Question

Statement (A): Inside a charged hollow metal sphere, \(E=0,\;V\neq0\). (\(E\) - electric field, \(V\) - electric potential)
Statement (B): The work done in moving a positive charge on an equipotential surface is zero.
Statement (C): When two like charges are brought closer, their mutual electrostatic potential energy will increase.

• A. A, B, C are true
• B. A, B are true, C is false
• C. A, C are true, B is false
• D. B, C are true, A is false

Hint:

Inside a conductor in electrostatic equilibrium: \[ E=0 \] but the potential remains constant and may be non-zero. Also, no work is done along an equipotential surface.

Answer 1

The Correct Option is A

Step 1: Test Statement (A) about the hollow sphere.
Inside a charged hollow metal sphere the field vanishes, $E = 0$, because no charge resides in the cavity, yet the potential is constant and equal to the non-zero surface value, so $V \neq 0$. Statement (A) is true.
Step 2: Justify why the potential is non-zero.
Potential is the work per unit charge to bring a charge from infinity, and that work is non-zero even though no further work is needed once inside, leaving $V$ at a finite constant.
Step 3: Test Statement (B) on an equipotential surface.
Moving a charge along an equipotential surface means $\Delta V = 0$, so the work done is \[ W = q\,\Delta V = 0 \] Statement (B) is true.
Step 4: Test Statement (C) on like charges.
For two like charges the potential energy is \[ U = \frac{1}{4\pi\varepsilon_0}\,\frac{q_1 q_2}{r} \] which is positive and rises as $r$ shrinks, so bringing like charges closer increases their mutual energy. Statement (C) is true.
Step 5: Collect the verdicts.
All three statements (A), (B) and (C) are true.
Step 6: Choose the matching option.
Hence the correct choice is \[ \boxed{\text{A, B, C are true}} \]