The field potential $V$ inside a charged ball depends only on the radial distance $r,\text{ as }V\left(r\right)=a{r}^2+b$ where $a and b$ are constants. If the space charge density$\rho$ inside the ball is proportional to $r^n$, then $n$ equals :

The electric field is given by

$E\left(r\right)=\frac{-dV}{dr}=-2ar$

Applying Gauss's theorem on a spherical surface of radius $r$, we have 

$\begin{array}{l}{\epsilon }_0\text{∮}\overrightarrow{E}\cdot d\overrightarrow{A}=Q_{\text{inside }A}\\ \Rightarrow {\epsilon }_0(-2ar)\left(4\pi r^2\right)={\int }_0^r \rho \left(x\right)4\pi x^{2}dx\end{array}$

Differentiating both sides with respect to $r,$we have 

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-8a\pi {\epsilon }_{o}3r^2=\rho \left(r\right)4\pi r^2\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\rho \left(r\right)=-6a{\epsilon }_o\Rightarrow \text{ Independent of }r\Rightarrow n=0\end{array}$