Point charges of \(+6\ \mu\mathrm{C}\) and \(-2\ \mu\mathrm{C}\) are held 20 cm apart. The work needed to separate them until they are very far apart is (Take \(\frac{1}{4\pi\varepsilon_0}=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\).)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(0.54\ \mathrm{J}\)
The potential energy of the pair is the work done in bringing the charges from infinity. For unlike charges it is negative. Far apart, the energy is zero, so the work needed to pull them apart is \( 0-U \).
\( \begin{aligned} U &= \frac{1}{4\pi\varepsilon_0}\,\frac{q_1q_2}{r} \\ &= \frac{9\times10^{9}\times(6\times10^{-6})\times(-2\times10^{-6})}{0.2} \\ &= -0.54\ \mathrm{J} \end{aligned} \)
\( \begin{aligned} W &= 0-U \\ &= 0.54\ \mathrm{J} \end{aligned} \)
NCERT (2026-27):
Potential Energy of a System of Charges – “Since electrostatic force is conservative, this work gets stored in the form of potential energy of the system. Thus, the potential energy of a system of two charges q1 and q2 is”
\( U=\frac{1}{4\pi\varepsilon_0}\,\frac{q_1q_2}{r_{12}} \)\( \qquad (2.22) \)
Potential Energy of a System of Charges – “For unlike charges (q1 q2 < 0), the electrostatic force is attractive. In that case, a positive amount of work is needed against this force to take the charges from the given location to infinity.”