Two charges \( 10\,\mu\mathrm{C} \) and \( 12\,\mu\mathrm{C} \) are 10 cm apart in vacuum. The work done in bringing them 4 cm closer will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 7.2 J
The work done in bringing two like charges closer equals the increase in their electrostatic potential energy \( U = k q_1 q_2/r \).
\( q_1 = 10\times10^{-6}\,\mathrm{C} \)
\( q_2 = 12\times10^{-6}\,\mathrm{C} \)
\( r_1 = 10\,\mathrm{cm} = 0.10\,\mathrm{m} \)
\( r_2 = 10 - 4 = 6\,\mathrm{cm} = 0.06\,\mathrm{m} \)
\( \displaystyle W = k q_1 q_2\left(\frac{1}{r_2} - \frac{1}{r_1}\right) \)
\( \displaystyle k q_1 q_2 = 9\times10^{9} \times 10\times10^{-6} \times 12\times10^{-6} = 1.08\,\mathrm{N\,m^{2}} \)
\( \displaystyle \frac{1}{0.06} - \frac{1}{0.10} = 16.67 - 10 = 6.67\,\mathrm{m^{-1}} \)
\( \displaystyle W = 1.08 \times 6.67 = 7.2\,\mathrm{J} \)