A charge of \(+3\ \mathrm{nC}\) is placed at the origin and a charge of \(-2\ \mathrm{nC}\) at the point (0.4 m, 0). The electric potential at the point (0.4 m, 0.3 m) is (take \(1/4\pi\varepsilon_0=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(-6\ \mathrm{V}\)
Potential is a scalar. The potential at a point due to several charges is the algebraic sum of the potentials due to each charge. The sign of each charge is kept.
The point is 0.3 m from the \( -2\ \mathrm{nC} \) charge. Its distance from the origin is \( \sqrt{0.4^{2}+0.3^{2}}=0.5\ \mathrm{m} \).
\( \begin{aligned} V &= \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}\right) \\ &= 9\times10^{9}\left(\frac{3\times10^{-9}}{0.5}-\frac{2\times10^{-9}}{0.3}\right) \\ &= 54-60 \\ &= -6\ \mathrm{V} \end{aligned} \)
NCERT (2026-27):
Potential Due to a System of Charges – “… By the superposition principle, the potential V at P due to the total charge configuration is the algebraic sum of the potentials due to the individual charges”
\( V=V_1+V_2+\dots+V_n \qquad (2.17) \)