Solving the Mystery of the 2: JD Jackson on Classical Electrodynamics

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The discussion focuses on the calculation of force on a small patch of a grounded conducting sphere with a charge outside it, as presented in Jackson's Classical Electrodynamics. The confusion arises regarding the factor of 2 in the force calculation, where the expected electric field is thought to be sigma/epsilon_nought, but is instead sigma/2epsilon_nought. The explanation involves Gauss's Law, noting that while typically half the electric field emerges from each side of a surface, in this case, the inner side has zero electric field, causing all the field to emerge from the outer side. This leads to the conclusion that the effective electric field at the surface patch is influenced by the entire surface, resulting in the observed factor of 2. The reasoning concludes that the electric field from the rest of the conductor reinforces the external field while canceling the internal field, confirming the factor of 2 in the calculation.
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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.
 
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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.
 
Bill_K said:
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.
 
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?
 
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?
 
Weeee!
 

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