A charge Q is placed at the centre of a cube. The flux through two opposite faces of the cube is ( ε₀ = permittivity of free space) (a) (Q)/(6ε₀) (b) (Q)/(3ε₀) (c) (Q)/(ε₀) (d) (Q)/(2ε₀)

Answer: (b). Total flux = (Q)/(ε₀); each of the 6 faces gets (Q)/(6ε₀) by symmetry. Two opposite faces: (2Q)/(6ε₀) = (Q)/(3ε₀) So the answer is (b).

A charge Q is placed at the centre of a cube. The flux through two… | MHT CET 2025 (22 Apr, Shift 2)

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Electric Flux & Gauss's Law

A charge $Q\ \mu\text{C}$ is placed at the centre of a cube. The flux through two opposite faces of the cube is ( $\varepsilon_0 =$ permittivity of free space)

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Answer: (b)

Total flux $= \dfrac{Q}{\varepsilon_0}$ ; each of the 6 faces gets $\dfrac{Q}{6\varepsilon_0}$ by symmetry.
Two opposite faces: $\dfrac{2Q}{6\varepsilon_0} = \dfrac{Q}{3\varepsilon_0}$
So the answer is (b).

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