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JEE PHYSICS MCQ: ELECTRIC FIELDS AND CHARGES

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Application +4 | -1
A positive charge of 17.7μC17.7 \mu \mathrm{C} is placed at the centre of a hollow sphere of radius 0.5 m0.5 \mathrm{~m}. Calculate the flux density through the surface of the sphere.
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Correct option: A
6.4×105 NC−16.4 \times 10^5 \mathrm{~NC}^{-1}
Option A (6.4 x 10⁵ NC⁻¹) is correct because it is calculated using Gauss's law, which relates the electric flux to the enclosed charge.

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