Question

A thin spherical shell of radius R and surface charge density \( \sigma \) is placed in a cube of side 5R with their centers coinciding. The electric flux through one face of the cube is (\( \epsilon_o \) = Permittivity of free space)

• A. \( \frac{2\pi R^2 \sigma}{3 \epsilon_o} \)
• B. \( \frac{\pi R^2 \sigma}{3 \epsilon_o} \)
• C. \( \frac{\sigma}{6 \epsilon_o} \)
• D. \( \frac{\sigma}{4\pi \epsilon_o R^2} \)

Hint:

Symmetry arguments are crucial in Gauss's Law problems. If a charge distribution is symmetric with respect to the enclosing surface faces, simply divide the total flux by the number of faces.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:

According to Gauss's Law, the total electric flux \( \phi_{\text{total}} \) through a closed surface enclosing a charge \( q \) is \( \frac{q}{\epsilon_o} \). Due to the symmetry of the cube with the sphere at its center, the flux is distributed equally among the 6 faces.
Step 2: Key Formula or Approach:

1. Total charge on shell: \( q = \text{Surface Area} \times \sigma = 4\pi R^2 \sigma \). 2. Total Flux: \( \phi_{\text{total}} = \frac{q}{\epsilon_o} \). 3. Flux through one face: \( \phi_{\text{face}} = \frac{1}{6} \phi_{\text{total}} \).
Step 3: Detailed Explanation:

Calculate the total charge \( q \): \[ q = \sigma (4\pi R^2) \] Total flux through the cube: \[ \phi_{\text{total}} = \frac{4\pi R^2 \sigma}{\epsilon_o} \] Since the sphere is centered in the cube, flux through each of the 6 faces is identical. \[ \phi_{\text{face}} = \frac{1}{6} \times \frac{4\pi R^2 \sigma}{\epsilon_o} \] \[ \phi_{\text{face}} = \frac{2\pi R^2 \sigma}{3 \epsilon_o} \]
Step 4: Final Answer:

The flux through one face is \( \frac{2\pi R^2 \sigma}{3 \epsilon_o} \).