A unit positive point charge is slowly moved through an infinitely thin tube inside a uniformly charged dielectric sphere of radius \(R\) and volume charge density \(\rho\).
The initial and final positions of the charge are \(B\) and \(A\), located at distances \(3R\) and \(2R\) respectively from the centre.
If the magnitude of work done on the charge is
\[
\frac{\rho R^2}{n\varepsilon_0}
\]
then find \(n\).
Show Hint
Outside a uniformly charged sphere, treat the entire charge as concentrated at the centre.
Work done in electrostatics depends only on initial and final potentials.
Concept:
Outside a uniformly charged sphere, the electric potential is the same as that of a point charge placed at the centre.
\[
V=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r}
\]
where
\[
Q=\frac{4}{3}\pi R^3\rho
\]
Step 1: Calculate potential at point A
\[
r_A=2R
\]
\[
V_A
=
\frac{1}{4\pi\varepsilon_0}
\frac{\frac{4}{3}\pi R^3\rho}{2R}
\]
\[
V_A
=
\frac{\rho R^2}{6\varepsilon_0}
\]
Step 2: Calculate potential at point B
\[
r_B=3R
\]
\[
V_B
=
\frac{1}{4\pi\varepsilon_0}
\frac{\frac{4}{3}\pi R^3\rho}{3R}
\]
\[
V_B
=
\frac{\rho R^2}{9\varepsilon_0}
\]
Step 3: Calculate work done
Since unit charge is moved,
\[
W=|V_A-V_B|
\]
\[
W=
\frac{\rho R^2}{\varepsilon_0}
\left(
\frac16-\frac19
\right)
\]
\[
W=
\frac{\rho R^2}{18\varepsilon_0}
\]
Comparing with
\[
W=
\frac{\rho R^2}{n\varepsilon_0}
\]
gives
\[
\boxed{n=18}
\]
