The potential at a point due to the charge of \( 2\times10^{-7} \) C located \( 6 \) cm away from it will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 3\times10^{4}\ \mathrm{V} \)
The potential due to a point charge q at a distance r is
\( \displaystyle V = \frac{1}{4\pi\varepsilon_{0}}\,\frac{q}{r} \)
Converting the distance to metres:
\( r = 6\ \mathrm{cm} = 0.06\ \mathrm{m} \)
\( \displaystyle V = \frac{(9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}) \times (2\times10^{-7}\ \mathrm{C})}{0.06\ \mathrm{m}} \)
\( \displaystyle V = \frac{1800}{0.06}\ \mathrm{V} = 3\times10^{4}\ \mathrm{V} \)
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 Point Charge – …
