Two point charges \( q_1 = 10 \times 10^{-8}\,\mathrm{C} \) and \( q_2 = -2 \times 10^{-8}\,\mathrm{C} \) are separated by a distance of 6 cm. The distance from \( q_1 \) where the electric potential is zero, can be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 5 cm and 7.5 cm
The potential is a scalar, so the potentials of the two charges simply add. It is zero where \( kq_1/r_1 = k|q_2|/r_2 \). Since the charges have opposite signs, such points exist on the line joining them, and they lie nearer the smaller charge \( q_2 \).
Take \( q_1 \) at \( x = 0 \) and \( q_2 \) at \( x = 6\,\mathrm{cm} \).
Point between the charges, at distance \( x \) from \( q_1 \):
\( \displaystyle \frac{10}{x} = \frac{2}{6 - x} \;\Rightarrow\; 60 - 10x = 2x \;\Rightarrow\; x = 5\,\mathrm{cm} \)
Point beyond \( q_2 \), at distance \( x \gt 6\,\mathrm{cm} \) from \( q_1 \):
\( \displaystyle \frac{10}{x} = \frac{2}{x - 6} \;\Rightarrow\; 10x - 60 = 2x \;\Rightarrow\; x = 7.5\,\mathrm{cm} \)
A point beyond \( q_1 \) is closer to the larger charge, so the two terms can never cancel there.
So the potential is zero at 5 cm and at 7.5 cm from \( q_1 \).
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Potential Due to a System of Charges – …
