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Electric field flux question

  • Thread starter ayestaran1
  • Start date
  • #1
For a uniform electric field the vector E = 5.0 kN/C in the i direction. What is the flux of this field thru a square of side 20 cm if the normal to its plane makes a 45 degree angle with the x axis?

How do I set this up? I thought according to Gauss's law E = q/epsilon naught so r didn't matter?
 

Answers and Replies

  • #2
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gauss' law only holds for closed Gaussian surfaces.

In this question you need only think about the dot product. Remember [tex] \phi = \int E.dA [/tex]
 
  • #3
Defennder
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Yes this isn't a closed surface. So you have to use the fundamental mathematical definition of electric flux to do this. No integration is required since the field is uniform and the plane is appropriately oriented.
 
  • #4
Thanks, so I did E*A_1 = E*A_2cos(theta) since it has to be normal to the surface, and as a result I got (5.0 k N/C) * (.2 m)^2 * cos(45) = 141 N*m^2 /C. Does that look right?

I know this is pretty basic - but how did you know it wasn't a closed surface? I thought squares can be Gaussian surfaces... why isn't it here?
 
  • #5
Defennder
Homework Helper
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5
Gaussian surfaces are closed surfaces, meaning to say they can enclose some volume. Your square isn't a closed surface.
 

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