The work done by an external force in bringing a charge of \(3\ \mathrm{nC}\) slowly from infinity to a point P is \(1.2\times10^{-5}\ \mathrm{J}\). The work needed to bring a charge of \(5\ \mathrm{nC}\) slowly from infinity to the same point P is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(2\times10^{-5}\ \mathrm{J}\)
The work done in bringing a charge to a point is proportional to the charge. The work per unit charge does not depend on the charge used. It is the potential at the point, \( V=W/q \). So first find the potential at P from the given work.
\( \begin{aligned} V_P &= \frac{W_1}{q_1} \\ &= \frac{1.2\times10^{-5}}{3\times10^{-9}} \\ &= 4000\ \mathrm{V} \end{aligned} \)
\( \begin{aligned} W_2 &= q_2V_P \\ &= 5\times10^{-9}\times4000 \\ &= 2\times10^{-5}\ \mathrm{J} \end{aligned} \)
NCERT (2026-27):
Electrostatic Potential – “This work is obviously proportional to q, since the force at any point is qE, where E is the electric field at that point due to the given charge configuration. It is, therefore, convenient to divide the work by the amount of charge q, so that the resulting quantity is independent of q.”
Electrostatic Potential – “Work done by an external force in bringing a unit positive charge from infinity to a point = electrostatic potential (V ) at that point.”