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Flux of electric field through the square

  • Thread starter leventa2
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  • #1
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Homework Statement



A small particle, with charge q = 8.85 μC, is located at (x,y,z) = (0,0,h). Calculate the flux of electric field through the square in the xy-plane, of size L × L, centered at the origin.

(B) What is the flux if h = 0.05 L?


Homework Equations



Flux=Charge/permittivity (in a closed surface)

The Attempt at a Solution



If h = 0.5 L, then I would just divide the flux by six. But I don't know where to go from here...
 

Answers and Replies

  • #2
gneill
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Contemplate the relevant equation you provided!
 
  • #3
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but how should I contemplate it? The surface is an open surface, and the flux is asymmetrical.
 
  • #4
gneill
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but how should I contemplate it? The surface is an open surface, and the flux is asymmetrical.
Ah, sorry. I jumped to the (incorrect) assumption that it was a square box, not just a square surface.

I suppose you'll have to integrate E.dS over the surface to determine the total flux.
 
  • #5
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Actually, I just solved it no integral necessary :D. Basically, I imagined the surface as part of a cube, and the cube surrounded by a sphere. Connecting the four corners of the surface to the electron in the center creates a square pyramid with base length L X L and height 0.05 L. I used the formula 4*arcsin((a^2)/(a^2+4*h^2)) to find the solid angle, and divided it by 4pi, the total solid angle. I multplied the resulting value by the total flux of the cube. It seems that solid angles play an important part in Gauss's law.

This was all tiring, but totally worth it when I got the correct answer.
 
  • #6
gneill
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Nicely done.
 

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