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

In summary, Gauss's Law is used to calculate the electric field at a point on a surface with a charge outside it. The E field is sigma/2epsilon, but is actually sigma/epsilon/2 because of the inside field.
  • #1
Berko
68
0
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.
 
Physics news on Phys.org
  • #2
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
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.
 
  • #4
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
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
Weeee!
 

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

1. What is the mystery of the 2 in JD Jackson's Classical Electrodynamics?

The mystery of the 2 in JD Jackson's Classical Electrodynamics refers to the inconsistency between the equations describing electromagnetic fields in free space and those in matter. In free space, the equations are symmetric with respect to electric and magnetic fields, with a factor of 1/2 present in front of the curl terms. However, in matter, this factor of 1/2 disappears, leading to an asymmetry between the equations.

2. Why is solving this mystery important?

Solving the mystery of the 2 is important because it is a fundamental problem in classical electrodynamics and has implications for our understanding of electromagnetism. It also has practical applications in various fields such as engineering and physics, where accurate calculations of electromagnetic phenomena are crucial.

3. What approaches have been used to solve this mystery?

Several approaches have been used to solve the mystery of the 2, including theoretical analyses, numerical simulations, and experimental studies. Some researchers have proposed modifications to the fundamental equations of electromagnetism, while others have focused on understanding the underlying physical mechanisms that lead to the discrepancy.

4. Has the mystery of the 2 been solved?

The mystery of the 2 is still an ongoing research topic, and while progress has been made, it has not been fully resolved. Different researchers have proposed their theories and explanations, but there is no consensus on a single solution yet. Further studies and experiments are needed to fully understand and solve this mystery.

5. What are the potential implications of solving this mystery?

If the mystery of the 2 is solved, it could lead to a better understanding of electromagnetism and potentially lead to new discoveries and advancements in various fields. It could also help reconcile conflicting theories and explanations, leading to a more unified understanding of physics. Additionally, it could have practical applications in technologies that rely on accurate calculations of electromagnetic phenomena.

Similar threads

Jackson Classical Electrodynamics: Deriving W_{int} from (1.57) to (1.58)
Replies
3
Views
2K
[E&M] Question on the Image charge method of a grounding sphere
Replies
3
Views
1K
Studying Preparing for Plasma Physics (Jackson)
Replies
4
Views
2K
Jackson's 6.4 - Uniformly magnetized conducting sphere
Replies
1
Views
3K
Potential inside a grounded conducting sphere
Replies
4
Views
3K
Understanding Jackson J.D.'s "Classical Electrodynamics" Retarded Fields
Replies
1
Views
1K
Going through Jackson's book (electrodynamics, method of images)
Replies
4
Views
2K
Method of images, charge q inside conducting sphere
Replies
7
Views
5K
Point Charge in the presence of charged, insulated, conducting sphere
Replies
1
Views
6K
Electrostatic force between 2 hemispheres
Replies
2
Views
4K

Hot Threads

Recent Insights

Back
Top