Question

A charge is uniformly distributed inside a spherical body of radius $r_1=2r_0$ having a concentric cavity of radius $r_2=r_0$ ( $\rho$ is charge density inside the sphere). The potential of a point $P$ at a distance $\frac{3r_0}{2}$ from the centre is

• A. $\frac{7\rho r_0^2}{6{ϵ}_0}$
• B. $\frac{101\rho r_0^2}{72{ϵ}_0}$
• C. $\frac{17\rho r_0^2}{72{ϵ}_0}$
• D. none of these

Answer 1

Answer: (B)

Electric field at a distance $\mathrm{r},\left({\mathrm{r}}_0<\mathrm{r}<2{\mathrm{r}}_0\right)$ from the center is given by, $\begin{array}{rl}\mathrm{E}& =\frac{1}{4\pi {\epsilon }_0}\cdot \frac{{\mathrm{q}}_{\text{enclosed }}}{{\mathrm{r}}^2}\\ & =\frac{1}{4\pi {\epsilon }_0}\cdot \frac{\rho \cdot \frac{4}{3}\pi \left({\mathrm{r}}^3-{\mathrm{r}}_0^3\right)}{{\mathrm{r}}^2}\\ & =\frac{\rho }{3{\epsilon }_0}\cdot \left(\mathrm{r}-\frac{{\mathrm{r}}_0^3}{{\mathrm{r}}^2}\right)\end{array}$ Potential at the outer surface, $\begin{array}{l}{\mathrm{V}}_{\mathrm{s}}=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{\mathrm{Q}}{2{\mathrm{r}}_0}\\ =\frac{1}{4\pi {\epsilon }_0}\cdot \frac{\rho \cdot \frac{4}{3}\pi \left({\left(2{\mathrm{r}}_0\right)}^3-{\mathrm{r}}_0^3\right)}{2{\mathrm{r}}_0}\\ =\frac{7\rho }{6{\epsilon }_0}\cdot {\mathrm{r}}_0^2\end{array}$ $\therefore$ If $\mathrm{V}$ is the required potential at $\mathrm{r}=\frac{3{\mathrm{r}}_0}{2}$, $\begin{array}{l}\mathrm{V}-{\mathrm{V}}_{\mathrm{s}}=-{\int }_{2{\mathrm{r}}_0}^{\frac{3{\mathrm{r}}_0}{2}}\mathrm{E}\cdot \mathrm{dr}\\ \mathrm{V}={\mathrm{V}}_{\mathrm{s}}-\frac{\rho }{3{\epsilon }_0}{\int }_{2{\mathrm{r}}_0}^{\frac{3{\mathrm{r}}_0}{2}}\left(\mathrm{r}-\frac{{\mathrm{r}}_0^3}{{\mathrm{r}}^2}\right)\mathrm{dr}\\ =\frac{7\rho }{6{\epsilon }_0}{\mathrm{r}}_0^2-{\frac{\rho }{3{\epsilon }_0}\left(\frac{{\mathrm{r}}^2}{2}+\frac{{\mathrm{r}}_0^3}{\mathrm{r}}\right)|}_{2{\mathrm{r}}_0}^{\frac{3{\mathrm{r}}_0}{2}}\\ =\frac{7\rho }{6{\epsilon }_0}{\mathrm{r}}_0^2-\frac{\rho }{3{\epsilon }_0}\left[\frac{1}{2}\left(\frac{9{\mathrm{x}}_0^2}{4}-4{\mathrm{r}}_0^2\right)-\left(\frac{2{\mathrm{r}}_0^3}{3{\mathrm{r}}_0}-\frac{{\mathrm{r}}_0^3}{2{\mathrm{r}}_0}\right)\right]\\ =\frac{101\rho {\mathrm{r}}_0^3}{72{\epsilon }_0}\end{array}$