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JenniferSmart1

  • one year ago

Gauss's Law anyone?

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  1. JenniferSmart1
    • one year ago
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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|

  2. JenniferSmart1
    • one year ago
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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?

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

  4. JenniferSmart1
    • one year ago
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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

  5. Jemurray3
    • one year ago
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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?

  6. JenniferSmart1
    • one year ago
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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|

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

  8. JenniferSmart1
    • one year ago
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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|

  9. Jemurray3
    • one year ago
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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.

  10. Jemurray3
    • one year ago
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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|

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

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

  13. Jemurray3
    • one year ago
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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.

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

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

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

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

  18. JenniferSmart1
    • one year ago
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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?

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

  20. Jemurray3
    • one year ago
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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.

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

  22. Jemurray3
    • one year ago
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    No problem

  23. JenniferSmart1
    • one year ago
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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.

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