Q146 / 160 · AP EAMCET 2022 (04 Jul Shift 2) · Physics

Statement (A): Inside a charged hollow metal sphere, E=0, V≠0. (E- electric field, V-eslectric 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.

  • Correct answer
    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

Solution

Inside a charged hollow, sphere electric field is constant. Now,

dV=-E→·dr→ therefore, dV=0 hence V is constant and not zero.

Potential difference is defined as the work done in moving per unit charge in moving it from one point to other. Therefore, work done in moving a charge on an equipotential surface is zero.

When two like charges are brought closer, work is done against electrostatic force. Therefore, their mutual electrostatic potential energy increases.

Hence, all the three statements are correct.

Hence, the correct answer is option: 'A'.

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