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. $16\, A\varepsilon _{0}$
• B. $-6 \,A \varepsilon_{0}$
• C. $20 \,A\varepsilon _{0}$
• D. $-15 \,A \varepsilon_{0}$

Answer 1

Answer: (B)

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

$E=\frac{K\, q r}{R^{3}}$
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,
$E=\frac{\rho r}{3 \varepsilon_{0}}$
$\Rightarrow \rho=\frac{3 E \varepsilon_{0}}{r}$
Now, electric field, $E=-\frac{d V}{d r}$
Here, $ V= A r^{2}+B$
$\therefore E=-\frac{d V}{d r}=-2 A r+0=-2 A r $
$\therefore \rho=\frac{3 E \varepsilon_{0}}{r} $
$ =\frac{3\left(-2 \,A r \cdot \varepsilon_{0}\right)}{r}=-6 A \varepsilon_{0}$