A point charge q moves from point A to point D along the path ABCD (as shown) in a uniform electric field E pointing parallel to negative direction of X-axis. The co-ordinates of A, B, C, D are (0, 0, 0), (a, b, 0), (3a, 0, 0) and (2a, -b, 0) respectively. The work done by the field in the above process is given by expression:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( -2aqE \)
The electrostatic force is a conservative force, so the work done by the field depends only on the starting and final points, not on the path ABCD.
Take \( +x \) as the positive direction. The field points along \( -x \), so the force on the charge is
\( \vec{F} = q\vec{E} = -qE\,\hat{\imath} \)
The displacement from A (0, 0, 0) to D (2a, \(-b\), 0) is
\( \vec{d} = 2a\,\hat{\imath} - b\,\hat{\jmath} \)
\( W = \vec{F}\cdot\vec{d} = (-qE)(2a) + 0 \times (-b) = -2aqE \)
The work is negative because the net displacement along \( x \) is opposite to the force; the \( y \)-part of the displacement is perpendicular to the field and does no work.