The electric field components due to a charge inside the cube of side 0.1m are as shown below: `E_(x)= ax`
where `alpha= 500N//C-m, E_(Y)= 0, E_(Z)= 0`
Calculate (a) the flux through the cube and (b) the charge inside the cube.

Explanation

Step I Since, the electric field has only x-component, `therefore phi_(E)= E.Delta S= 0` for each of the four faces of cube perpendicular to Y -axis and Z-axis. So, electric flux is only non-zero for left and right faces perpendicular to X-axis.
Step II Electric field at the left face situated at x=a is given by, `E_(L)= alpha a`
`therefore phi_(L) = E_(L). Delta S = E_(L) Delta S cos 180^(@)`
`= alpha a.a^(2) xx -1 = -alpha a^(3)`
Electric field at the right face, `x= a + a= 2a` is
`E_(R )= alpha (2a)`
`rArr phi_(R )= E_(R ).Delta S`
`= alpha (2a) a^(2) cos 0^(@)= 2alpha a^(3)`
Step (III) (a) `therefore` Net flux through the cube
`=phi_(L) + phi_(R)`
`= -alpha a^(3) + 2alpha a^(3)`
`=alpha a^(3)= 500 xx (0.1)^(3)`
`=0.5 N-m^(2) C^(-1)`
(b) By Gauss.s law, `q= epsi_(0) phi`
`=8.85 xx 10^(-12) xx 0.5`
`=4.425 xx 10^(-12)C`