A point charge of \(+3\ \mathrm{nC}\) is at point A and a point charge of \(-1\ \mathrm{nC}\) is at point B, with AB = 10 cm. Point P lies on the line BA extended beyond A, 5 cm from A. The electric potential at P 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) → \(480\ \mathrm{V}\)
Potential is a scalar. The potential at P is the algebraic sum of the potentials of the two charges. Each is taken with its own sign and its own distance from P. P is 5 cm from A and \( 10+5=15 \) cm from B.
\( \begin{aligned} V_A &= \frac{9\times10^{9}\times3\times10^{-9}}{0.05} \\ &= 540\ \mathrm{V} \end{aligned} \)
\( \begin{aligned} V_B &= \frac{9\times10^{9}\times(-1\times10^{-9})}{0.15} \\ &= -60\ \mathrm{V} \end{aligned} \)
\( \begin{aligned} V &= V_A+V_B \\ &= 540-60 \\ &= 480\ \mathrm{V} \end{aligned} \)
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”
\( V=V_1+V_2+\ldots+V_n \qquad (2.17) \)