A point charge of \( +2\,\mu\mathrm{C} \) is placed at the centre of a thick conducting spherical shell of inner radius 5 cm and outer radius 10 cm. The shell carries a net charge of \( -5\,\mu\mathrm{C} \). The electric field at a point 30 cm from the centre is (Take \( \frac{1}{4\pi\varepsilon_0}=9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}} \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 3\times10^{5}\,\mathrm{N\,C^{-1}} \), directed radially inward
The point lies outside the shell, so the field there follows from Gauss’s law applied to a concentric spherical Gaussian surface of radius 30 cm.
By symmetry, the field on this surface is radial and has the same magnitude at every point.
The flux through it depends only on the total charge enclosed. This is the point charge together with the whole charge of the shell. The shell charge settles as \( -2\,\mu\mathrm{C} \) on its inner surface and \( -3\,\mu\mathrm{C} \) on its outer surface. Both lie inside the Gaussian surface.
The net charge enclosed is:
\( \begin{aligned} q_{\mathrm{enc}} &= (+2\,\mu\mathrm{C})+(-5\,\mu\mathrm{C}) \\ &= -3\,\mu\mathrm{C} \end{aligned} \)
From \( E\times4\pi r^{2}=q_{\mathrm{enc}}/\varepsilon_0 \), the magnitude of the field at \( r=0.30\,\mathrm{m} \) is:
\( \begin{aligned} E &= \frac{1}{4\pi\varepsilon_0}\,\frac{|q_{\mathrm{enc}}|}{r^{2}} \\ &= \frac{9\times10^{9}\times3\times10^{-6}}{(0.30)^{2}}\,\mathrm{N\,C^{-1}} \\ &= 3\times10^{5}\,\mathrm{N\,C^{-1}} \end{aligned} \)
The enclosed charge is negative, so the field points towards the centre.
NCERT (2026-27):
Summary – “15. Gauss’s law: The flux of electric field through any closed surface S is 1/ε0 times the total charge enclosed by S. …”