Question

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 the charge density in the sphere is

• A. $16A{\in }_0$
• B. $-6A{\in }_0$
• C. $20A{\in }_0$
• D. $-15A{\in }_0$

Answer 1

The correct option is B $-6A{\in }_0$

Electric field, $E=-\frac{d\varphi }{dt}=-2ar$

By Gauss's theorem

$E\left(4\pi r^2\right)=\frac{q}{{\in }_0}$

$q=-8\pi {\in }_{0}a{r}^2$

Now, the charge density

$P=\frac{q}{V}=\frac{-8\pi {\in }_{0}a{r}^{-3}}{\frac{4}{3}{\pi r}^2}=-6{\in }_{0}a$

where $a=A$

$=-6{\in }_{0}A$