Physics - Electrostatics Question with Solution | TestHub

Potential inside $(\varphi )=a{r}^2+b$ $$\therefore {\mathrm{E}}_{\mathrm{r}}=-\frac{\delta \mathrm{v}}{\delta \mathrm{r}}=-2\mathrm{ar}$$ Electric field inside uniformly charged solid volume varies with ' $r$ '. So charge density is constant $$\begin{array}{rl}& {\varphi }_{\text{net }}=(-2\mathrm{ar})4\pi {\mathrm{r}}^2=-8\pi a{r}^3\\ & -8\pi {\mathrm{ar}}^3=\frac{\sigma \times \frac{4}{3}\pi {\mathrm{r}}^3}{{\epsilon }_0}\\ & \therefore \sigma =-6\mathrm{a}{\epsilon }_0\end{array}$$