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

To find the electric field using Gauss's law: \[ \oint \vec{E} \cdot d\vec{A} = \frac{Q}{\varepsilon_0}, \] where \( Q \) is the total charge enclosed and \( \varepsilon_0 \) is the permittivity of free space. Step 1: Relate charge to surface charge density. The total charge \( Q \) on the spherical shell is related to the surface charge density \( \sigma \) and the surface area of the shell (which is \( 4\pi R^2 \)): \[ Q = \sigma \cdot 4\pi R^2. \] Step 2: Simplify the electric field. By symmetry, the electric field on the surface of the spherical shell is uniform, and the flux through the Gaussian surface (a sphere of radius \( R \)) is: \[ \oint \vec{E} \cdot d\vec{A} = E \cdot 4\pi R^2, \] where \( E \) is the magnitude of the electric field. Substituting into Gauss's law: \[ E \cdot 4\pi R^2 = \frac{Q}{\varepsilon_0}. \] Substitute the expression for \( Q \): \[ E \cdot 4\pi R^2 = \frac{\sigma \cdot 4\pi R^2}{\varepsilon_0}. \] Canceling \( 4\pi R^2 \) from both sides: \[ E = \frac{\sigma}{\varepsilon_0}. \] Final Answer: \[ \boxed{\frac{\sigma}{\varepsilon_0}}. \]