anonymous
  • anonymous
Gauss's Law anyone?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
A very large (infinite), uniformly charged slab of plastic of thickness 2a occupies the region between the z=-a plane and the z=+a plane. |dw:1359854942958:dw|
anonymous
  • anonymous
In the problem they drew an object in the shape of a can, why do we assume that that's the shape?
anonymous
  • anonymous
I meant to say "In the solution"

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anonymous
  • anonymous
It further says Find the Electric field everywhere due to this charge configuration. The charge per unit volume is \[\rho\] The solution is \[\vec{E}=E_z\hat{k}=\] and then it gives us a piecewise solution I can write them all out if you want me to
anonymous
  • anonymous
The crux of the argument is that the only possible orientation of the electric field is toward or away from the plate. That is, there can be no component other than either up or down. Do you understand why that is?
anonymous
  • anonymous
I don't quite understand it. Wouldn't the electric field point away from that slab of plastic? To the right and left|dw:1359856455623:dw|
anonymous
  • anonymous
Oh, I apologize, I misread the question. Yes, left and right. So that's clear, then?
anonymous
  • anonymous
uhm yes. but why do we assume that the shape is the following? It's supposed to look like a symmetrically shaped can. |dw:1359856604911:dw|
anonymous
  • anonymous
What you're trying to do is to use the symmetry argument to determine the electric field strength. A "can" shape is a good choice because the only contribution to the electric flux is through the top and bottom of the can. The sides don't contribute, because the electric field is tangential to them.
anonymous
  • anonymous
So the total flux is equal to the flux through the top of the can plus the flux through the bottom:|dw:1359856720464:dw|
anonymous
  • anonymous
Why is \[Q_{inside}=\rho A2a?\] I understand that the charge inside (Q) is \[\phi_{net}=\frac{Q_{inside}}{\epsilon_0}\]
anonymous
  • anonymous
\[\rho\] is the charge per unit volume of the plastic
anonymous
  • anonymous
rho is equal to the charge per unit volume. Therefore, the total charge is equal to rho times the volume of plastic contained inside the can. 2a* A is equal to the volume contained in the can.
anonymous
  • anonymous
So A is the area of ......?
anonymous
  • anonymous
|dw:1359857331358:dw| A is the cross-sectional area of the can
anonymous
  • anonymous
are we calculating the charge inside of that can? in the area that I have highlighted? |dw:1359857447106:dw|
anonymous
  • anonymous
The only charge inside the can is in the volume of plastic
anonymous
  • anonymous
yeah that makes sense, haha.....sorry about the silly questions. one last question: Why is plastic charged? Wouldn't it make more sense to charge a metal?
anonymous
  • anonymous
I meant to say, why is this slab of plastic charged?
anonymous
  • anonymous
You can put charge on whatever you'd like, be it plastic or metal. However, the behavior of charge on metals is very different from that of plastics. Specifically, you can't have uniform charging of metals -- all the charge flows to the outer surfaces.
anonymous
  • anonymous
I see. Thanks @Jemurray3 !!!
anonymous
  • anonymous
No problem
anonymous
  • anonymous
To recap: We have an infinitely large uniformly charged slab of plastic of thickness 2a. We used the shape of a can to determine the charge inside that plastic. Aha! and that charge should be consistent throughout that infinitely large slab of plastic.

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