Two positive point charges of \( 0.2\,\mu\mathrm{C} \) and \( 0.01\,\mu\mathrm{C} \) are placed 10 cm apart. The work done in reducing the distance to 5 cm will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1.8 \times 10^{-4}\,\mathrm{J} \)
The work done in bringing the charges closer equals the increase in their electrostatic potential energy, \( U = k q_1 q_2/r \).
\( \displaystyle W = U_2 - U_1 = k q_1 q_2\left(\frac{1}{r_2} - \frac{1}{r_1}\right) \)
\( q_1 = 0.2\,\mu\mathrm{C} = 2\times10^{-7}\,\mathrm{C} \)
\( q_2 = 0.01\,\mu\mathrm{C} = 1\times10^{-8}\,\mathrm{C} \)
\( r_1 = 0.10\,\mathrm{m},\ r_2 = 0.05\,\mathrm{m} \)
\( k = 9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}} \)
\( \displaystyle k q_1 q_2 = 9\times10^{9} \times 2\times10^{-7} \times 1\times10^{-8} = 1.8\times10^{-5}\,\mathrm{J\,m} \)
\( \displaystyle W = 1.8\times10^{-5}\left(\frac{1}{0.05} - \frac{1}{0.10}\right) = 1.8\times10^{-5} \times 10 = 1.8\times10^{-4}\,\mathrm{J} \)
The work is positive because like charges repel, so work must be done against the repulsion to bring them closer.
Note: A similar question was also asked in CUET (UG) 2026 on 30-05-2026 (Slot 2).