Question

A hollow metal sphere of radius \(r\) contains a charge \(+Q\). Consider an imaginary circle of radius \(R\,(R\gt r)\) concentric with the charged sphere. A point charge \(q\) is carried from (I) \(A\) to \(B\) and then from (II) \(A\) to \(C\). Choose the correct answer from the given alternatives.

• A. The work done in case (I) is less than that done in case (II).
• B. The work done in case (I) is greater than that done in case (II).
• C. Identical finite amount of work is done in both the cases.
• D. No work is done in either of the cases.

Hint:

Moving a charge along an equipotential surface requires no work: \[ W=q\Delta V. \] If \[ \Delta V=0, \] then \[ W=0. \] All points at the same distance from a charged conducting sphere are at the same potential.

Answer 1

The Correct Option is D

Concept: Outside a charged conducting sphere, the electric potential depends only on the distance from the centre. \[ V=\frac{1}{4\pi\varepsilon_0}\frac{Q}{R} \] for every point situated at the same radial distance \(R\). Hence all points on a concentric circle (or spherical surface) are equipotential.

Step 1: Compare the potentials at points \(A\), \(B\) and \(C\). From the figure, \[ OA=OB=OC=R. \] Therefore, \[ V_A=V_B=V_C = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R}. \]

Step 2: Calculate work done in moving charge \(q\). Work done in moving a charge between two points is \[ W=q(V_i-V_f). \] For motion from \(A\) to \(B\), \[ W_{AB} = q(V_A-V_B) = 0. \] Similarly, for motion from \(A\) to \(C\), \[ W_{AC} = q(V_A-V_C) = 0. \]

Step 3: Interpret the result. Since \(A\), \(B\) and \(C\) lie on the same equipotential surface, \[ \Delta V=0 \] for both paths. Hence no work is required in either case. \[\begin{aligned} \boxed{W_{AB}=W_{AC}=0} \end{aligned}\] Therefore, \[ \boxed{\text{No work is done in either of the cases.}} \] Hence, option \(\mathbf{(D)}\) is correct.