Explanation:
The charge q is placed at the center of the cube of side L
According to Gauss's law the flux through any closed surface is q/ε₀
here q is the charge enclosed .
In this case cube has the six faces . The flux through each face = q/6ε₀
In the second case The cube has the face with length L₁
The flux through each face = q/6ε₀
Thus flux through the cube does not depend upon the size of the cube .
a) The electric flux through the six faces of the cube is [tex]\phi_{E} = \frac{q}{\epsilon_{o}}[/tex].
b) The electric flux through a face of the cube is [tex]\phi_{E'} = \frac{q}{6\cdot \epsilon_{o}}[/tex].
By the Gauss's Law, the net electric flux of a given close surface is equal to the electric charge within divided by electric permitivity. The definition of the Gauss's Law is described below:
[tex]\phi_{E} = \oint_{S}\,\vec E\,\bullet\,d\vec A = \frac{q}{e_{o}}[/tex] (1)
Where:
[tex]\phi_{E}[/tex] - Electric flux, in newton-square meters per coulomb.
[tex]\vec E[/tex] - Electric field, in newtons per coulomb.
[tex]d \vec A[/tex] - Infinitesimal vector area, in square meters.
[tex]q[/tex] - Electric charge enclosed in the cube, in coulombs.
[tex]\epsilon_{o}[/tex] - Electric permitivity, in farads per meter.
As we know that point charge is centered and cube has six faces with same surface area and same configuration, we can expand (1) by principle of symmetry:
[tex]\phi_{E} = 6\cdot \oint_{S} \vec E \,\bullet \,d\vec A = \frac{q}{\epsilon_{o}}[/tex] (2)
a) The electric flux through the six faces of the cube is [tex]\phi_{E} = \frac{q}{\epsilon_{o}}[/tex].
b) The electric flux through a face of the cube is [tex]\phi_{E'} = \frac{q}{6\cdot \epsilon_{o}}[/tex].
Related question: https://brainly.com/question/2664005
