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Electric Flux Gauss's Law

  • Thread starter Alex G
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  • #1
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Homework Statement



A particle with charge Q is located immediately above the center of the flat face of a hemisphere of radius R as shown in the figure below.

https://www.webassign.net/serpop/p19-33.gif

(a) What is the electric flux through the curved surface? (Use any variable or symbol stated above along with the following as necessary: ε0.)

(b) What is the electric flux through the flat face? (Use any variable or symbol stated above along with the following as necessary: ε0.)


Homework Equations


Gauss's Law states
The Integral of (E)dot(dA) = qenclosed/Epsilon0



The Attempt at a Solution


I'm not sure what the picture is stating with the symbol that I suppose goes to 0. But from my understanding the point is outside the surface or so I thought and the flux would be 0 but that is wrong. So I thought that the flux given out is Q/Epsilon0 for the curved surface, but apparently it's not enclosed either. I'm not sure what I'm missing. It's probably something simple.
 

Answers and Replies

  • #2
gneill
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If the shape had been a sphere rather than a hemisphere, with the charge located at its center, what would the flux through the entire surface have been?
 
  • #3
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If it was just a sphere would it be 4(pi)*kq?
 
  • #4
gneill
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Consider a spherical Gaussian surface surrounding a charge Q. What does Gauss' law say?
 
  • #5
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Ah that is qenclosed/Epsilon0 correct? So for a hemisphere would it be half that? Or am I getting ahead of myself?
 
  • #6
gneill
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I think you've just caught up with yourself!
 
  • #7
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Thanks for helping me sort that out so I have:
qenclosed/Epsilon0*(1/2)

However what can I say about the flat surface, I'm assuming that formula applies to the curved surface.
 
  • #8
gneill
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Thanks for helping me sort that out so I have:
qenclosed/Epsilon0*(1/2)

However what can I say about the flat surface, I'm assuming that formula applies to the curved surface.
It is stated that the charge is situated "immediately above" the flat surface, with the implication that the distance above the surface is negligibly small. Suppose you were to draw flux lines from the point charge through the flat surface and beyond through the hemisphere. Every possible such line that you would draw would pass through both, right?
 
  • #9
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Ahhh yes it would, I see the significance of that now. So the lines to pass through are of equivalence since flux can be considered as the number of electric lines passing through an area? So both of them use the same equation then?
It really through me off because I assumed the point to be outside the surface so it wouldn't contribute a flux.
 
  • #10
gneill
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Ahhh yes it would, I see the significance of that now. So the lines to pass through are of equivalence since flux can be considered as the number of electric lines passing through an area? So both of them use the same equation then?
It really through me off because I assumed the point to be outside the surface so it wouldn't contribute a flux.
The method works because of the symmetry of the particular problem. If the charge had not been at the center of the spherical Gaussian surface that also centered on the hemisphere's radius, then the flux through the actual hemisphere would not have been half of the total. The charge "happens" to be in a location where the symmetry does us a big favor!
 
  • #11
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Well the qenclosed/2Epsilon0 was correct for the curved surface, however the flat surface it does not apply for. I'm in full understanding of the curved surface and I know the point is in the center and symmetric.

But why is the flat surface giving me a hard time here. Perhaps plane symmetry? No ... this is a circular surface, augh.
 
  • #12
gneill
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Hmm. The problem does state that the charge is above the flat face, and not somehow embedded in it. As long as the distance above the surface is negligible as they indicate, I don't see how the flux through that face could be anything but the same as that through the hemisphere.
 
  • #13
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Wow, I guessed using the idea that the total flux should be 0 because of symmetry (not sure if that's an okay assumption) and also figured that since the normal of the flat surface is 180 degrees of the electric lines coming in that the answer would be the negative of the curved surface because the curved surface's normals are parallel to the electric lines.

So the answer to the flat surface was
-Q/2Epsilon0

:) Thank you for the help to getting all the way there!

Edit: Thanks Sammy forgot to put the 2, however, the answer was negative rather than positive for some reason (tried both)
 
Last edited:
  • #14
SammyS
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What happened to the 2 in the denominator?

The flat surface by itself doesn't need to point out of the hemi-sphere. No need for the negative sign.

Original question: "What is the electric flux through the flat face?"
 
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