$$\sigma$$ 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 :

We need to find the electric field at any point on the surface of a thin spherical shell with uniform surface charge density $$\sigma$$ and radius $$R$$.

The total charge on the shell is $$Q = \sigma \cdot 4\pi R^2$$.

By Gauss's law, the electric field just outside the shell (at distance $$R$$ from center) is:

$$E_{\text{outside}} = \frac{Q}{4\pi\epsilon_0 R^2} = \frac{\sigma \cdot 4\pi R^2}{4\pi\epsilon_0 R^2} = \frac{\sigma}{\epsilon_0}$$

The electric field just inside the shell is $$E_{\text{inside}} = 0$$ (no enclosed charge for a Gaussian surface inside).

For a conducting shell or a charged surface, the electric field AT the surface is taken as the value just outside the surface, which is $$\frac{\sigma}{\epsilon_0}$$. This is the standard convention used in JEE for the field "on the surface" of a spherical shell.

The correct answer is Option B: $$\sigma/\epsilon_0$$.