JEE Main 2024 — Electrostatics Question with Solution

The most appropriate answer from the options given would be Option A: Both (A) and (R) are correct, and (R) is the correct explanation of (A).

Here is the reasoning for this answer:

Assertion (A) states that the work done by an electric field on moving a positive charge on an equipotential surface is always zero. This statement is true because by definition, an equipotential surface is a surface over which the electric potential is constant. When a charge moves along an equipotential surface, there is no change in its electric potential energy since potential difference $$\Delta V$$ is zero. Work done ($$W$$) is defined as the product of charge ($$q$$), potential difference ($$\Delta V$$), and the cosine of the angle between the field and direction of motion ($$\cos \theta$$), which can be written as:

$$W=q\Delta V\cos \theta$$

Because $$\Delta V=0$$ on an equipotential surface, irrespective of the value of $$\cos \theta$$, the work $$W$$ will be zero. Hence, the Assertion (A) is correct.

Reason (R) says that electric lines of forces are always perpendicular to equipotential surfaces. This statement is also correct as the electric field lines, by definition, are directed such that they are tangent to the electric field vector at any point in space. Since the electric potential is constant on an equipotential surface, there can be no component of the electric field parallel to the surface, as that would imply a force and potential change along the surface. The electric field thus must be perpendicular to the equipotential surface, which means that the electric field lines must also be perpendicular to the equipotential surface. In other words, the electric field does no work when a charge moves along an equipotential surface because the motion is perpendicular to the force.

Therefore, Reason (R) is not only correct, but it is also the correct explanation for Assertion (A), making Option A the right choice.