A system consists of two charges \( 5\ \mu\mathrm{C} \) and \( -4\ \mu\mathrm{C} \) placed at \( \vec{r}_{1}\ (-10\ \mathrm{cm},\ 0,\ 0) \) and \( \vec{r}_{2}\ (10\ \mathrm{cm},\ 0,\ 0) \), respectively in an external electric field \( \mathrm{E} = \displaystyle\frac{A}{r^{2}} \), \( A = 10^{6}\ \mathrm{N\,C^{-1}m^{2}} \). The electrostatic potential energy of the configuration will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 9.1\ \mathrm{J} \)
In an external field, the potential energy of a system of charges is the energy of each charge in the external potential plus the mutual potential energy of the pair.
\( \displaystyle U = q_{1}V(r_{1}) + q_{2}V(r_{2}) + \frac{1}{4\pi\varepsilon_{0}}\frac{q_{1}q_{2}}{r_{12}} \)
The external field \( E = A/r^{2} \) is radial, like the field of a point charge, so its potential (zero at infinity) is \( V(r) = A/r \).
\( r_{1} = r_{2} = 10\,\mathrm{cm} = 0.1\,\mathrm{m} \)
\( \displaystyle V = \frac{A}{r} = \frac{10^{6}\,\mathrm{N\,C^{-1}m^{2}}}{0.1\,\mathrm{m}} = 10^{7}\,\mathrm{V} \)
\( q_{1}V = (5\times10^{-6}\,\mathrm{C})(10^{7}\,\mathrm{V}) = 50\,\mathrm{J} \)
\( q_{2}V = (-4\times10^{-6}\,\mathrm{C})(10^{7}\,\mathrm{V}) = -40\,\mathrm{J} \)
The charges are \( r_{12} = 20\,\mathrm{cm} = 0.2\,\mathrm{m} \) apart:
\( \displaystyle U_{12} = \frac{(9\times10^{9})(5\times10^{-6})(-4\times10^{-6})}{0.2}\,\mathrm{J} = -0.9\,\mathrm{J} \)
\( U = 50\,\mathrm{J} - 40\,\mathrm{J} - 0.9\,\mathrm{J} = 9.1\,\mathrm{J} \)
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 – “… Example 2.5 (a) Determine the electrostatic potential energy of a system consisting of two charges 7 μC and –2 μC (and with no external field) placed at (–9 cm, 0, 0) and (9 cm, 0, 0) respectively. (b) How much work is required to separate the two charges infinitely away from each other? (c) Suppose that the same system of charges is now placed in an external electric field E = A (1/r2); A = 9 × 105 NC–1 m2. What would the electrostatic energy of the configuration be?”