(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) Since the electric field has only an x component. For faces perpendicular to x direction, the angle between E and `DeltaS` is `pm pi//2`. Therefore the flux `phi=E, DeltaS` is separately zero for each face of the cube except the two shaded ones. Now th magnitude of the electric the two shaded ones. Now the magnitude of the electric at the left face is
`E_(L)=ax^(3//2)=aa^(1//2)`
(x=`alpha` at the left face)
The magnitude of electric field at the night face is
`E_(R)=ax^(1//2)=alpha(2a)^(1//2)`
(x=2a at the right face).
The coresponding fluxes are
`phi_(L)=E_(L),DeltaS =DeltaSE_(L).hatn_(L)=E_(L)DeltaS cos theta=-E_(L) DeltaS, "since" theta=180^(@)=-E_(L)a^(2)`
`phi_(R)=E_(R).DeltaS =E_(R)DeltaS costheta=E_(R) DeltaS, "since" theta=0^(@)=E_(R)a^(2)`
Net flux through the cube
`phi_(R)=phi_(L)=E_(R)a^(2)-E_(L)a^(2)=alphaa^(3)[(2a)^(1//2)-a^(1//2)]`
`=alphaa^(5//2)(sqrt2-1)`
`=800(0.1)^(5//2) (sqrt2-1)`
`=1.05Nm ^(2) C^(-1)`
(b) We can use Gauss's law to find th total charge q inside the cube We have `phi=q//epsilon_(0) or q= phiepsilon_(0)` therefore
`q=1.05xx8.854xx10.^(-12)C=9.27xx10^(-12)C.`.