The electric field due to a charged conducting sphere of radius \( 10 \) cm at a distance of \( 20 \) cm from its centre is \( 3\times10^{3} \) N/C and the direction of the field is radially inward. The net charge on the sphere is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( -13.33 \) nC
By Gauss’s law, outside a charged conducting sphere the field is the same as that of a point charge at its centre, so the radius of the sphere does not enter the calculation.
\( r = 20\,\mathrm{cm} = 0.20\,\mathrm{m} \)
\( \displaystyle E = \frac{1}{4\pi\varepsilon_{0}}\frac{|Q|}{r^{2}} \)
\( \displaystyle |Q| = \frac{Er^{2}}{9\times10^{9}} = \frac{(3\times10^{3}\,\mathrm{N\,C^{-1}})(0.20\,\mathrm{m})^{2}}{9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}}} \)
\( \displaystyle |Q| = 1.33\times10^{-8}\,\mathrm{C} = 13.33\,\mathrm{nC} \)
The field is radially inward, so the field lines end on the sphere and its charge is negative:
\( Q = -13.33\,\mathrm{nC} \)
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 – …
