Question

An insulating solid sphere of the radius $R$ is charged in a non-uniform manner such that the volume charge density $\rho =\frac{A}{r}$ , where $A$ is a positive constant and $r$ is the distance from the centre. The potential difference between the centre and surface of the sphere is

• A. $\frac{A R}{8 \epsilon _{0}}$
• B. $\frac{A R}{4 \epsilon _{0}}$
• C. $\frac{A R}{ \epsilon _{0}}$
• D. $\frac{A R}{2 \epsilon _{0}}$

Answer 1

Answer: (D)

$E \, \times \, 4\pi r^{2}=\frac{q}{\epsilon _{0}}$
$E\times 4\pi r^{2}=\displaystyle \int _{0}^{r} \frac{A}{r}\times 4\pi r^{2}dr$
$E=\frac{A}{2 \epsilon _{0}}$
Since electric field is constant, we get
$\Delta V=E\times d$
$=\frac{A}{2 \epsilon _{0}}\times R=\frac{A R}{2 \epsilon _{0}}$