QJEEM24Phba04: JEE Main 2024 Physics - Class 12 JEEM24Phba04 Solution: is the uniform surface charge density of a thin spherical shell …

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 ( $\epsilon_0$ ), mathematically represented as:

$\Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0}$

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:

$E \cdot 4\pi R^2 = \frac{\sigma \cdot 4\pi R^2}{\epsilon_0}$

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

$E = \frac{\sigma}{\epsilon_0}$

Therefore, the correct answer is Option B:

$\frac{\sigma}{\epsilon_0}$