Question:

A point charge \( q \) coulomb is placed at the centre of a cube of side length \( L \). Then the electric flux linked with each face of the cube is:

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Gauss's law states that the total electric flux through a closed surface is proportional to the charge enclosed within the surface. For a symmetrical surface like a cube, the flux is evenly distributed over all the faces.
Updated On: May 17, 2025
  • \( \frac{q}{\epsilon_0} \)
  • \( \frac{q}{L^2 \epsilon_0} \)
  • \( \frac{q}{6L^2 \epsilon_0} \)
  • \( \frac{q}{6 \epsilon_0} \)
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The Correct Option is D

Approach Solution - 1

To find the electric flux linked with each face of the cube when a point charge \( q \) is placed at the center, we can use Gauss's Law. The total electric flux \( \Phi \) through a closed surface enclosing a charge \( q \) is given by:
\[\Phi = \frac{q}{\epsilon_0}\]
For a cube with a point charge in the center, the charge is symmetrically enclosed, and the electric flux distributes equally through all six faces of the cube. Therefore, the flux through one face of the cube is:
\[\Phi_{\text{face}} = \frac{\Phi_{\text{total}}}{6} = \frac{q}{6\epsilon_0}\]
This gives us the flux linked with each face of the cube as:
\[\frac{q}{6\epsilon_0}\]
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Approach Solution -2

We are given a point charge \( q \) placed at the centre of a cube with side length \( L \). We are asked to find the electric flux linked with each face of the cube. 
Step 1: We use Gauss's law to find the total electric flux through a closed surface. According to Gauss's law: \[ \Phi_{{total}} = \frac{q}{\epsilon_0} \] where \( \Phi_{{total}} \) is the total electric flux and \( \epsilon_0 \) is the permittivity of free space. 
Step 2: The cube has 6 faces, and the point charge \( q \) is located at the centre of the cube. Since the electric flux is symmetric, the flux through each face of the cube is the same. Thus, the flux linked with each face of the cube is: \[ \Phi_{{face}} = \frac{\Phi_{{total}}}{6} = \frac{q}{6 \epsilon_0} \] Thus, the electric flux linked with each face of the cube is \( \frac{q}{6 \epsilon_0} \).

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