The electrostatic potential inside a charged sphere is given as \[V = A{r^2} + B\], where r is the distance from the center of the sphere; A and B are constants. Then charge density in the sphere is.

Hint: To solve this question we used the poison equation. In the terms of potential and the charge density because in the question potential and charge density are given to us after that, we use the equation in spherical coordinates and solve the question.

Complete step by step solution:
According to question,
The electrostatic potential inside a charged sphere is given by,\[V = A{r^2} + B\]
Where A and B are constants.
Applying the poisson's equation,\[{\Delta ^2}V = - \dfrac{\rho }{{{ \in _0}}}\]----(1)
Where V = potential inside the sphere,\[\rho \]= charge density
First we need find out\[{\Delta ^2}V\],
\[\Rightarrow V = A{r^2} + B\]
In spherical coordinates \[{\Delta ^2}V = \dfrac{1}{{{r^2}\sin \theta }}\left[ {\dfrac{\partial }{{\partial r}}\left( {{r^2}\sin \theta \cdot\dfrac{{\partial V}}{{\partial r}}} \right)} \right]\]
\[\Rightarrow \dfrac{1}{{{r^2}\sin \theta }}\dfrac{\partial }{{\partial r}}\left[ {{r^2}\sin \theta \cdot\dfrac{\partial }{{\partial r}}\left( {A{r^2} + B} \right)} \right]\]
\[ \Rightarrow \dfrac{1}{{{r^2}\sin \theta }}\dfrac{\partial }{{\partial r}}\left[ {{r^2}\sin \theta \cdot \left( {2Ar} \right)} \right]\]
\[ \Rightarrow \dfrac{1}{{{r^2}\sin \theta }}\left( {2A\sin \theta } \right)\dfrac{\partial }{{\partial r}}\left[ {{r^3}} \right]\]
\[ \Rightarrow \dfrac{1}{{{r^2}}}\left( {2A} \right) \times 3{r^2}\]
\[\Rightarrow {\Delta ^2}V = 6A\]
Put the value of \[{\Delta ^2}V\]in equation(1)
\[\Rightarrow {\Delta ^2}V = - \dfrac{\rho }{{{ \in _0}}}\]
\[\Rightarrow 6A = -\dfrac{\rho }{{{ \in _0}}}\]
\[\Rightarrow \boxed{\rho = - 6A{ \in _0}}\]
Where\[\rho \] is the charge density.

Note: Maximum times, students do not use the spherical coordinates but we should the spherical coordinates, because we have a charged sphere.