Electric Flux of a Cube – Rankers Physics
Topic: Electrostatics
Subtopic: Gauss's Law

Electric Flux of a Cube

A charge \(Q \mu\text{C}\) is placed at the centre of a cube. The flux coming out from any one of its faces will be (in SI unit)
\(\frac{Q}{6\epsilon_0} \times 10^{-3}\)
\(\frac{Q}{6\epsilon_0} \times 10^{-6}\)
\(\frac{Q}{\epsilon_0} \times 10^{-6}\)
\(\frac{2Q}{3\epsilon_0} \times 10^{-3}\)

Solution:

According to Gauss's Law, total flux through the cube is \(\phi = \frac{q}{\epsilon_0}\). Since a cube has 6 identical faces, the flux through one face is \(\phi' = \frac{\phi}{6} = \frac{Q \times 10^{-6}}{6\epsilon_0}\).

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