Two positive point charges of 0.2 μC and 0.01 μC are placed 10 cm apart. The work done in reducing their separation to 5 cm is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1.8 × 10−4 J
The work done in bringing the charges closer equals the increase in their electrostatic potential energy. Both charges are positive, so they repel and positive work has to be done.
\( \displaystyle W = U_f - U_i = \frac{q_1q_2}{4\pi\varepsilon_0}\left(\frac{1}{r_2} - \frac{1}{r_1}\right) \)
\( q_1 = 0.2\,\mu\mathrm{C} = 2\times10^{-7}\,\mathrm{C} \)
\( q_2 = 0.01\,\mu\mathrm{C} = 1\times10^{-8}\,\mathrm{C} \)
\( r_1 = 10\,\mathrm{cm} = 0.10\,\mathrm{m} \)
\( r_2 = 5\,\mathrm{cm} = 0.05\,\mathrm{m} \)
\( \displaystyle W = 9\times10^{9} \times 2\times10^{-7} \times 1\times10^{-8} \times \left(\frac{1}{0.05} - \frac{1}{0.10}\right)\,\mathrm{J} \)
\( \displaystyle W = 1.8\times10^{-5} \times (20 - 10)\,\mathrm{J} \)
\( \displaystyle W = 1.8\times10^{-4}\,\mathrm{J} \)
Note: A similar question was also asked in CUET (UG) 2025 on 03-06-2025 (Slot 2).