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]

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 \()\)
\(\begin{aligned}
& 4 \pi \mathrm{r}^2 \times(-2 \mathrm{ar})=\frac{4 \rho}{3 \varepsilon_0} \pi \mathrm{r}^3 \\
& -2 \mathrm{a}=\frac{\rho}{3 \varepsilon_0}
\end{aligned}\)
\(\rho=-6 a \varepsilon_0\)