- #1

WrongMan

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## Homework Statement

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)

## Homework Equations

flux=∫E.dA

## 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 ?