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anonymous
 3 years ago
Gauss's Law anyone?
anonymous
 3 years ago
Gauss's Law anyone?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0A 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
 3 years ago
Best ResponseYou've already chosen the best response.0In the problem they drew an object in the shape of a can, why do we assume that that's the shape?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I meant to say "In the solution"

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It 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
 3 years ago
Best ResponseYou've already chosen the best response.0The 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
 3 years ago
Best ResponseYou've already chosen the best response.0I don't quite understand it. Wouldn't the electric field point away from that slab of plastic? To the right and leftdw:1359856455623:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh, I apologize, I misread the question. Yes, left and right. So that's clear, then?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0uhm 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
 3 years ago
Best ResponseYou've already chosen the best response.0What 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
 3 years ago
Best ResponseYou've already chosen the best response.0So the total flux is equal to the flux through the top of the can plus the flux through the bottom:dw:1359856720464:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Why is \[Q_{inside}=\rho A2a?\] I understand that the charge inside (Q) is \[\phi_{net}=\frac{Q_{inside}}{\epsilon_0}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\rho\] is the charge per unit volume of the plastic

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0rho 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
 3 years ago
Best ResponseYou've already chosen the best response.0So A is the area of ......?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359857331358:dw A is the crosssectional area of the can

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0are we calculating the charge inside of that can? in the area that I have highlighted? dw:1359857447106:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The only charge inside the can is in the volume of plastic

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah 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
 3 years ago
Best ResponseYou've already chosen the best response.0I meant to say, why is this slab of plastic charged?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You 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
 3 years ago
Best ResponseYou've already chosen the best response.0I see. Thanks @Jemurray3 !!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0To 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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