The electric field components in Figure are `E_(x)=alpha x^(1/2), E_y=E_(z)=0` in which `alpha=800 N//Cm^(1/2)`. Calculate (a) the flux through the cube, and (b) the charge within the cube. Assume that a=0.1m.

Explanation

a. Since the electric field has only an x component, for faces perpendicular to x direction, the angle between E and `triangleS" is "+pi/2`. Therefore, the flux `phi=E triangleS` is separately zero for each face of the cube except the two shaded ones. Now the magnitude of the electric field at the left face is `E_L= ax^(1/2)=alpha a^(1/2)` (x= a at the left face).
The magnitude of electric field at the right face is `E_a=alpha x^((-1)/2) = alpha (2a)^(1/2)`
(x=2 a at the right face).
The corresponding fluxes are `phi_(L)=E_(L). triangleS=triangleS E_(L). hatn_(L)=E_(L) triangleS cos theta=-E_(L) triangleS," since "theta=180^(@), phi_(R)=E_(R). triangleS=E_(R) triangleS cos theta=E_(R) triangleS,"" sin ce"theta=0^(@)`
Net flux through the cube `=phi_(R)+phi_(L)=E_(R)a^(2)-E_(L)a^(2)=a^(2) (E_(R)-E_(L)) =alpha a^(2) [(2a)^(1/2)-a^(1/2)]`
`=alpha a^(5/2) (sqrt2-1) =800 (0.1)^(5/2) (sqrt2-1)=1.05 Nm^(2) C^(-1)`
b. We can use Gauss. law to find the total charge q inside the cube.
We have `phi=(q)/(epsi_(0)) or q=phi epsi_(0). " Therefore ",q=1.05 xx 8.854 xx 10^(-12) C=9.27 xx 10^(-12)C`.