This article aims to find the electric flux in a cube having six sides. This article uses the concept of electric flux. For a closed gaussian surface electric flux is given by the formula
\[\Phi_{e} = \dfrac{Q}{xi_{o}}\]
Expert Answer
Consider a cube having side length $ L $ in which a size $ q $ charge is placed at the center. Consider a closed Gaussian surface, which is a cube whose electric flux is $\Phi $, which is given by:
\[\Phi=\dfrac{ q } {\xi_{o}}\]
The number of lines of force arising from the charge will be divided into six walls. So the electric flux is given by:
\[\Phi =\dfrac{q}{6\xi_{o}}\]
Part (A)
The electric flux of each of the six faces of the cube is $\Phi = \dfrac{ q } { 6 \xi _{ o } } $.
Electric flux is number of field lines passing per unit area. The flux through any face of the cube is equal to the total flux of the cube divided by six.
Consider the sides of the cube $ L_{1}$.
Since the electric flux depends only on the enclosed charge $ q $, the flux through each surface would be the same as the previous part, even if the cube’s dimension changes. That is, the electric flux of each of the six walls of the cube, the length $ L_{ 1 } $ of which
\[\Phi _{1}=\dfrac{q}{6\xi_{o}}\]
Part (B)
The electric flux of each of the six faces of the cube is $\Phi _{ 1 }=\dfrac{q}{6\xi _{o}}$.
Since the flux depends on the charge inside the closed surface, the flux through each surface would be same as in the previous section, even if the dimension changes.
Numerical Result
(a) Electric flux $\Phi $ across each of the six faces of the cube is equal to $ \dfrac{ q } { 6 \xi _{ o } }$.
(b) Flux $ \Phi _{1} $ over the face of the cube if its sides were $ L_{1} $ long is equal to $\dfrac{ q } { 6 \xi _{ o } }$.
Example
A point charge of magnitude $Q$ is at the cube’s center with sides of length $x$. What is the electric flux $\Phi $ across each of the cube’s six faces? What would be the flux $ \Phi $ over the face of the cube if its sides were long $ x_{1}$?
Solution
Consider a closed Gaussian surface, which is a cube whose electric flux is $\Phi $ which is given by
\[\Phi =\dfrac{Q}{\xi _{o}}\]
The number of lines of force arising from the charge will be divided into six walls. So the electric flux is given by
\[\Phi =\dfrac{Q}{6\xi _{o}}\]
Part (A)
The electric flux of each of the six faces of the cube is $\Phi = \dfrac{Q}{6\xi _{ o }}$.
Consider the sides of the cube $ x_{1}$. That is, the electric flux of each of the six walls of the cube, the length $L_{1}$ of which
\[\Phi _{1}=\dfrac{Q}{6\xi _{o}}\]
Part (B)
The electric flux of each of the six faces of the cube is $\Phi _{1}=\dfrac{Q}{ 6 \xi _{o}}$.