Two point charges, \(+4\ \mathrm{nC}\) and \(-2\ \mathrm{nC}\), are 9 cm apart in an external electric field. The external potential is 300 V at the position of the \(+4\ \mathrm{nC}\) charge. It is 100 V at the position of the \(-2\ \mathrm{nC}\) charge. The total electrostatic potential energy of the system is (Take \(\frac{1}{4\pi\varepsilon_0}=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(2\times10^{-7}\ \mathrm{J}\)
In an external field, the energy of a pair of charges has two parts. Each charge has energy \( qV \) due to the external potential at its position. The pair also has its own mutual energy, \( \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r_{12}} \).
\( \begin{aligned} q_1V(\mathbf{r}_1) &= 4\times10^{-9}\times300 \\ &= 1.2\times10^{-6}\ \mathrm{J} \end{aligned} \)
\( \begin{aligned} q_2V(\mathbf{r}_2) &= (-2\times10^{-9})\times100 \\ &= -2\times10^{-7}\ \mathrm{J} \end{aligned} \)
\( \begin{aligned} U_{12} &= \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r_{12}} \\ &= \frac{9\times10^{9}\times4\times10^{-9}\times(-2\times10^{-9})}{0.09} \\ &= -8\times10^{-7}\ \mathrm{J} \end{aligned} \)
\( \begin{aligned} U &= q_1V(\mathbf{r}_1)+q_2V(\mathbf{r}_2)+U_{12} \\ &= (12-2-8)\times10^{-7} \\ &= 2\times10^{-7}\ \mathrm{J} \end{aligned} \)
NCERT (2026-27):
Potential energy of a system of two charges in an external field – “… Thus, Potential energy of the system = the total work done in assembling the configuration”
\( =q_1V(\mathbf{r}_1)+q_2V(\mathbf{r}_2)+\frac{q_1q_2}{4\pi\varepsilon_0r_{12}} \)\( \qquad (2.29) \)