Two point charges, \( +2\,\mu\mathrm{C} \) and \( -3\,\mu\mathrm{C} \), are separated by a distance of 20 cm in the air. At what point between the two charges and on the line joining the two charges is the electric potential zero? (Take the potential at infinity to be zero)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 8 cm from charge \( +2\,\mu\mathrm{C} \)
Potential is a scalar, so the potentials of the two charges simply add. Between the charges they have opposite signs, and the total is zero where they are equal in size. Let the point be \( x \) from the \( +2\,\mu\mathrm{C} \) charge.
\( \displaystyle \frac{k\,(+2\,\mu\mathrm{C})}{x} + \frac{k\,(-3\,\mu\mathrm{C})}{20 - x} = 0 \;\Rightarrow\; \frac{2}{x} = \frac{3}{20 - x}\quad (x \text{ in cm}) \)
\( 2(20 - x) = 3x \)
\( 40 = 5x \)
\( x = 8\,\mathrm{cm} \)
The point is 8 cm from the \( +2\,\mu\mathrm{C} \) charge (and 12 cm from the \( -3\,\mu\mathrm{C} \) charge), nearer the smaller charge as expected.
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 – …
