(a) Define electric flux. Write its SI units.
(b) The electric field components due to a charge inside the cube of side 0.1 m are as shown :

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

Explanation

(a) Electric flux : Electric flux over an area in an electric field represents the total number of electric field lines crossing this area. It is denoted by `phi_(E)`.
Electric flux is a scalar quantity, its SI unit is `Nm^(2) c^(-1)`.
(b) Here, `E_(x)=alpha x, E_(y) = 0, E_(z)=0`.
`alpha = 500 N//C - m, ` side of cube `a=0.1 m`
Since the electric field has only x component,
`:. phi_(E) = vec(E). vec(Delta S) = 0` for each of four faces of cube `_|_` to y-axis and z-axis.
`:.` Electric field at the left face, x=a is
`:. E_(L) = alpha a `
`phi_(L)=vec(E_(L)).vec(Delta S)= alpha a a^(2) cos 180^(@) = - alpha a^(3)` and electric field at the right face, `[:' E=alpha x]`
`x=a+a=2a` is
`:. E_(R)=alpha(2a)`
`phi_(R)=vec(E_(R)). vec(Delta s)=alpha (2a) a^(2) cos 0^(@) = 2 alpha a^(3)`
`:.` Net flux through the cube `= phi_(L) + phi_(L)`
`=-alpha a^(3)+2 alpha a^(3)=alpha a^(3)=500xx(0.1)^(3)`
`=0.5 Nm^(2)C^(-1)`
(ii) By Gauss's law
`q=in_(0) phi`
`=8.85 xx 10^(-12) xx 0.5=4.425 xx 10^(-12)C`.