Answer:
[tex]\phi_E=\frac{q}{6\epsilon_o}[/tex]
Explanation:
Given: A charge q is enclosed within the cube
The length of the one side of cube = a
Now, according to the Gauss Law the net flux ([tex]\phi_E [/tex]) through the closed surface containing the charge q is:
[tex]\phi_E=\oint \vec{E}.\vec{ds}=\frac{q_o}{\epsilon_o}[/tex]
where, [tex]\epsilon_o}[/tex] is the permittivity
This electric flux is for the 6 faces of the cube
thus, for the one face of the cube the electric flux will be [tex]\phi_E=\frac{q}{6\epsilon_o}[/tex]
