A charge is placed at the centre of cube of side a then flux linked with one of its given faces will be

Options

A. `(Q)/(epsi_(0))`
B. `(Q)/(6epsi_(0))`
C. `(Q)/(epsi_(0)a^(2))`
D. `(Q)/(4piepsi_(0)a^(2))`

Answer

B

Explanation

Gauss's law tells that the total flux through an area enclosing a charge Q is `(Q)/(epsi_(0))` . Now as the cube is having is faces and as we can assume a symmetrical distribution of fluxes among its faces, the flux associated with one of its faces is `(Q)/(6epsi_(0))`