Work done in moving q coulomb charge along a closed path of length \( l \) by an electric field \( \vec{E} \) is given by
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Zero
The electrostatic field is a conservative field: the work it does on a charge moving between two points depends only on the potentials of those points, \( W = q(V_A - V_B) \), and not on the path taken.
On a closed path the charge returns to its starting point, so the start and end potentials are the same and
\( W = q(V_A - V_A) = 0 \)
Equivalently, \( \oint\vec{E}\cdot d\vec{l} = 0 \) for any electrostatic field, whatever the length of the path.
NCERT (2026-27):
Introduction – “… It means that the work done by an electrostatic field in moving a charge from one point to another depends only on the initial and the final points and is independent of the path taken to go from one point to the other. This is the fundamental characteristic of a conservative force.”