The electrostatic potential inside a charged spherical ball is given by potential inside a charged spherical ball is given by \[\phi =a{{r}^{2}}+b\]where r is the distance from the centre and a b are constants. The charge density inside the ball is?
A. \[-6a{{\varepsilon }_{0}}r\]
B. \[-24\pi a{{\varepsilon }_{0}}r\]
C. \[-6a{{\varepsilon }_{0}}\]
D. \[-24\pi a{{\varepsilon }_{0}}\]

Hint: We are given electrostatic potential and we need to find the charge density inside the ball. We know there exists a relationship between electric field and electric potential. We can use that relation to find out the field. Then by using the Gauss theorem we can find out the charge enclosed by the sphere. Once, we had found out the charge enclosed we can find the charge density easily.

Complete step by step answer:
Given electric potential is \[\phi =a{{r}^{2}}+b\]. We know electric field is given by \[E=-\dfrac{dV}{dr}\]
\[
\Rightarrow E=-\dfrac{d[a{{r}^{2}}+b]}{dr} \\
\Rightarrow E=-2ar \\
\]
According to gauss law, the electric flux through a closed surface is equal to \[\dfrac{1}{{{\varepsilon }_{0}}}\]the total charge enclosed by that surface.
\[\oint{\overrightarrow{E}.\overrightarrow{dS}=}\dfrac{{{q}_{enc}}}{{{\varepsilon }_{0}}}\], since the electric field is perpendicular always to the surface of the conductor, the angle between the field vector and the area vector is \[{{0}^{0}}\].
\[\Rightarrow \oint{EdS=}\dfrac{{{q}_{enc}}}{{{\varepsilon }_{0}}}\]
$\Rightarrow ES=\dfrac{{{q}_{enc}}}{{{\varepsilon }_{0}}} \\
\Rightarrow -2ar\times 4\pi {{r}^{2}}=\dfrac{{{q}_{enc}}}{{{\varepsilon }_{0}}} \\
\Rightarrow -8a\pi {{r}^{3}}=\dfrac{{{q}_{enc}}}{{{\varepsilon }_{0}}} \\
\therefore {{q}_{enc}}=-8a{{\varepsilon }_{0}}\pi {{r}^{3}} \\ $
This is the charge enclosed by the surface. Now, we need to find the charge density, in order to do that we just divide the total charge by the volume.
$\rho =\dfrac{-8a{{\varepsilon }_{0}}\pi {{r}^{3}}}{\dfrac{4}{3}\pi {{r}^{3}}} \\
\therefore \rho =-6a{{\varepsilon }_{0}} $

So, the correct option is A.

Note: Gauss law can only be used when the surface is closed. If the charge is placed in an open surface, then we assume a closed gaussian surface to find the answer. In Gauss law we have to take the charge enclosed. Also, the electric field lines are always perpendicular to the surface. If they are not then for moving charge from one point to another on an equipotential surface we will have to do work.