# Homework Help: Finding flux through surface where E field is function of x

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1. Sep 17, 2015

### Jen2114

1. The problem statement, all variables and given/known data
In a region of space there is an electric field E(vector E) that is in the x-direction and that has magnitude E=(663 N/(C*m))x

Find flux for this field through a square in xy plane at z=0 and with side length 0.330m . One side of the square is along the +x-axis and another side is along the +y-axis.

I understand setting up the integral but I do not understand why there is a y in the answer

2. Relevant equations

3. The attempt at a solution

ΦE= ∫Ecos∅dA , where phi is the angle between the normal area vector and the E field. In this case E-field and A are parallel so cos∅=1 . Thus integral reduces to ΦE=∫EdA. dA of a small piece of the square is determined by dy*dx.

Then, integral becomes ∫663x*dydx (from 0 to 0.330)

Final answer should be

(y*663x^2)/2 (from 0 to 0.330) so final answer is 11.91.

I understand anti derivative of x is x^2/2 but I am not understanding/seeing where the y factor comes from.

Thanks

2. Sep 17, 2015

### axmls

You have to integrate with respect to $y$ first. Since the term $663x$ does not depend on $y$, you can pull it out, and you have to deal with $\int \ dy$, which is...?

3. Sep 17, 2015

### Jen2114

Hi,

ok so bc A=dxdy and that's part of the integral and like you said 663x does not depend on y you have 2 integrals ∫dy=y and ∫663x= 663*(x^2)/2 (from 0 to 0.330). Ok got it. Thank you for the clarification

4. Sep 17, 2015

### rude man

You sure the problem doesn't say the E field is in the z direction?

5. Sep 17, 2015

### Jen2114

Hello, yes I am sorry the problem statement is in z-direction and I solved it with the z-direction in mind but I failed to revise my problem statement submission here. thanks for pointing that out