Question

A point charge q is located at the center of a cube whose sides are of length a. If there are no other charges in this system, what is the electric flux through one face of the cube?

Answer 1

`\phi_E=(q)/(6\epsilon_o)`

Step-by-step 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 (
`\phi_E`) through the closed surface containing the charge q is:

`\phi_E=\oint \vec{E}.\vec{ds}=(q_o)/(\epsilon_o)`

where,
`\epsilon_o}` 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
`\phi_E=(q)/(6\epsilon_o)`