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A charge q is placed at the centre of a cube, the magnitude of flux through one of its faces will be: ans. q q / 6 6qq / 3  Previous Marked for Review Next

Question

A charge q is placed at the centre of a cube, the magnitude of flux through one of its faces will be:

  • ans.
  • q
  • q / 6
  • 6qq / 3
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Solution

The electric flux through a closed surface surrounding a charge is given by Gauss's law, which states that the total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. In this case, the charge is enclosed in a cube.

The total flux Φ through the cube is given by Φ = q/ε, where ε is the permittivity.

However, the cube has 6 faces and the charge is symmetrically placed, so the flux will be evenly distributed through each face.

Therefore, the flux through one face of the cube will be Φ/6 = q/(6ε).

So, the correct answer is q/6ε.

This problem has been solved

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