What is the flux through a cube of side `'a'` if a point charge of `q` is at one of its corner :

Options

A. `(q)/(2in_(0))6a^(2)`
B. `(2q)/(in_(0))`
C. `(q)/(8in_(0))`
D. `(q)/(in_(0))`

Answer

c

Explanation

We can imagine a charged particle is placed at the centre of a cube of side `2l`
The flux enclosed with the cube `phi=(q)/(epsilon_(0))`.
The flux passing through one of the face of the cube,
`bcgf = (phi')/(4)=(q)/(24 epsilon_(0))`.
Each of the face `(efgh), (bcgf)` and `(dcgh)` are symmetrical with respect to charge hence the flux passing through each of the face is `q//24 epsilon_(0)`


The electric filed lines of the faces `(efgh), (bcfg)` and `(dchg)` are away from the faces hence the flux associated with each of the faces will be positive
`(i.e., +(q)/(24 epsilon_(0)))`
Hence, total flux through this side `=(3xxq)/(24 epsilon_(0))= (q)/(8epsilon_(0))`