The surface charge density of a thin spherical shell placed in an air medium is 88.54 c/m² The intensity of the electric field measured 12 mm outside the shell from the centre of the shell is 5.625 × 10¹² N/C. The thin spherical shell has a radius of:

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AAI ATC Junior Executive 21 Feb 2023 Shift 1 Official Paper

Answer 1

Given:

Surface charge density \(\sigma=88.54C/m^2\)

Electric field \(E=5.625\times10^{12}N/C\)

Distance measured outside the shell, \(x=12mm=12\times10^{-3}m\)

Concept:

For a thin uniformly charged spherical shell, the Electric field points outside the shell at a distance x from the centre is

\(E=\frac{1}{4\pi \epsilon_o}.\frac{Q}{x^2}\)
If the radius of sphere is R, \(Q=\sigma4\pi r^2\)
\(E=\frac{1}{4\pi \epsilon_o}.\frac{\sigma.4\pi R^2}{x^2}=\frac{\sigma}{\epsilon_o}\frac{R^2}{x^2}\)

Explanation:

Put all the given values in above equation to get the value of radius, we get,

\(R^2=\frac{E.x^2.\epsilon_0}{\sigma}=\frac{5.625\times10^{12}\times8.85\times 10^{-12}\times 12\times12\times10^{-6}}{88.54}\)

\(R^2=80.963\times10^{-6}m\)

\(\implies R=\sqrt{80.963\times10^{-6}}m=8.99\times10^{-3}m=9.0mm\)

Hence, the correct answer is \(R=9.0mm\).

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