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JenniferSmart1

  • one year ago

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  1. JenniferSmart1
    • one year ago
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    |dw:1361317472233:dw|

  2. JenniferSmart1
    • one year ago
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    |dw:1361317624407:dw|

  3. JenniferSmart1
    • one year ago
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    \[W=-\triangle U=-\int F\cdot dl=-\int_{x_1}^{x_2} Fdx=-kq_1q_2\int_{x_1}^{x_2}\frac{dr}{r^2}=-\frac{-kq_1q_2}{r}\] I figured that \[q=\sigma A\] I'm told that the charge densities are equal and opposite on the two large plates.

  4. JenniferSmart1
    • one year ago
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    Oh and we're moving an electron from the negatively charged plate to the positively charged plate.

  5. JenniferSmart1
    • one year ago
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    @TuringTest

  6. JenniferSmart1
    • one year ago
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    I calculated the electric field already

  7. TuringTest
    • one year ago
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    what is the question?

  8. JenniferSmart1
    • one year ago
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    I have to calculate the work done to move an electron from the negative plate to the positive plate.

  9. JenniferSmart1
    • one year ago
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    |dw:1361318208383:dw| This is the force I need to overcome correct?

  10. JenniferSmart1
    • one year ago
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    |dw:1361318290447:dw|

  11. JenniferSmart1
    • one year ago
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    My class is about to start...

  12. TuringTest
    • one year ago
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    there is no resistance force to overcome, just the inertia of the electron moving due to the E field

  13. JenniferSmart1
    • one year ago
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    but the electric field is pointing in the opposite direction of where the electron is moving to

  14. TuringTest
    • one year ago
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    but the E-field points in the direction that a *positive* test charge would move. After all, an electron wants to go from negative to positive, right? Like repels like and so forth...

  15. JenniferSmart1
    • one year ago
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    Oh I see...

  16. JenniferSmart1
    • one year ago
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    In that case: \[W=-\triangle U=-q\triangle\] q= is the charge of an electron correct?

  17. JenniferSmart1
    • one year ago
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    @phi

  18. JenniferSmart1
    • one year ago
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    oops! \[W=-\triangle U=-q\triangle V\]

  19. TuringTest
    • one year ago
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    I would say the elementary charge is \(q=-e\)\[W=-\int(-e)\vec E\cdot d\vec\ell=-e\int Edx\]Note that three negatives multiply to give a negative result. One negative from the charge, one from the fact that a positive charge moving along the E field lines goes down in energy, and one from the fact that E and the direction of motion are antiparallel.

  20. JenniferSmart1
    • one year ago
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    so shouldn't use the relationship I wrote earlier

  21. JenniferSmart1
    • one year ago
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    I'm given the potential difference between the plates

  22. JenniferSmart1
    • one year ago
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    I mean shouldn't *I use the relationship I wrote earlier

  23. JenniferSmart1
    • one year ago
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    sorry typo

  24. JenniferSmart1
    • one year ago
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    I mean if I'm given delta V and I know e ....I should be able to just multiply those to get delta U

  25. TuringTest
    • one year ago
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    right, then that's all

  26. JenniferSmart1
    • one year ago
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    oh and http://en.wikipedia.org/wiki/Joule are \[C\cdot V=Joules \] correct?

  27. TuringTest
    • one year ago
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    coulombs*volts=jouls, yeah

  28. JenniferSmart1
    • one year ago
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    Thanks Max!

  29. TuringTest
    • one year ago
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    Welcome "Jennifer/sophya"

  30. JenniferSmart1
    • one year ago
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    LOL!!!!!! Not even close haha

  31. JenniferSmart1
    • one year ago
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    since KE=U the kinetic energy would be the same correct?

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