Two small balls, each with a charge of \(+2\ \mu\mathrm{C}\), are held 6 cm apart and then released. Ignore gravity. When they are very far apart, the total kinetic energy of the two balls is (Take \(\frac{1}{4\pi\varepsilon_0}=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(0.6\ \mathrm{J}\)
The two like charges repel and fly apart. The electrostatic force is conservative, so the stored potential energy changes into kinetic energy. When the balls are very far apart, \( U \) is zero. So the total kinetic energy equals the starting potential energy of the pair.
\( \begin{aligned} U &= \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r} \\ &= \frac{9\times10^{9}\times(2\times10^{-6})^{2}}{0.06} \\ &= 0.6\ \mathrm{J} \end{aligned} \)
Therefore, the two balls together gain 0.6 J of kinetic energy.
NCERT (2026-27):
Introduction – “… When the external force is removed, the body moves, gaining kinetic energy and losing an equal amount of potential energy. The sum of kinetic and potential energies is thus conserved. …”
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”
\( U=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r_{12}} \)\( \qquad (2.22) \)