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What is the flux through a cube of side 'a' if a point charge q is at one of its corners?

A

2q/ε₀

B

q/8ε₀

C

q/ε₀

D

q/2ε₀

Step-by-Step Solution

According to Gauss's Law, the total flux through a closed surface enclosing a charge qq is ϕ=q/ϵ0\phi = q/\epsilon_0. A charge placed at the corner of a cube is shared equally by 8 identical cubes arranged around that corner (4 in the lower layer and 4 in the upper layer). Therefore, the portion of the charge effectively enclosed within one specific cube is qenclosed=q/8q_{enclosed} = q/8. Consequently, the electric flux through this cube is ϕ=qenclosedϵ0=q8ϵ0\phi = \frac{q_{enclosed}}{\epsilon_0} = \frac{q}{8\epsilon_0}.

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