Question

What is the electric flux through a cube of side $1\mathrm{cm}$ which encloses an electric dipole?

Answer 1

Step 1: Definition

The number of field lines which thus travel through a given surface area defines the electric flux through such a region.

Electric flux is a measurement of the passage of an electric field through a substrate.

Step 2: Given parameters

The side of the cube = $1\mathrm{cm}$

Step 3: Finding the electric flux through a cube of the side $1\mathrm{cm}$ which encloses an electric dipole

According to Gauss’ law of electrostatics, electric flux through any closed surface is given by

$\mathrm{\varphi E}=\oint E.dS=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}...\left(1\right)$

Here, E is the electrostatic field, q is the total charge, ${\mathrm{\epsilon }}_0$ is the permittivity of free space.

As a result, in the aforementioned example, the cube contains an electric dipole. The net charge contained within the cube equivalent to zero is $\mathrm{q}=0$.

From equation (1),

$\mathrm{\varphi E}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}=0$

Therefore, the electric flux is zero.