Question

The electrostatic potential inside a charged spherical ball is given by $\varphi =a{r}^2+b$ where r is the distance from the centre a, b are constants. Then the charge density inside the ball is

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

Answer 1

Answer: (C)

Electric field, $E=-\frac{d\varphi }{dt}=-2ar$ ...(i)
By Gauss's theorem $E\left(4\pi r^2\right)=\frac{q}{{\epsilon }_0}$ ...(ii)
From (i) and (ii),
$\Rightarrow q=-8\pi {\epsilon }_{0}a{r}^3$
$\rho =\frac{dq}{dV}=\frac{dq}{dr}\times \frac{dr}{dV}$
$=\left(-24\pi {\epsilon }_{0}a{r}^2\right)\left(\frac{1}{4\pi r^2}\right)=-6{\epsilon }_{0}a$