A charge of magnitude \( 3 \times 10^{-7}\,\mathrm{C} \) is located at a distance of \( 0.09\,\mathrm{m} \) from a point P. Obtain the work done in bringing a charge of \( 2 \times 10^{-9}\,\mathrm{C} \) from infinity to the point P.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 6 \times 10^{-5}\,\mathrm{J} \)
The work done in bringing a charge from infinity to a point is the charge times the electric potential at that point, because the potential at infinity is zero.
\( \displaystyle W = Q\,V, \qquad V = \frac{kq}{r} \)
\( \displaystyle V = \frac{9\times10^{9} \times 3\times10^{-7}}{0.09} = 3\times10^{4}\,\mathrm{V} \)
\( \displaystyle W = 2\times10^{-9} \times 3\times10^{4} = 6\times10^{-5}\,\mathrm{J} \)
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 – …
