Finding flux through surface where E field is function of x

In summary, the electric field E(vector E) has magnitude E=(663 N/(C*m))x and flux through a square in xy plane at z=0 is found to be (y*663x^2)/2 (from 0 to 0.330).
  • #1
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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  • #2
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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  • #3
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
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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  • #5
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
 

1. What is flux through a surface?

Flux through a surface is a measure of the flow or movement of a vector field through that surface. It is represented by the symbol Φ and is typically measured in units of Newtons per meter squared (N/m²).

2. How is flux calculated?

Flux can be calculated by taking the dot product of the vector field and the unit normal vector of the surface, and then integrating this value over the entire surface. This is represented by the formula Φ = ∫∫S E · dA, where E is the electric field and dA is an infinitesimal area element on the surface.

3. What is the significance of the electric field being a function of x?

When the electric field is a function of x, it means that the strength and direction of the field vary at different points on the surface. This can affect the calculation of flux, as the dot product will also vary at different points.

4. How does the orientation of the surface affect the flux calculation?

The orientation of the surface is important in calculating flux, as the dot product between the electric field and the unit normal vector will be negative if the surface is not parallel to the field. Therefore, the orientation of the surface must be carefully considered when calculating flux.

5. What are some real-world applications of finding flux through a surface?

Finding flux through a surface has many practical applications, such as calculating the flow of fluids through pipes, determining the amount of light passing through a window, or analyzing the movement of particles through a membrane. It is also an important concept in understanding electromagnetic fields and their effects on objects and systems.

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