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. $\frac{109 ρr_{0}^{2}}{72 \epsilon _{0}}$

Answer 1

Answer: (B)

Electric field at a distance $r$ , $\left(r_{0} < r < 2 r_{0}\right)$ from the centre is given by,
$E=\frac{1}{4 \pi ε_{0}}.\frac{q_{enclosed}}{r^{2}}$
$=\frac{1}{4 \left(\pi ε\right)_{0}}.\frac{\rho . \frac{4}{3} \pi \left(\right. r^{3} - r_{0}^{3} \left.\right)}{r^{2}}$
$=\frac{\rho }{3 \left(\epsilon \right)_{0}}.\left(\right. r - \frac{r_{0}^{3}}{r^{2}} \left.\right)$
Potential at the outer surface,
$V_{s}=\frac{1}{4 \pi ε_{0}}.\frac{Q}{2 r_{0}}$
$=\frac{1}{4 \left(\pi ε\right)_{0}}.\frac{\rho . \frac{4}{3} \pi \left(\right. \left(\right. 2 r_{0} \left.\right)^{3} - r_{0}^{3} \left.\right)}{2 r_{0}}$
$=\frac{7 \rho }{6 \epsilon _{0}}. \, r_{0}^{2}$
$\therefore $ If $V$ is the required potential at $r=\frac{3 r_{0}}{2}$ ,
$V-V_{s}=-\displaystyle \int _{2 r_{0}}^{\frac{3 r_{0}}{2}}E. \, dr$
$V=V_{s}-\frac{\rho }{3 \left(\epsilon \right)_{0}}\displaystyle \int _{2 r_{0}}^{\frac{3 r_{0}}{2}}\left(\right. r - \frac{r_{0}^{3}}{r^{2}} \left.\right)dr$
$\left(= \frac{7 \rho }{6 \left(\epsilon \right)_{0}} r_{0}^{2} - \frac{\rho }{3 \left(\epsilon \right)_{0}} \left(\right. \frac{r^{2}}{2} + \frac{r_{0}^{3}}{r} \left.\right) \left|\right.\right)_{2 r_{0}}^{\frac{3 r_{0}}{2}}$
$=\frac{7 \rho }{6 \left(\epsilon \right)_{0}}r_{0}^{2}-\frac{\rho }{3 \left(\epsilon \right)_{0}}\left[\frac{1}{2} \left(\frac{9 r_{0}^{2}}{4} - 4 r_{0}^{2}\right) + \left(\frac{2 r_{0}^{3}}{3 r_{0}} - \frac{r_{0}^{3}}{2 r_{0}}\right)\right]$
$=\frac{101 ρr_{0}^{2}}{72 \epsilon _{0}}$