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

Answer 1

Answer: (B)

$-6A{ϵ}_0$

Electric field at distance $r$ is $-2Ar$.

Using gauss law:
$-2Ar\times 4\pi r^2=\frac{q}{{ϵ}_0}\Rightarrow \frac{q}{V}=\frac{q}{\frac{4\pi r^3}{3}}=-6A{ϵ}_0$