The electric potential at point A is 120 V and at point B is 40 V. A charge of \(-3\ \mu\mathrm{C}\) is moved slowly from A to B. The work done by the external force is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(2.4\times10^{-4}\ \mathrm{J}\)
The potential difference \( V_B-V_A \) is the work done by an external force per unit positive charge in moving it from A to B. So for a charge \( q \), the work is \( W=q(V_B-V_A) \), with the sign of \( q \) kept.
\( \begin{aligned} W &= q(V_B-V_A) \\ &= (-3\times10^{-6})\times(40-120) \\ &= 2.4\times10^{-4}\ \mathrm{J} \end{aligned} \)
The work is positive. A negative charge is pulled towards the higher potential at A, so the external force must push it towards B.
NCERT (2026-27):
Electrostatic Potential – “Work done by external force in bringing a unit positive charge from point R to P”
\( =V_P-V_R=\frac{U_P-U_R}{q} \)\( \qquad (2.4) \)
“where VP and VR are the electrostatic potentials at P and R, respectively.”