Two point charges q and -3q are kept 12 cm apart. The distance of the point from q on the line between two charges at which potential due to this system of charges is zero will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3 cm
The potential of a system of point charges is the algebraic sum of \( \frac{1}{4\pi\varepsilon_0}\frac{q_i}{r_i} \). Let the point be at a distance x from q, so it is \( (12 - x)\,\mathrm{cm} \) from \( -3q \).
\( \displaystyle V = \frac{1}{4\pi\varepsilon_0}\left[\frac{q}{x} + \frac{(-3q)}{12 - x}\right] = 0 \)
\( \displaystyle \frac{1}{x} = \frac{3}{12 - x} \Rightarrow 12 - x = 3x \Rightarrow x = 3\,\mathrm{cm} \)
Shortcut: The potential is zero where the distances are in the inverse ratio of the charge magnitudes. The distances are in the ratio 1 : 3, so the point divides the 12 cm in that ratio: \( \frac{12}{4} = 3\,\mathrm{cm} \) from q.
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
