Question

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 \(n\).

• A. 18
• B. 2
• C. 6
• D. 9

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.

Answer 1

The Correct Option is A

Step 1: Note both points are outside.
The charge moves from $B$ at $3R$ to $A$ at $2R$, both beyond the sphere of radius $R$. Outside a uniform sphere the potential equals that of all the charge at the centre.
Step 2: Total charge of the sphere.
\[ Q = \frac{4}{3}\pi R^3 \rho \]
Step 3: Potential at A ($r = 2R$).
\[ V_A = \frac{1}{4\pi\varepsilon_0}\frac{Q}{2R} = \frac{1}{4\pi\varepsilon_0}\frac{\frac{4}{3}\pi R^3 \rho}{2R} = \frac{\rho R^2}{6\varepsilon_0} \]
Step 4: Potential at B ($r = 3R$).
\[ V_B = \frac{1}{4\pi\varepsilon_0}\frac{\frac{4}{3}\pi R^3 \rho}{3R} = \frac{\rho R^2}{9\varepsilon_0} \]
Step 5: Work on a unit charge.
For a unit positive charge, the work done equals the potential difference in magnitude.
\[ W = |V_A - V_B| = \frac{\rho R^2}{\varepsilon_0}\left(\frac{1}{6} - \frac{1}{9}\right) = \frac{\rho R^2}{\varepsilon_0}\cdot\frac{1}{18} \]
Step 6: Compare to find $n$.
Matching $W = \dfrac{\rho R^2}{n\varepsilon_0}$ gives $n = 18$.
\[ \boxed{n = 18} \]