The electrostatic potential inside a charged spherical ball is given by $\phi=\alpha \rho^2+b$ where $r$ is…
The electrostatic potential inside a charged spherical ball is given by $\phi=\alpha \rho^2+b$ where $r$ is the distance from the centre; $a, b$ are constants. Then the charge density inside ball is
$-6 a \varepsilon_0 r$
$-24 \pi a \varepsilon_0 r$
$-6 a \varepsilon_0$
$-24 \pi a \varepsilon_0 r$
Solution
Potential inside $(\phi)=a r^2+b$
$
\therefore \mathrm{E}_{\mathrm{r}}=-\frac{\delta \mathrm{v}}{\delta \mathrm{r}}=-2 \mathrm{ar}
$
Electric field inside uniformly charged solid volume varies with ' $r$ '. So charge density is constant
$
\begin{aligned}
& \phi_{\text {net }}=(-2 \mathrm{ar}) 4 \pi \mathrm{r}^2=-8 \pi a r^3 \\
& -8 \pi \mathrm{ar}^3=\frac{\sigma \times \frac{4}{3} \pi \mathrm{r}^3}{\varepsilon_0} \\
& \therefore \sigma=-6 \mathrm{a} \varepsilon_0
\end{aligned}
$