Q23P - Introduction to Electrodynamics

Question

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

Step-by-Step Solution

Answer

The potential at the center of sphere is $V_{center}=\frac{k}{{\epsilon }_0}\left[-In\left(\frac{b}{a}\right)\right].$

Step 1: Determine charge density

Consider the shadow of hollow of spherical shell with inner radius is a and 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, k is constant,  r is the radius of the Gaussian surface and p Volume charge density 

Step 2: Determine electrical potential

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,

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