Electric Charges And Fields

Question

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

$\frac{4 πq}{6 (4 {πε}_{o})}$

$\frac{πq}{6 (4 {πε}_{o})}$

$\frac{q}{6 (4 {πε}_{o})}$

$\frac{2 πq}{6 (4 {πε}_{o})}$

Solution

$\frac{4 πq}{6 (4 {πε}_{o})}$

According to Gauss's theorem, electric flux through the cube (closed surface)

$ϕ = \frac{q}{ε_{o}}$

q = total charged enclosed by surface S

The law implies that the total electric flux through a closed surface is zero if no change is enclosed by the surface.

Since, cube has six surfaces and all the faces are symmetrical, therefore electric flux through any face

$ϕ' = \frac{q}{6 ε_{o}}$

$ϕ' = \frac{4 πq}{6 (4 {πε}_{o})}$