Finding flux through surface where E field is function of x

Click For Summary
The discussion revolves around calculating the electric flux through a square in the xy-plane due to an electric field defined as E = (663 N/(C*m))x. The integral for flux is set up as ΦE = ∫Ecos∅dA, simplifying to ΦE = ∫EdA since the electric field and area vector are parallel. The confusion arises regarding the presence of a y factor in the final answer, which is clarified by recognizing that the area element dA is expressed as dx*dy, leading to two integrals: one with respect to y and another for the x-dependent term. Ultimately, the integration confirms that the y factor is necessary due to the area calculation. The discussion concludes with a resolution of the misunderstanding regarding the direction of the electric field.
Jen2114
Messages
23
Reaction score
0

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]
 
Physics news on Phys.org
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...?
 
  • Like
Likes 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
 
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?
 
  • Like
Likes Jen2114
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
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 14 ·
Replies
14
Views
4K
Replies
9
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
5
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K