For the charge configuration of Prob. 2.15, find the potential at the center, using infinity as your reference point.

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.

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

Write the expression for potential at the center.

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

Consider the formulas are used for simplification.

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

Apply the limits and solve the integration.

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

Solve further as,

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

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,

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