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

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Discussion Overview

The discussion revolves around a problem presented in JD Jackson's Classical Electrodynamics regarding the method of images applied to a grounded conducting sphere with an external charge. Participants explore the calculation of the force on a small patch of the sphere's surface and the factors influencing the electric field in this context.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the factor of 2 in the force calculation, suggesting it should be the electric field at that point times the charge of the patch.
  • Another participant explains that the electric field from a surface charge density is typically symmetrical, but in this case, all the electric field emerges from the outer side, leading to a doubled value.
  • A participant reiterates the point about the expected electric field being sigma/epsilon_nought, but notes it appears to be sigma/2epsilon_nought in this scenario.
  • One participant proposes that the electric field might be influenced by the rest of the surface of the conductor, which could be producing the field at the patch under consideration.
  • Another participant summarizes their reasoning, suggesting that the electric field is generally sigma/2epsilon, but due to the conductor's properties, the field is sigma/epsilon, leading to a specific contribution from the rest of the conductor.
  • A later reply expresses agreement with the reasoning presented.

Areas of Agreement / Disagreement

Participants express differing views on the calculation of the electric field and the resulting force, with some agreeing on aspects of the reasoning while others raise questions about the factors involved. The discussion remains unresolved regarding the exact nature of the electric field and the factor of 2.

Contextual Notes

There are assumptions regarding the symmetry of the electric field and the contributions from the surface charge density that are not fully explored. The discussion also reflects uncertainty about the implications of the conductor's properties on the electric field calculations.

Berko
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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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