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.
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.
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.
Show 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.
Updated On: Jun 22, 2026
- A, B, C are true
- A, B are true, C is false
- A, C are true, B is false
- B, C are true, A is false
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Analyze Statement (A).
Inside a charged hollow conducting sphere in electrostatic equilibrium, \[ E=0 \] because electric field inside a conductor is zero.
However, electric potential inside the conductor is constant and equal to the surface potential, which is generally non-zero. Hence, \[ V\neq 0 \] Therefore, Statement (A) is true.
Step 2: Analyze Statement (B).
An equipotential surface has constant electric potential throughout.
Work done in moving a charge between two points is \[ W=q\Delta V \] Since \[ \Delta V=0 \] on an equipotential surface, \[ W=0 \] Hence, Statement (B) is true.
Step 3: Analyze Statement (C).
Electrostatic potential energy of two point charges is \[ U=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r} \] For like charges, \[ q_1q_2\gt 0 \] When the charges are brought closer, \(r\) decreases. Therefore, \[ U\propto \frac{1}{r} \] so the potential energy increases.
Hence, Statement (C) is also true.
Step 4: Final conclusion.
All the three statements are correct.
Therefore, \[ \boxed{\text{A, B, C are true}} \]
Inside a charged hollow conducting sphere in electrostatic equilibrium, \[ E=0 \] because electric field inside a conductor is zero.
However, electric potential inside the conductor is constant and equal to the surface potential, which is generally non-zero. Hence, \[ V\neq 0 \] Therefore, Statement (A) is true.
Step 2: Analyze Statement (B).
An equipotential surface has constant electric potential throughout.
Work done in moving a charge between two points is \[ W=q\Delta V \] Since \[ \Delta V=0 \] on an equipotential surface, \[ W=0 \] Hence, Statement (B) is true.
Step 3: Analyze Statement (C).
Electrostatic potential energy of two point charges is \[ U=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r} \] For like charges, \[ q_1q_2\gt 0 \] When the charges are brought closer, \(r\) decreases. Therefore, \[ U\propto \frac{1}{r} \] so the potential energy increases.
Hence, Statement (C) is also true.
Step 4: Final conclusion.
All the three statements are correct.
Therefore, \[ \boxed{\text{A, B, C are true}} \]
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