JEE Main 2024 — Electrostatics Question with Solution

The question is about calculating the electric field at the surface of a thin spherical shell with a uniform surface charge density denoted by $$\sigma$$. To determine the electric field at any point on the surface of the shell, we can use Gauss's law, which is particularly useful for systems with high symmetry like a spherical shell.

Gauss's law in its integral form states that the electric flux through a closed surface is equal to the charge enclosed by the surface divided by the permittivity of free space ($${ϵ}_0$$), mathematically represented as:

$${\Phi }_E=\frac{Q_{\text{enc}}}{{ϵ}_0}$$

In the case of a spherical shell of radius $$R$$, the total charge $$Q_{\text{enc}}$$ on the shell can be written in terms of the surface charge density $$\sigma$$ as:

$$Q_{\text{enc}}=\sigma \times 4\pi R^2$$

Now, we apply Gauss's law using a Gaussian surface that coincides with the surface of the spherical shell. The electric field $$E$$ at the surface of the shell is uniform over the Gaussian surface, and the area of the Gaussian surface (which is also the area of the spherical shell) is $$4\pi R^2$$. Thus, the electric flux $${\Phi }_E$$ through the Gaussian surface is:

$${\Phi }_E=E\cdot 4\pi R^2$$

Using Gauss's law:

$$E\cdot 4\pi R^2=\frac{\sigma \cdot 4\pi R^2}{{ϵ}_0}$$

Simplifying this equation gives us the electric field at the surface of the spherical shell:

$$E=\frac{\sigma }{{ϵ}_0}$$

Therefore, the correct answer is Option B:

$$\frac{\sigma }{{ϵ}_0}$$