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. \( \frac{\sigma}{\varepsilon_0 R} \)
• B. \( \frac{\sigma}{2 \varepsilon_0} \)
• C. \( \frac{\sigma}{\varepsilon_0} \)
• D. \( \frac{\sigma}{4 \varepsilon_0} \)

Hint:

For a spherical shell, the electric field on its surface depends solely on the surface charge density \( \sigma \) and the permittivity of free space \( \varepsilon_0 \), and is independent of the radius \( R \).

Answer 1

The Correct Option is C

Gauss's law, \( \oint \vec{E} \cdot d\vec{A} = \frac{Q}{\varepsilon_0} \), is used to determine the electric field. Here, \( Q \) represents the total enclosed charge, and \( \varepsilon_0 \) is the vacuum permittivity.

Step 1: Relate charge to surface density. The total charge \( Q \) on a spherical shell is calculated using the surface charge density \( \sigma \) and the shell's surface area, \( 4\pi R^2 \): \[ Q = \sigma \cdot 4\pi R^2. \] 

Step 2: Simplify the electric field calculation. Due to symmetry, the electric field at the spherical shell's surface is uniform. The flux through a spherical Gaussian surface of radius \( R \) is given by: \[ \oint \vec{E} \cdot d\vec{A} = E \cdot 4\pi R^2, \] where \( E \) is the electric field's magnitude. Substituting this into Gauss's law yields: \[ E \cdot 4\pi R^2 = \frac{Q}{\varepsilon_0}. \] Replacing \( Q \) with its expression: \[ E \cdot 4\pi R^2 = \frac{\sigma \cdot 4\pi R^2}{\varepsilon_0}. \] By canceling \( 4\pi R^2 \) from both sides: \[ E = \frac{\sigma}{\varepsilon_0}. \] 

Final Answer: \[ \boxed{\frac{\sigma}{\varepsilon_0}}. \]