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B
q6ϵo
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C
q12ϵo
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D
q24ϵo
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Solution
The correct option is Dq24ϵo
Let's make a surface such that the charge is at the centre of the surface.(as shown below) Let us consider that this surface is a face of the cube. So, the flux through the cube will be : qϵ0 So, flux through the face of the cube will be : q6ϵ0 Four squares fit on face of the cube. Hence, ϕsquare=14q6ϵ0 ϕsquare=q24ϵ0