What is electric flux assoacited with one of faces of the cube, when a charge `(q)` is enclosed in a cube ?

Options

A. `(6q)/(epsilon_(0))`
B. `(3q)/(epsilon_(0))`
C. `(q)/(3epsilon_(0))`
D. `(q)/(6epsilon_(0))`

Answer

D

Explanation

According to Gauss's law electric flux associated with any closed surface is

`:. Phi = int E. dA = (q)/(epsilon_(0))`
Since cube consists of six faces of equal area, hence, flux linked with each face is
`(1)/(6) xx (q)/(epsilon_(0)) = (q)/(6epsilon_(0))`