$\sigma$ 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 :

<p>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.</p> <p>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:</p> <p>$\Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0}$</p> <p>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:</p> <p>$Q_{\text{enc}} = \sigma \times 4\pi R^2$</p> <p>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:</p> <p>$\Phi_E = E \cdot 4\pi R^2$</p> <p>Using Gauss's law:</p> <p>$E \cdot 4\pi R^2 = \frac{\sigma \cdot 4\pi R^2}{\epsilon_0}$</p> <p>Simplifying this equation gives us the electric field at the surface of the spherical shell:</p> <p>$E = \frac{\sigma}{\epsilon_0}$</p> <p>Therefore, the correct answer is <strong>Option B</strong>:</p> <p>$\frac{\sigma}{\epsilon_0}$</p>