Question

A charge q is placed at the centre of a cube of side l what is the electric flux passing through two opposite faces of the cube

Answer 1

Given :

Side length of the cube = l

The magnitude of the charge placed at the center of cube = q

To Find :

Magnitude of electric flux passing through two opposite faces of the cube = ?

Solution :

According to Gauss's law , we know that formula for electric flux is given as :

$\phi = \frac{q}{\epsilon_o}$  

Here , q is total charge enclosed within the closed surface

And $\epsilon_0$ is permitivity of free space

Now for a charge q inside a cube the flux through  each face  is :

=  $\frac{q}{6\epsilon_o}$

So flux passing through two opposite faces will be :

$= 2 \times \frac{q}{6 \epsilon_o}$

$= \frac{q}{3 \epsilon_o}$

So the electric flux passing through two opposite faces of the cube is $= \frac{q}{3 \epsilon_o}$