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. $\frac{109\rho r_0^2}{72{ϵ}_0}$

Answer 1

Answer: (B)

Electric field at a distance $r$ , $\left(r_0<r<2r_0\right)$ from the centre is given by,
$E=\frac{1}{4\pi {\epsilon }_0}.\frac{q_{enclosed}}{r^2}$
$=\frac{1}{4{\left(\pi \epsilon \right)}_0}.\frac{\rho .\frac{4}{3}\pi (r^3-r_0^3)}{r^2}$
$=\frac{\rho }{3{\left(ϵ\right)}_0}.(r-\frac{r_0^3}{r^2})$
Potential at the outer surface,
$V_s=\frac{1}{4\pi {\epsilon }_0}.\frac{Q}{2r_0}$
$=\frac{1}{4{\left(\pi \epsilon \right)}_0}.\frac{\rho .\frac{4}{3}\pi ((2r_0{)}^3-r_0^3)}{2r_0}$
$=\frac{7\rho }{6{ϵ}_0}.r_0^2$
$\therefore$ If $V$ is the required potential at $r=\frac{3r_0}{2}$ ,
$V-V_s=-{\int }_{2r_0}^{\frac{3r_0}{2}}E.dr$
$V=V_s-\frac{\rho }{3{\left(ϵ\right)}_0}{\int }_{2r_0}^{\frac{3r_0}{2}}(r-\frac{r_0^3}{r^2})dr$
${\left(=\frac{7\rho }{6{\left(ϵ\right)}_0}{r}_0^2-\frac{\rho }{3{\left(ϵ\right)}_0}(\frac{r^2}{2}+\frac{r_0^3}{r})|\right)}_{2r_0}^{\frac{3r_0}{2}}$
$=\frac{7\rho }{6{\left(ϵ\right)}_0}{r}_0^2-\frac{\rho }{3{\left(ϵ\right)}_0}\left[\frac{1}{2}\left(\frac{9r_0^2}{4}-4r_0^2\right)+\left(\frac{2r_0^3}{3r_0}-\frac{r_0^3}{2r_0}\right)\right]$
$=\frac{101\rho r_0^2}{72{ϵ}_0}$