Question

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

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

Answer 1

The correct option is C $-6a{\epsilon }_0$

Given that,
Potential inside a charged spherical ball, $V=a{r}^2+b$
The relation between electric field $\left(E\right)$ and electric potential $\left(V\right)$ is
$E=-\frac{dV}{dr}$
$\Rightarrow E=-\frac{d}{dr}(a{r}^2+b)$

$\Rightarrow E=-2ar\dots \left(i\right)$

Let's draw a spherical Gaussian surface $\left(S\right)$ of radius $r$ from the centre of the sphere and the charge enclosed by the surface $S$ is $q$.
By using Gauss law

$\overrightarrow{E}.\overrightarrow{S}=\frac{q}{{ϵ}_0}$
$\Rightarrow ES=\frac{q}{{ϵ}_0}$

$\Rightarrow 4\pi r^{2}E=\frac{q}{{ϵ}_0}\dots \left(ii\right)$

Let's assume that the charge density of the spherical ball is $\rho$.
The total charge enclosed by Gaussian surface of radius $r$ is $q$.
$\Rightarrow q=\frac{4\rho }{3}\pi r^3$
Substitute $E$ from equation $\left(i\right)$ and $q$ in equation $\left(ii\right)$

$\Rightarrow 4\pi r^2\times (-2ar)=\frac{4\rho }{3{ϵ}_0}\pi r^3$

$\Rightarrow -2a=\frac{\rho }{3{ϵ}_0}$
$\Rightarrow \rho =-6a{ϵ}_0$

Hence, option (c) is correct.