Question:medium

A charge q is located at the centre of a cube. The electric flux through any face is

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In some formulations, electric flux is expressed using solid angle \(4\pi\) factor.
Updated On: Jun 19, 2026
  • \(\frac{q}{6\epsilon_0}\)
  • \(\frac{q}{6\pi\epsilon_0}\)
  • \(\frac{2\pi q}{6\epsilon_0}\)
  • \(\frac{4\pi q}{6\epsilon_0}\)
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The Correct Option is D

Solution and Explanation

To solve this problem, we need to use Gauss's Law, which relates the electric flux passing through a closed surface to the charge enclosed by the surface. Gauss's Law is mathematically expressed as:

\(\Phi = \frac{q}{\epsilon_0}\)

where:

  • \(\Phi\) is the total electric flux through the closed surface.
  • \(q\) is the total charge enclosed.
  • \(\epsilon_0\) is the permittivity of free space.

When a charge \(q\) is placed at the center of a cube, it is symmetrically enclosed by the cube. The total flux through the entire cube is given by Gauss's Law as:

\(\frac{q}{\epsilon_0}\)

The cube has six faces, and because of symmetry, the electric flux through each face of the cube will be equal. Thus, the flux through one face is:

\(\frac{1}{6} \times \frac{q}{\epsilon_0} = \frac{q}{6\epsilon_0}\)

However, the electric flux through any face of the cube is not affected by \(4\pi\) as the correct factor stems directly from the symmetry and division by the six faces of the cube. Therefore, no additional factors such as \(4\pi\) are needed. The initial correct answer given was incorrect upon review.

The correct electric flux through any face, given symmetry and following from Gauss's Law, is indeed:

\(\frac{q}{6\epsilon_0}\)

Thus, the answer aligns with the physics principles of symmetry and Gauss's Law.

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