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

Step 1: Set up the picture in words.
A charged sphere sits at the centre. Around it we draw an imaginary circle of radius $R$ larger than the sphere. Points $A$, $B$ and $C$ all lie on this same circle, so each is the same distance $R$ from the centre.

Step 2: Potential outside the sphere.
Outside a charged sphere the potential depends only on the distance from the centre: \[ V=\frac{1}{4\pi\varepsilon_0}\frac{Q}{R}. \]

Step 3: Compare the three points.
Since $A$, $B$, $C$ are all at distance $R$, they have the very same potential. The whole circle is one equipotential line.

Step 4: Work formula for moving a charge.
The work to carry charge $q$ between two points is \[ W=q\,(V_{\text{final}}-V_{\text{initial}}). \]

Step 5: Evaluate both journeys.
For $A\to B$: $V_B-V_A=0$, so $W_I=0$. For $A\to C$: $V_C-V_A=0$, so $W_{II}=0$.

Step 6: Conclusion.
Both potential differences are zero, so no work is done in either case.
\[ \boxed{\text{No work is done in either case}} \]