Question

A charge $Q$ is present inside a cube, then find out flux due to cube and flux due to each face of a cube?

Answer 1

Step 1: Given data

Charge inside the cube = $Q$

The dielectric constant of the vacuum = ${\epsilon }_0$

Faces of a cube = $6$

Step 2: Concept applied

Gauss's law

Gauss's law for electricity states that the electric flux $\varphi$ across any closed surface is proportional to the net electric charge $q$enclosed by the surface.

i.e., $\varphi =\frac{q}{{\epsilon }_0}$

1. Flux ($\varphi$) is defined as the electric field flowing out through a specific area.
2. Consider a surface element $d\overrightarrow{S}=\hat{n}dS$ in an electric field $\overrightarrow{E}$ , where $\hat{n}$ is the outward unit vector normal to the surface element.
3. The quantity $d\varphi =\overrightarrow{E}.d\overrightarrow{S}=E\cos \theta dS$ is called the flux of $\overrightarrow{E}$ through $d\overrightarrow{S}$.
4. If we consider that $\frac{q}{{\epsilon }_0}$ number of lines of force emanate isotropically from a point charge $q$ then the idea of electric flux becomes meaningful.
5. If we use this picture of an electric field then $d\varphi =\overrightarrow{E}.d\overrightarrow{S}$ becomes equal to the number of lines of force passing through the area $d\overrightarrow{S}$.
6. The flux of $\overrightarrow{E}$ over any arbitrary surface $S$ is given by the integral $\Phi =\int d\varphi ={\int }_S\overrightarrow{E}.d\overrightarrow{S}$.

Step 3: Calculation and conclusion

1. Since Charge $Q$ is present inside the cube, the flux through the cube will be$\varphi =\frac{q}{{\epsilon }_0}$
2. The flux will be one-sixth of the total flux=$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.$(since a cube has 6 faces)
3. Flux through each face will be = $\frac{q}{6{\epsilon }_0}$

Hence, flux due to cube is $\varphi =\frac{q}{{\epsilon }_0}$ and flux due to each face of a cube is $\frac{q}{6{\epsilon }_0}$