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Derive an expression for the flux through a sphere

  • Thread starter phosgene
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  • #1
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Homework Statement



Three charges with values + 20 μC, -10 μC and -10 μC are placed at the corners of a 0.10m square, as shown.

c) Consider a sphere of radius 0.3m which totally encloses all three charges. Using Gauss's law derive an expression for the flux through this surface.

Homework Equations



Electric flux = ƩQ/ε0

The Attempt at a Solution



I don't get it, isn't Gauss's law by itself an expression for finding the electric flux (which I think should be 0)? I think it may have something to do with the non-uniform distribution of charge making Gauss's law by itself inadequate for the calculation, but I really have absolutely no idea what I'm supposed to do.
 

Answers and Replies

  • #2
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The beauty of Gauss's law is that it doesn't care about where the charge is, only that you have the right total charge. So yes, you got it right, the flux is 0. If your total charge was not 0, however, the problem would be a bit trickier (trickier in that you'd have to think about what Gaussian surface to use).
 
  • #3
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The beauty of Gauss's law is that it doesn't care about where the charge is, only that you have the right total charge. So yes, you got it right, the flux is 0. If your total charge was not 0, however, the problem would be a bit trickier (trickier in that you'd have to think about what Gaussian surface to use).
Thanks for the reply, but are you sure there's nothing else I need to do? The wording of the question made me think that I had to do more than just calculate a value for the flux.
 
  • #4
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[tex]\Phi = \oint \mathbf{E} \cdot d \mathbf{A} = Q_{enc}/\epsilon_0[/tex]
Though I'm sure it can be done by making holes or something, we don't really want to think about that dot product, and we can just completely ignore evaluating the left hand side of this equation because the right hand side is most definitely 0. I dunno, I guess the question was ill-formed or it was trying to mess with your mind.
 

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