Q 12-02-198JEE MainJEE Main 2017 (8 Apr)Easy
The energy stored in the electric field produced by a metal sphere is $4.5$ J. If the sphere contains $4\ \mu$C charge, its radius will be: [Take $\dfrac{1}{4\pi\varepsilon_0} = 9\times10^9\ \text{N m}^2\text{C}^{-2}$]
Answer: (B) $16$ mm
For an isolated sphere $C = 4\pi\varepsilon_0R$, so
$$U = \frac{Q^2}{2C} = \frac{1}{4\pi\varepsilon_0}\frac{Q^2}{2R}$$
$$R = \frac{9\times10^9\times(4\times10^{-6})^2}{2\times4.5} = \frac{0.144}{9} = 0.016\ \text{m} = 16\ \text{mm}$$
Solution by Sreeraj P, M.Sc Physics