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Asked in GUJCET 2020 · Flux through a cube
Idea: a charge at a corner is shared by the eight cubes that meet there. Imagine them stacked around the charge: together they enclose it completely.
Total flux from the charge is q/(ε₀), and by symmetry each of the eight cubes gets 1/8×q/(ε₀)=q/(8ε₀).
The three faces that meet at the charge contain it in their own planes; the field runs along them, so they carry no flux.
The other three faces share the cube's flux equally:
φ_face=1/3×q/(8ε₀)=q/(24ε₀)
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