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Asked in GUJCET 2026 · Flux through a cube
Given: a point charge q at one corner of a cube; we want the flux through that cube alone.
Idea: Gauss's law only gives the flux through a surface that closes around the charge, so build one - stack 8 identical cubes around that corner and together they make one big cube with q exactly at its centre.
The big closed cube encloses q, so its total flux is q/(ε₀), and by symmetry the 8 small cubes take equal shares.
Flux through one cube = 1/8×q/(ε₀) = q/(8ε₀)
Check on the faces: the three faces meeting at the charge lie along its field lines and carry no flux, so the other three share it equally at q/(24ε₀) each, which adds back to q/(8ε₀).
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