The electric field components due to a charge inside the cube of side 0.1 m are as shown

where `alpha = 500` N/C - m , `E_(y) = 0` and `E_(z) = 0`
Calculate (i) the flux through the cube and (ii) the charge inside the cube .

Explanation

Step 1 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 for left and right face perpendicular to X-axis
Step 2 Electric field at the left face situated at x =a is
`E_(L) = alpha a `
`therefore phi_(L) = E_(L) * Delta S = alpha a a^(2) cos 180^(@) = - alpha a^(3)`
Electric field at the right face x = a + a = 2a is `E_(R) = alpha (2a)`
`therefore phi_(R) = E_(R) * Delta S = alpha (2a) a^(2) cos 0^(@) = 2 alpha a^(3)`
Step 3
(i) `therefore` Net flux through the cube
`= phi_(L) + phi_(R) = - alpha a^(3) + 2 alpha a^(3)`
`= alpha a^(3) = 500 xx (0.1)^(3)`
`= 0.5 N - m^(2) C^(-1)`
(ii) By Gauss.s law , q = `epsi_(0) phi`
= `8.845 xx 10^(-12) xx 0.5`
`= 4.425 xx 10^(-12) C`