The electrostatic potential inside a charged spherical ball is given by $V=a r^2+\mathrm{b}$ where ' r ' is…
The electrostatic potential inside a charged spherical ball is given by $V=a r^2+\mathrm{b}$ where ' r ' is the distance from its centre and ' $a$ ' and ' $b$ ' are constants. The volume charge density of the ball is $\left[\varepsilon_0=\right.$ permittivity of free space]
$-24 \pi \mathrm{a} \varepsilon_0 \mathrm{r}$
$-6 \mathrm{a} \varepsilon_0 \mathrm{r}$
$-24 \pi \mathrm{a} \varepsilon_0$
$-6 a \varepsilon_0$
Solution
Given: $V=a^2+b$
Given : $\mathrm{V}=\mathrm{ar}^2+\mathrm{b}$
$\begin{aligned}
E & =-\frac{d V}{d r} \\
& =-\frac{d}{d r}\left(a r^2+b\right) \\
E & =-2 a r
...(i)\end{aligned}$
By Gauss' law,
$\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{~S}}=\frac{\mathrm{q}}{\varepsilon_0}$
$4 \pi r^2 E=\frac{q}{\varepsilon_0}...(ii)$ Total charge $(q)$ enclosed $=\frac{4 p}{3} \pi r^3$.
From (i) and (ii) $\ldots(\rho=$ charge density of ball $)$
$\begin{aligned}
& 4 \pi \mathrm{r}^2 \times(-2 \mathrm{ar})=\frac{4 \rho}{3 \varepsilon_0} \pi \mathrm{r}^3 \\
& -2 \mathrm{a}=\frac{\rho}{3 \varepsilon_0}
\end{aligned}$
$\rho=-6 a \varepsilon_0$