Question

The electrostatic potential inside a charged spherical ball is given by $\Phi =a{r}^2+b$, where $r$ is the distance from the centre and $a,b$ are constant. Then the charge density inside the ball is
( ${\epsilon }_0=$ Permitivity in free space)

• A. $-6a{\epsilon }_{0}r$
• B. $-6a{\epsilon }_0$
• C. $-24\pi a{\epsilon }_0$
• D. $-24\pi a{\epsilon }_{0}r$

Answer 1

Answer: (B)

Potential function given is
$\phi =a{r}^2+b$
Electric field in a region is
$E=\frac{-d\phi }{dr}=\frac{-d}{dr}\left(a{r}^2+b\right)$
$\Rightarrow E=-2ar$
$\Rightarrow \frac{kq}{r^2}=-2ar$
$\Rightarrow \frac{q}{r^3}=\frac{-2a}{k}\dots (i)$
Charge density inside the ball $=\rho =\frac{q}{\frac{4}{3}\pi r^3}$
$=\frac{3}{4\pi }\cdot \left(\frac{q}{r^3}\right)=\frac{3}{4\pi }\left(\frac{-2a}{k}\right)$ [ from Eq. (i)]
$=\frac{-6a}{4\pi }\cdot 4\pi {\epsilon }_0\left[\because k=\frac{1}{4\pi {\epsilon }_0}\right]$
$=-6a{\epsilon }_0$