Question

$\sigma$ 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:

• A. $\frac{\sigma}{\epsilon_0 R}$
• B. $\frac{\sigma}{2\epsilon_0}$
• C. $\frac{\sigma}{\epsilon_0}$
• D. $\frac{\sigma}{4\epsilon_0}$

Answer 1

Answer: (C)

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

Answer 2

By Gauss's Law:

$\oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{in}}}{\varepsilon_0}.$

For a spherical shell, the electric field $E$ is constant across the surface area:

$E \cdot 4\pi R^2 = \frac{\sigma \cdot 4\pi R^2}{\varepsilon_0}.$

$E = \frac{\sigma}{\varepsilon_0}.$

Thus, the electric field at the surface of the spherical shell is:

$\boxed{\frac{\sigma}{\varepsilon_0}}.$