JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Electrostatics

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Given below are two statements : one is labelled a Assertion (A) and the other is labelled as Reason(R)
Assertion (A) : Work done by electric field on moving a positive charge on an equipotential surface is always zero.
Reason (R) : Electric lines of forces are always perpendicular to equipotential surfaces.

In the light of the above statements, choose the most appropriate answer from the options given below :
  1. (A)Both (A) and (R) are correct but (R) is not the correct explanation of (A)
  2. (B)((A) is correct but (R) is not correct
  3. (C)(A) is not correct but (R) is correct
  4. (D)Both (A) and (R) are correct and (R) is the correct explanation of (A)✓ Correct

Answer: (D)

Solution

On an equipotential surface the potential is constant, and the work done moving a charge between two points is ; with , — Assertion (A) is correct. The electric field is the negative gradient of potential and has no component along a surface of constant potential, so field lines are always perpendicular to equipotential surfaces — Reason (R) is correct. This perpendicularity is exactly why the force on the charge is perpendicular to any displacement along the surface, making ; (R) is the correct explanation of (A).

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Single correct option

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