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
( $\varepsilon_{0}=$ Permitivity in free space)

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

Answer 1

Answer: (B)

Potential function given is
$\varphi=a r^{2}+b$
Electric field in a region is
$E=\frac{-d \varphi}{d r}=\frac{-d}{d r}\left(a r^{2}+b\right) $
$\Rightarrow E=-2 a r$
$ \Rightarrow \frac{k q}{r^{2}}=-2 a r$
$\Rightarrow \frac{q}{r^{3}}=\frac{-2 a}{k} \ldots ( i )$
Charge density inside the ball $=\rho=\frac{q}{\frac{4}{3} \pi r^{3}}$
$=\frac{3}{4 \pi} \cdot\left(\frac{q}{r^{3}}\right)=\frac{3}{4 \pi}\left(\frac{-2 a}{k}\right)$ [ from Eq. (i)]
$=\frac{-6 a}{4 \pi} \cdot 4 \pi \varepsilon_{0}\left[\because k=\frac{1}{4 \pi \varepsilon_{0}}\right]$
$=-6 a \varepsilon_{0}$