Question

A charge q is enclosed in a cube. What is the electric flux associated with one of the faces of a cube?

Answer 1

Hint : In order to answer this question, to know about the electric flux associated with one of the faces of the cube we will first know about the electric flux. Also we will further discuss the Gauss-theorem and apply the formula of it to find out the desired answer.

Complete step-by-step solution:
Electric flux is the measurement of the electric field through a particular surface in electromagnetism, despite the fact that an electric field cannot flow. It's a means of defining the strength of an electric field at any distance from the charge that's creating it.
Now, coming to the question:
The total charge enclosed in a closed surface is proportional to the total flux enclosed by the surface, according to the Gauss theorem. As a result, the total electric charge $q$ enclosed by the surface is; if is total flux and ${\varepsilon _0}$ is electric constant.
Since charge $q$ is placed inside the cube, according to Gauss-theorem the total electric flux through the cube is:
$\phi = \dfrac{q}{{{\varepsilon _0}}}$
Now, a cube has six face with equal area Hence, flux through each face will be same and total flux through all faces is:
$= \left( {\dfrac{1}{6} \times \phi } \right) \\\ = \left( {\dfrac{1}{6} \times \dfrac{q}{{{\varepsilon _0}}}} \right) \\\ = \dfrac{q}{{6{\varepsilon _0}}} \\\$
Therefore, the electric flux associated with one of the faces of the cube is $\dfrac{q}{{6{\varepsilon _0}}}$ .

Note: The electric field is the most fundamental idea in understanding electricity. In general, the electric field of a surface is computed using Coulomb's equation, however understanding the notion of Gauss law is required to calculate the electric field distribution in a closed surface. It explains how an electric charge is enclosed in a closed surface or how an electric charge is present in a closed surface that is contained.