Physics - Electrostatics Question with Solution | TestHub

Since the electrostatic field is conservative, the work done in moving a charge depends only on the potential difference between the initial and final positions, regardless of the path taken.

The electric potential $V(r)$ at a distance $r$ ($r\ge R$) from the centre of a uniformly charged solid sphere of radius $R$ and charge density $\rho$ is given by:
$V(r)=\frac{1}{4\pi {ϵ}_0}\frac{Q}{r}$

The total charge $Q$ of the sphere is:
$Q=\frac{4}{3}\pi R^3\rho$

Substituting $Q$ into the potential formula, we get:
$V(r)=\frac{1}{4\pi {ϵ}_0}\frac{\frac{4}{3}\pi R^3\rho }{r}=\frac{\rho R^3}{3{ϵ}_{0}r}$

The initial position $A$ is at a distance $r_A=2R$ from the centre. The potential at $A$ is:
$V_A=\frac{\rho R^3}{3{ϵ}_0(2R)}=\frac{\rho R^2}{6{ϵ}_0}$

The final position $B$ is at a distance $r_B=3R$ from the centre. The potential at $B$ is:
$V_B=\frac{\rho R^3}{3{ϵ}_0(3R)}=\frac{\rho R^2}{9{ϵ}_0}$

The work done by the external agent in moving a unit positive charge ($q=1$) slowly from $A$ to $B$ is equal to the change in its potential energy:
$W=q(V_B-V_A)=1\times \left(\frac{\rho R^2}{9{ϵ}_0}-\frac{\rho R^2}{6{ϵ}_0}\right)$

$W=\frac{\rho R^2}{{ϵ}_0}\left(\frac{2-3}{18}\right)=-\frac{\rho R^2}{18{ϵ}_0}$

The magnitude of the total work done is:
$|W|=\frac{\rho R^2}{18{ϵ}_0}$

Comparing this with the given expression $\frac{\rho R^2}{n{ϵ}_0}$, we find:
$n=18$

Answer: $18$