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Homework Help: Rank magnitudes of the electron's accerlation.

  1. Jul 12, 2005 #1
    Hello everyone, i'm lost as usual. Because they are two sheets, infinite and nonconducting. I thought I would use this equation: E = [tex] \delta [/tex]/(2Eo). But they give the separation which i don't see how that fits into this equation. I figured I could find the accerlation using F = MA. But anyways, here is the question. An electron is rleleased between two infinite nonconducting sheets that are horiztonal and have uniform surface charge densities [tex] \delta[/tex](+) and [tex] \delta [/tex](-). The electron is subjected to the following three situations involving surface charge densities and sheet separations. Rank the magnitudes of the electron's acceleration, greatest first. The answer is they all tie. Here is a link to the image:
    http://img333.imageshack.us/img333/4792/image8pt.jpg [Broken]
    Am I approaching this problem totally wrong? :bugeye:
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jul 13, 2005 #2


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    Homework Helper

    I think you have the right idea. If you know what the electric field of one sheet looks like you'll see that seperation is indeed not important. Just add the electric fields from both sheets. Mind the direction.
  4. Jul 13, 2005 #3
    I think you can disregard the separation, as the plates are infinitely long.
  5. Jul 15, 2005 #4
    okay somthing isn't right....
    I used E = [tex] \delta [/tex]/2Eo. So I plugged the numbers given and then added up each situation. Like so:

    1. E = -4[tex] \delta [/tex]/2Eo; E = +4[tex] \delta [/tex]/2Eo; I added those and got 0.
    2. E = -[tex] \delta [/tex]/2Eo; E = +7[tex] \delta [/tex]/2Eo; I added those and got 3[tex] \delta [/tex]/Eo;
    3. E = -5[tex] \delta [/tex]/2Eo; E = +3[tex] \delta [/tex]/2Eo; I added those and got -[tex] \delta [/tex]/Eo;

    What did i do wrong exactly? :confused:
  6. Jul 15, 2005 #5

    Doc Al

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    The positive charge does not cancel the negative charge; just the opposite, since they are on opposite sides of the space between the plates. (Remember the parallel plate capacitor? Is the field between the plates zero? No!)

    Realize that both the positively charged plate and the negatively charged plate both contribute fields pointing down at any point in between the plates.
  7. Jul 15, 2005 #6
    ohh thats interesting...why would you add them? I see you get the correct answer if you do. I know the E-field goes from + to -. But doens't the negative and + tell the direction or is it just showing the charge?
  8. Jul 15, 2005 #7

    Doc Al

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    Staff: Mentor

    You add the fields because they point in the same direction.

    Sure the charge shows you the direction of the field, if you know how to interpret the formula. The field from a plane of charge is [itex]E = \sigma / (2 \epsilon_0)[/itex]. When [itex]\sigma[/itex] is positive, the field points away from the plane; when negative, towards the plane.

    It does not mean that the field from a positive charge is always "positive", whatever that would mean. If you use signs to indicate the direction of the field, the fields between the plates in this problem are all negative (assuming negative means "down").
  9. Jul 15, 2005 #8
    ohh of course, once i drew the 2 plates and looked at the directions i realized what i did wrong, thanks for the explanation!!
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