Q 12-01-051JEE MainJEE Main 2024 (6 Apr, Shift 1)Easy
$\sigma$ is the uniform surface charge density of a thin spherical shell of radius $R$. The electric field at any point on the surface of the spherical shell is
Answer: (B) $\sigma/\varepsilon_0$
By Gauss's law, at the surface (just outside the shell):
$$E = \frac{Q}{4\pi\varepsilon_0R^2} = \frac{\sigma\,4\pi R^2}{4\pi\varepsilon_0R^2} = \frac{\sigma}{\varepsilon_0}$$
Solution by Sreeraj P, M.Sc Physics