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JenniferSmart1 Group Title

Gauss's Law anyone?

  • one year ago
  • one year ago

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  1. JenniferSmart1 Group Title
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    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|

    • one year ago
  2. JenniferSmart1 Group Title
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    In the problem they drew an object in the shape of a can, why do we assume that that's the shape?

    • one year ago
  3. JenniferSmart1 Group Title
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    I meant to say "In the solution"

    • one year ago
  4. JenniferSmart1 Group Title
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    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

    • one year ago
  5. Jemurray3 Group Title
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    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?

    • one year ago
  6. JenniferSmart1 Group Title
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    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|

    • one year ago
  7. Jemurray3 Group Title
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    Oh, I apologize, I misread the question. Yes, left and right. So that's clear, then?

    • one year ago
  8. JenniferSmart1 Group Title
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    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|

    • one year ago
  9. Jemurray3 Group Title
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    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.

    • one year ago
  10. Jemurray3 Group Title
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    So the total flux is equal to the flux through the top of the can plus the flux through the bottom:|dw:1359856720464:dw|

    • one year ago
  11. JenniferSmart1 Group Title
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    Why is \[Q_{inside}=\rho A2a?\] I understand that the charge inside (Q) is \[\phi_{net}=\frac{Q_{inside}}{\epsilon_0}\]

    • one year ago
  12. JenniferSmart1 Group Title
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    \[\rho\] is the charge per unit volume of the plastic

    • one year ago
  13. Jemurray3 Group Title
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    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.

    • one year ago
  14. JenniferSmart1 Group Title
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    So A is the area of ......?

    • one year ago
  15. Jemurray3 Group Title
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    |dw:1359857331358:dw| A is the cross-sectional area of the can

    • one year ago
  16. JenniferSmart1 Group Title
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    are we calculating the charge inside of that can? in the area that I have highlighted? |dw:1359857447106:dw|

    • one year ago
  17. Jemurray3 Group Title
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    The only charge inside the can is in the volume of plastic

    • one year ago
  18. JenniferSmart1 Group Title
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    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?

    • one year ago
  19. JenniferSmart1 Group Title
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    I meant to say, why is this slab of plastic charged?

    • one year ago
  20. Jemurray3 Group Title
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    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.

    • one year ago
  21. JenniferSmart1 Group Title
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    I see. Thanks @Jemurray3 !!!

    • one year ago
  22. Jemurray3 Group Title
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    No problem

    • one year ago
  23. JenniferSmart1 Group Title
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    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.

    • one year ago
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