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 $\frac{3 r_{0}}{2}$ from the centre is

• A. $\frac{7 \rho r_{0}^{2}}{6 \epsilon _{0}}$
• B. $\frac{101 \rho r_{0}^{2}}{72 \epsilon _{0}}$
• C. $\frac{17 \rho r_{0}^{2}}{72 \epsilon _{0}}$
• 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{aligned} \mathrm{E} & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}_{\text {enclosed }}}{\mathrm{r}^2} \\ & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\rho \cdot \frac{4}{3} \pi\left(\mathrm{r}^3-\mathrm{r}_0^3\right)}{\mathrm{r}^2} \\ & =\frac{\rho}{3 \varepsilon_0} \cdot\left(\mathrm{r}-\frac{\mathrm{r}_0^3}{\mathrm{r}^2}\right) \end{aligned} $ Potential at the outer surface, $ \begin{array}{l} \mathrm{V}_{\mathrm{s}}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{Q}}{2 \mathrm{r}_0} \\ =\frac{1}{4 \pi \varepsilon_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 \varepsilon_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 \varepsilon_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 \varepsilon_0} \mathrm{r}_0^2-\left.\frac{\rho}{3 \varepsilon_0}\left(\frac{\mathrm{r}^2}{2}+\frac{\mathrm{r}_0^3}{\mathrm{r}}\right)\right|_{2 \mathrm{r}_0} ^{\frac{3 \mathrm{r}_0}{2}} \\ =\frac{7 \rho}{6 \varepsilon_0} \mathrm{r}_0^2-\frac{\rho}{3 \varepsilon_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 \varepsilon_0} \end{array} $