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Electrons and Parallel capacitors: after special relativity.

  1. Mar 28, 2008 #1
    1. [​IMG]



    2. Relevant equations: are too big to post here, meaning, I will give you a link to my work:
    http://farm4.static.flickr.com/3092/2369876018_5866bfb312_o.jpg




    3. The attempt at a solution: that's above as well, I know what the solution IS, it's at the top righthand corner of the page, boxed in.

    I have been looking everywhere for homework help, since I'm doing independent study and my teacher is not always available (it's Spring Break, neither are the tutors).

    I'm doing special relativity so far, and now I'm looking at electrons and positrons, parallel capacitors, etc. I have issues specifically with the easy stuff (for some reason I get the hard stuff just fine...).

    If you can't read the problem above it let me know I will transcribe it, now, I have worked out that somehow, thanks to the book's example, I have to set up a right triangle. As you can see by my work (ignore the top figurings, that's just me setting up, what you want to look at are the formulas and the stuff that's worked out fully at the end of the page; also I know it's E instead of B in the last equation there, I think I was losing it by that time).

    I don't know what I am doing wrong because I am lost in one sense and the book is unhelpful, I just want to understand it so I can move on to other things and other problems, btw this site rules :D.
     
  2. jcsd
  3. Mar 28, 2008 #2

    Doc Al

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    For the electron not to be deflected, what must the net force on it be?

    What's the electric force on the electron?
    What's the magnetic force on the electron?
     
  4. Mar 28, 2008 #3
    Is it really that simple? I just have to calculate force? Hmm...this does make sense...
     
  5. Mar 28, 2008 #4

    Doc Al

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    Yes, it's that simple. Give it another shot.
     
  6. Mar 28, 2008 #5
    Ok, I am assuming I have to use these:

    F = qvB

    F[E] = qE

    But that doesn't exactly work for me, meaning, I get: 5.5544 E-16 for F and 4 E-15 for F[E] (sorry about units it's been about a year since electromagnetism, if anyone is kind enough to let me know what they are I would be grateful, V/m? ) that's nowhere near the final answer if I subtract them from each other assuming that's how I will find the net force.
     
  7. Mar 28, 2008 #6

    Doc Al

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    Good. (Express E in terms of the voltage.)


    I have no idea what you're doing here. You have to solve for the B which will make the forces equal (and opposite, of course), so if the forces aren't equal you've done something wrong.

    Don't plug in numbers until the last step. Set those forces equal and solve for B algebraically.
     
  8. Mar 28, 2008 #7
    *Smacks forehead* yeah, it's been about two years since electromagnetism for me, it makes sense now, thanks!
     
  9. Mar 28, 2008 #8
    Wait, why is it out of the page? Right hand rule right? But how is it setup if you're not using a cross-product?
     
  10. Mar 28, 2008 #9

    Doc Al

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    Actually you are using a cross product. That magnetic force is really [itex]\vec{F} = q\vec{v}\times\vec{B}[/itex]. (The magnitude is qvB; the direction is given by the right hand rule.)
     
  11. Mar 28, 2008 #10
    Yes, I know I am, I just wanted to get a picture of it in my head, at any rate, thanks.

    Oh, and your help IS appreciated, but I suggest you don't treat people like they're stupid, asking for help doesn't mean someone warrants disrespect, if you didn't mean it to be this way (after all it is the internet) then no offense is taken.
     
    Last edited: Mar 28, 2008
  12. Mar 28, 2008 #11

    Doc Al

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    Please point out where I gave you "disrespect".
     
  13. Mar 28, 2008 #12
    It's your tone in your writing, if you can't see it you probably didn't mean it, like I said, if you didn't, then no offense is taken.
     
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