Electric flux theory.

  • Thread starter maiad
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  • #1
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https://www.smartphysics.com/Content/Media/Images/EM/03/h3_lineD.png [Broken]

In the image above, i was asked to find the electric field at point P. since the y-components cancel due to symmetry, i used he equation [itex]\Phi[/itex]=[itex]\int[/itex]E dA=Qenclosed/[itex]\epsilon[/itex] .

I found q1 and q2 by multiplying (charge density x h). then from that, i added the charges up to get Q(enclosed). I found my E by [itex]\Phi[/itex]/(2πah).
This method was wrong apparently but i don't know why. can someone explain?
Is it because the electric field through the surface is not uniform?

I later used superposition instead and i got the right answer.
 
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  • #2
Doc Al
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Is it because the electric field through the surface is not uniform?
Exactly.
 
  • #3
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There were also examples of point charges spread abritraily in a guassin space. Would that not be uniformed also? I would think that the electric flux should not change if the points charges shifted withing the guassin space, but the electric field should also change when the point charges move. Is my assumption correct?
 
  • #4
Doc Al
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There were also examples of point charges spread abritraily in a guassin space. Would that not be uniformed also?
No reason to think that an arbitrary distribution of charges within a Gaussian surface would produce a uniform field at the surface. Only in cases of sufficient symmetry would the the field be uniform.
I would think that the electric flux should not change if the points charges shifted withing the guassin space, but the electric field should also change when the point charges move. Is my assumption correct?
Yes, you are correct.
 

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