The electric field components in Fig. are `E_(x)=ax^(1//2), E_(y)=E_(z)=0`, in which `alpha=800N//C m^(1//2)`. Calculate the flux through the cube

Explanation

Since the electric field has only an x component, for faces perpendicular to x direction, the angle between E and `Delta S` is `pm pi//2`. Therefore, the flux `phi=E. Delta S` 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)=x^(1//2)=alpha a^(1//2)`
(x = a at the left face).
The magnitude of electric field at the right face is `E_(R)=alpha x^(1//2)=alpha (2a)^(1//2)`
(x = 2a at the right face).
The corresponding fluxes are
`phi_(L)=E_(L).Delta S=Delta SE_(L). hat(n)_(L)=E_(L) Delta S cos theta=-E_(L) Delta S`, since `theta=180^(@)`
`=-E_(L) a^(2)`
`phi_(R)=E_(R).Delta S=E_(R) Delta S cos theta =E_(R)Delta S`, since `theta=0^(@)`
`=-E_(R)a^(2)`
Net flux through the cube.
`=phi_(R)+phi_(L)=E_(L)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) (sqrt(2)-1)`
`=800(0.1)^(5//2) (sqrt(2)-1)` ,brgt `=1.05N m^(2) C^(-1)`