Finding flux through surface where E field is function of x

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Homework Help Overview

The discussion revolves around calculating the electric flux through a square surface in the xy-plane, given an electric field that varies with the x-coordinate. The electric field is defined as E = (663 N/(C*m))x, and the square has a side length of 0.330 m.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of the integral for calculating flux, questioning the presence of a y variable in the final answer. There is an exploration of integrating with respect to y and the implications of the electric field's dependence on x.

Discussion Status

Some participants have provided clarifications regarding the integration process and the treatment of the electric field in relation to the area element. There is acknowledgment of a potential misunderstanding regarding the direction of the electric field, which has led to further examination of the problem statement.

Contextual Notes

One participant noted a discrepancy in the problem statement regarding the direction of the electric field, suggesting it may be in the z-direction instead of the x-direction, which has implications for the flux calculation.

Jen2114
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Homework Statement


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

Homework Equations

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 [/B]
 
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You have to integrate with respect to [itex]y[/itex] first. Since the term [itex]663x[/itex] does not depend on [itex]y[/itex], you can pull it out, and you have to deal with [itex]\int \ dy[/itex], which is...?
 
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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
 
Jen2114 said:

Homework Statement


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
You sure the problem doesn't say the E field is in the z direction?
 
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rude man said:
You sure the problem doesn't say the E field is in the z direction?
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
 

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