A 1.0 cm x 1.0 cm x 1.0 cm box with its edges aligned with the xyz axes is in the electric field E=<350x+150>i N/C, where x is in meters. What is the net electric flux through the box.
I’m lost. :/
The question is badly worded. You can’t have ‘net … flux through the box’. You can only have net flux into (or out of) the box.
Area of each face A = 0.01m x 0.01m = 1×10⁻⁴m²
For convenience, take the +x (i) direction to be to the right.
If the left face is at x = X, then the right face is at x = X+0.01 (metres).
The electric field has only an x-component, so flux passes through only the left and right faces.
Flux passing through left face is Φₗ = EA = (350X+150)*0.0001
This is in the +x direction so it enters the box.
Flux passing through right face is Φᵣ = EA = (350(X+0.01)+150)*10⁻⁴
This is also in the +x direction so it leaves the box.
The flux leaving is greater than the flux entering, so:
Net flux, Φnet, *leaving* box is
Φnet = Φᵣ – Φₗ
. . . . .= (350(X+0.01)+150)*10⁻⁴ – (350X+150)*10⁻⁴
When you evaluate this, most terms cancel leaving:
Φnet = 350*0.01*10⁻⁴
. . . . .= 3.5×10⁻⁴ Vm (the unit of electric flux is volt metre).
(If you use the convention that flux entering the box is positive, then the answer would be -3.5×10⁻⁴ Vm.)
Two ways of doing it here:
The field is in the x-direction only, so the flux is nonzero through the squares of the cube at x=0 and x=1.
At x=0 the field is a constant 150 N/C, and over the area of 1 cm^2 the flux therefore is (negative, into the oriented surface) -150 N cm^2/C
At x=1 the field is a constant 500N/c, and so the flux through that surface is 500 N cm^2/C .
The total flux therefore is 350 N cm^2/C, or 3.50*10^-4 N m^2 / C.
Second method: the flux integral equals the volume integral of the divergence. The divergence is ∂Ex / ∂x = 350 N C / m, and integrating this constant over the volume (10^-6 m^3) gives 3.50*10^-4 N m^2/C