JD Jackson example

  • Thread starter Berko
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  • #1
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On page 60 of his 3rd edition of Classical Electrodynamics, he discusses the method of images applied to a grounded conducting sphere with a single charge q outside it.

Near the end of the problem, he calculates the force on a small patch of area da as (sigma^2/2epsilon_nought)da.

Now, it seems to me that the force should be the E field at that point times the charge of the patch....i.e.

dF = (sigma/epsilon_nought)*(sigma da).

Where does the factor of 2 come from?

Thank you.
 

Answers and Replies

  • #2
Bill_K
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The E field of a surface charge density is calculated by applying Gauss's Law. The usual situation is symmetrical -- half the E field comes out of each side of the surface. However in this case E=0 on the inner side, so all of the E has to come out the outer side, and is consequently twice the usual value.
 
  • #3
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The E field of a surface charge density is calculated by applying Gauss's Law. The usual situation is symmetrical -- half the E field comes out of each side of the surface. However in this case E=0 on the inner side, so all of the E has to come out the outer side, and is consequently twice the usual value.

But, that's exactly my point. The E field SHOULD be sigma/epsilon_nought....but it IS sigma/2epsilon_nought.
 
  • #4
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I'm having a thought.

Can it be due to the fact that the electric field (sigma/epsilon_nought) is due to the entire surface and we are only interested in the field the REST of the surface produces at the surface patch under consideration?
 
  • #5
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So, here's what I came up with. Would love confirmation.

The E field is generally sigma/2epsilon. Since it's a conductor, the E field is sigma/epsilon. So, the rest of the conductor must be supplying the needed E field to cancel the inside E field and reinforce the outside E field. How much cancellation and reinforcing is there? You guessed it. E/2epsilon.

So, the E field provided by the rest of the conductor is E/2epsilon.

Tada?
 
  • #7
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Weeee!
 

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