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Electromagnetic Field Question

  • #1

Homework Statement


Problem is as follows:
(Figure 1) shows two very large slabs of metal that are parallel and distance l apart. The top and bottom surface of each slab has surface area A. The thickness of each slab is so small in comparison to its lateral dimensions that the surface area around the sides is negligible. Metal 1 has total charge Q1=Q and metal 2 has total charge Q2=2Q. Assume Q is positive.
28_P50.jpg

Homework Equations


Gauss's Law
Int(E*dA)=Q/ep_0

The Attempt at a Solution


So for working through this problem, I've already determined 2 & 4, those were zero which is pretty easy, and I think I know the general path to follow for this

Only been working on one so far, just need some guidance on where to move but so far its
E*dA=Q/ep_0
Then I imposed a gaussian surface over the upper half of the top slab in order to get the E-field at 1, I chose a rectangle for this
E*(A*L)=Q/ep_0
E=Q/((.5A*L)*ep_0)
I know its the right path, and sorry if I didn't write it out perfectly, just not sure how to handle the area of the rectangle I imposed over 1.

Thanks!
 

Answers and Replies

  • #2
cnh1995
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Figure 1) shows two very large slabs of metal that are parallel and distance l apart. The top and bottom surface of each slab has surface area A. The thickness of each slab is so small in comparison to its lateral dimensions that the surface area around the sides is negligible. Metal 1 has total charge Q1=Q and metal 2 has tota
This is same as a parallel plate capacitor. Electric field will be uniform in the space between the plates. I believe formula for electric field between two capacitor plates will be useful here.
 
  • #3
Ray Vickson
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Homework Statement


Problem is as follows:
(Figure 1) shows two very large slabs of metal that are parallel and distance l apart. The top and bottom surface of each slab has surface area A. The thickness of each slab is so small in comparison to its lateral dimensions that the surface area around the sides is negligible. Metal 1 has total charge Q1=Q and metal 2 has total charge Q2=2Q. Assume Q is positive.
28_P50.jpg

Homework Equations


Gauss's Law
Int(E*dA)=Q/ep_0

The Attempt at a Solution


So for working through this problem, I've already determined 2 & 4, those were zero which is pretty easy, and I think I know the general path to follow for this

Only been working on one so far, just need some guidance on where to move but so far its
E*dA=Q/ep_0
Then I imposed a gaussian surface over the upper half of the top slab in order to get the E-field at 1, I chose a rectangle for this
E*(A*L)=Q/ep_0
E=Q/((.5A*L)*ep_0)
I know its the right path, and sorry if I didn't write it out perfectly, just not sure how to handle the area of the rectangle I imposed over 1.

Thanks!
What is the question? You have described the setup, but have not stated what you were asked to find/compute/explain.
 
  • #4
rude man
Homework Helper
Insights Author
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I assume you want to know the charge distributions on the four surfaces, which of course will also give you the E field between the plates, since the two inner surfaces will have to have the same E field just above them ...

So how many Gaussian surfaces can you come up with to solve for thefour surface charge densities?

Actually, this is a very interesting problem and I would like to see it addressed by such as mfb, vela or tsny!
EDIT:
I won't let go of this! Think I have it:
OK, two Gaussian surfaces and 3 unknowns: let σ = surface charge, then
σ1 - σ3 = Q/A
σ5 + σ3 = 2Q/A
and now, forcing the D fields = 0 in 2 and 4, a third equation, left as exercise for the student!
NOTE: σ on inside surface of 2 = -σ on inside of surface 4 = σ3. (Why?).
Now you have the D and E fields in all 5 regions.
 
Last edited:

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