Gauss law

  • Thread starter Suziii
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  • #1
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Homework Statement




In case of having a conductor (spherically shaped and uncharged) and an irregularly shaped hole inside it with a charge at an arbitrary point, I am asked to find the electric field right outside the conductor sphere.


Homework Equations



Gauss's law ∫EdA= Q(enclosed) / (a constant)

The Attempt at a Solution


I assumed that the electric field outside of the conductor is uniform so I had:
E*(4πr^2)= q/ the constant - we can get E from this, but it turns out to be a wrong answer, maybe just taking E to be constant is not correct...
 

Answers and Replies

  • #2
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Your answer is almost correct. However, you must remember that E is a vector.
Also, you should note that the electric field is not constant but falls off as [itex]\frac{1}{r^2}[/itex].
You can see this from your answer.
 
  • #3
ehild
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The magnitude of E is the same all over the outer surface of the sphere, and E*(4πr^2)= q/ε0 according to Gauss' Law. Haven't you forgotten ε0?
As the electric field is zero inside the metal, the electric field lines starting from the inner charge do not go through the metal wall. Therefore the charges accumulated on the outer surface "do not know" anything about the field inside, so they arrange themselves evenly, as far from each other as possible. The surface charge density is homogeneous and equal to σ=q/(4πr2). The electric field can be expressed by the surface charge density as E=σ/ε0.

ehild
 
Last edited:

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