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

Answer 1

The correct option is D $-6a{ϵ}_0$

Potential inside $=a{r}^2+b$

$\therefore E_r=-\frac{\delta v}{\delta r}=-2ar$

Electric field inside uniformly charged solid volume varies with ${}^{\prime }{r}^{\prime }$. So, charge density is constant.

${\varphi }_{net}=(-2ar)4\pi r^2$

$-8a\pi r^3=\frac{\sigma \times \frac{4}{3}\pi r^3}{{ϵ}_0}$

$\sigma =-6a{ϵ}_0$