Two charges \( q_1=4\times10^{-7}\,\mathrm{C} \) and \( q_2=-6\times10^{-7}\,\mathrm{C} \) are located 3 m apart. At what point the line joining the two charges is the electric potential zero? (Consider potential zero at infinity.)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1.2 meter from charge \( q_1 \)
Potential is a scalar, so the potential at a point is the algebraic sum of the potentials of the two charges, \( V = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}\right) \). Let the point lie on the line between the charges at a distance x from \( q_1 \), so that it is \( (3-x) \) from \( q_2 \).
\( \displaystyle \frac{4\times10^{-7}}{x} + \frac{-6\times10^{-7}}{3-x} = 0 \)
\( \displaystyle \frac{4}{x} = \frac{6}{3-x} \Rightarrow 4(3-x) = 6x \Rightarrow 12 = 10x \)
\( x = 1.2\,\mathrm{m} \) from \( q_1 \) (which is \( 1.8\,\mathrm{m} \) from \( q_2 \)).
Shortcut: The potential cancels only where the larger \( 1/r \) of the smaller charge makes up for its smaller magnitude, so the point must lie nearer \( q_1 \) (\( 4\times10^{-7}\,\mathrm{C} \)) than \( q_2 \). This removes Options (2) and (4) at once, and Option (1), the mid-point, fails because \( 4\neq6 \).
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 – …
