The electric field at 20 cm from the center of a conducting sphere of radius 2 cm is \( 4.5\times10^{3}\,\mathrm{N\,C^{-1}} \). The magnitude of charge on the sphere is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2\times10^{-8}\,\mathrm{C} \)
Outside a charged conducting sphere the field is the same as that of a point charge at its centre (Gauss's law), \( E = kQ/r^{2} \) with \( k = 1/(4\pi\varepsilon_{0}) \). The point at 20 cm lies outside the sphere of radius 2 cm, so the radius of the sphere is not needed.
\( \displaystyle Q = 4\pi\varepsilon_{0}\,E\,r^{2} = \frac{E\,r^{2}}{9\times10^{9}} \)
\( \displaystyle Q = \frac{4.5\times10^{3} \times (0.20)^{2}}{9\times10^{9}} = \frac{180}{9\times10^{9}} = 2\times10^{-8}\,\mathrm{C} \)
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 – …
