A cube side length a=2 has one vertice at the origin of the referencial.
an electric field is present described as follows:
E(x,y,z)= (2xz) i + (x+2) j + (y(z^2-3) k
Find the flux through the cube
My cube's front face is at (0 to 2, 0, 0 to 2)
The Attempt at a Solution
My first thought here is, for the face of the cube situated in x=2, the flux in the y and z directions is 0, so i only consider the "i" part of the field (same for the opposing face) so
E= 4i bring e out of the integral and calculate (same for opposing face but since x=0 its 0). Correct?
What about for the face at y=0?there can only be field in the y direction, but the field there deppends only of the x position, how do i integrate?
i don't know how to do ∫(x-2)dA i imagine i have to integrate from 0 to 2 but how do i do it? do i say dA is (x*z) dx? and do ∫(x-2)(x*2) dx ?