First slide

Gauss's Law

Question

A charge q is located at the centre of a cube. The electric flux through any face is

Easy

A

$\frac{4 πq}{6 (4 {πε}_{0})}$
B

$\frac{π q}{6 (4 π ε_{0})}$
C

$\frac{q}{6 (4 {πε}_{0})}$
D

$\frac{2 πq}{6 (4 {πε}_{0})}$

Solution

This problem can be solved by symmetry consideration and Gauss law.
The total flux passing through the closed cube $ϕ = \frac{q}{ε_{0}}$
All the six surfaces are symmetrical with respect to charge, hence they will have equal contribution of the flux. So, flux through any one face is
$ϕ^{'} = \frac{ϕ}{6} = \frac{q}{6 ε_{0}}$
$ϕ_{face} = \frac{q}{6 ε_{0}} = \frac{4 πq}{6 (4 {πε}_{0})}$

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