Question

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

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

Answer 1

Answer: (B)

Let the volume charge density be $\rho$.
The electric field inside the charged sphere,

$E=\frac{Kqr}{R^3}$
where, $R=$ Radius of charged sphere
$r=$ Distance of the point $(P)$ inside the sphere
As volume charge density is $\rho$, then in terms of charge density,
$E=\frac{\rho r}{3{\epsilon }_0}$
$\Rightarrow \rho =\frac{3E{\epsilon }_0}{r}$
Now, electric field, $E=-\frac{dV}{dr}$
Here, $V=A{r}^2+B$
$\therefore E=-\frac{dV}{dr}=-2Ar+0=-2Ar$
$\therefore \rho =\frac{3E{\epsilon }_0}{r}$
$=\frac{3\left(-2Ar\cdot {\epsilon }_0\right)}{r}=-6A{\epsilon }_0$