}E\left(r\right)dr-{∫}_{∞}^{b}E\left(r\right)dr \end{gathered}$$

Consider the formulas are used for simplification.

$$\begin{gathered} ∫X^{n}dx=x^{n+1} \\ ∫\frac{1}{x}dx=In\left(x\right) \end{gathered}$$

Apply the limits and solve the integration.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{center}}=-{∫}_{∞}^{\mathrm{b}}\frac{\mathrm{k}\left(\mathrm{b}-\mathrm{a}\right)}{{\mathrm{Î\mu }}_0{\mathrm{r}}^2}\mathrm{dr}-{∫}_{\mathrm{b}}^{\mathrm{a}}\frac{\mathrm{k}\left(\mathrm{r}-\mathrm{a}\right)}{{\mathrm{e}}_0{\mathrm{r}}^2}\mathrm{dr}-{∫}_{\mathrm{b}}^{\mathrm{a}}0\mathrm{dr} \\ =-{∫}_{∞}^{\mathrm{b}}\frac{\mathrm{k}\left(\mathrm{b}-\mathrm{a}\right)}{{\mathrm{Î\mu }}_0{\mathrm{r}}^2}\mathrm{dr}-\frac{\mathrm{k}}{{\mathrm{Î\mu }}_0}{∫}_{\mathrm{b}}^{\mathrm{a}}\left(\frac{1}{\mathrm{r}}-\frac{\mathrm{a}}{{\mathrm{r}}^2}\right)\mathrm{dr}-{∫}_{\mathrm{b}}^{\mathrm{a}}0\mathrm{dr} \\ =\frac{\mathrm{k}\left(\mathrm{b}-\mathrm{a}\right)}{{\mathrm{Î\mu }}_0{\mathrm{r}}^2}\left(\frac{1}{\mathrm{b}}-0\right)-\frac{\mathrm{k}}{{\mathrm{Î\mu }}_0}\left[\left(\mathrm{In}\mathrm{a}-\mathrm{In}\mathrm{b}\right)+\left(\frac{\mathrm{a}}{\mathrm{b}}-\frac{\mathrm{a}}{\mathrm{b}}\right)\right] \end{gathered}$$

Solve further as,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{center}}=\frac{\mathrm{k}}{{\mathrm{Î\mu }}_0}\left\{\left(1-\frac{\mathrm{a}}{\mathrm{b}}\right)-\mathrm{In}\left(\frac{\mathrm{a}}{\mathrm{b}}\right)-\left(1-\frac{\mathrm{a}}{\mathrm{b}}\right)\right\} \\ =\frac{\mathrm{k}}{{\mathrm{Î\mu }}_0}\left[-\mathrm{In}\left(\frac{\mathrm{a}}{\mathrm{b}}\right)\right] \end{gathered}$$

Here, Logarithm formula is used $\mathrm{In}\left(\frac{\mathrm{a}}{\mathrm{b}}\right)=-\mathrm{In}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)$

Rewrite the equation as,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{center}}=\frac{\mathrm{k}}{{\mathrm{Î\mu }}_0}\left[-\mathrm{In}\frac{\mathrm{b}}{\mathrm{a}}\right] \\ \mathrm{Therefore},\mathrm{the}\mathrm{potential}\mathrm{at}\mathrm{the}\mathrm{center}\mathrm{of}\mathrm{sphere}\mathrm{is}{\mathrm{V}}_{\mathrm{center}}=\frac{\mathrm{k}}{{\mathrm{Î\mu }}_0}\left[-\mathrm{In}\frac{\mathrm{b}}{\mathrm{a}}\right] \end{gathered}$$