A system consisting of two point charges \( 7\,\mu\mathrm{C} \) and \( -4\,\mu\mathrm{C} \) are placed at \( (-9, 0, 0)\,\mathrm{cm} \) and \( (9, 0, 0)\,\mathrm{cm} \) respectively. The electrostatic potential energy of the system is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( -1.4\,\mathrm{J} \)
The electrostatic potential energy of a system of two point charges is the work done in bringing them from infinity to their positions.
\( \displaystyle U = \frac{1}{4\pi\varepsilon_0}\,\frac{q_1 q_2}{r} \)
\( r = 9 - (-9) = 18\,\mathrm{cm} = 0.18\,\mathrm{m} \)
\( q_1 = 7\times10^{-6}\,\mathrm{C} \)
\( q_2 = -4\times10^{-6}\,\mathrm{C} \)
\( \displaystyle U = \frac{9\times10^{9} \times 7\times10^{-6} \times (-4\times10^{-6})}{0.18} = \frac{-0.252}{0.18} = -1.4\,\mathrm{J} \)
The energy is negative because the charges are unlike and attract each other.
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Potential Energy of a System of Two Charges in an External Field – …
