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Find the electric field

  1. Feb 1, 2008 #1

    tony873004

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    [​IMG]
    I know I can express the distance from a point on the z-axis from the charge 2q as sqr(2a2+z2).

    But I don't know how to express the distances to the other charges. I guess I could start by saying:

    Let A be the point where charge q sits, B the point where charge 2q sits. Then the distance from a point on the z-axis to charge q would be sqrt((|AB|-a) 2+a2+z2)

    and then do the same for the other charges. But this seems like a lot of work. I wouldn't imagine I'm supposed to start making up names for the points. Am I barking up the right tree with this method?
     
  2. jcsd
  3. Feb 2, 2008 #2
    Aren't they all just the same? The four corners of the square are sqrt{2}a units away from its centre.

    For example, the location of -3q is (a,-a). So the distance to the origin from that point would be

    [tex]\sqrt{\left(a-0\right)^2 + \left(-a-0\right)^2} = \sqrt{2a^2}[/tex]
     
    Last edited: Feb 2, 2008
  4. Feb 2, 2008 #3

    Dick

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    So what's AB? It looks to me from the picture like all of the charges are supposed to be the same distance from a point on the z axis. I.e. the origin O is the center of the square.
     
  5. Feb 2, 2008 #4

    tony873004

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    That's what I thought, but I measured it. It's clearly off-center by a few millimeters. It even looks off-center which is why I double checked. This problem has 3 diamonds next to it which means it's supposed to be a butt-kicker. If the z-axis were in the middle, then it would be easy.
     
  6. Feb 2, 2008 #5

    Dick

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    If that were true, wouldn't they label the off-center distance?
     
  7. Feb 2, 2008 #6

    tony873004

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    It's the lack of labeling that has me concerned, and makes me think this is a trick question. If it were centered, I would expect them to throw in a few more <-a-> 's to remove all ambiguity. Even if I measured it, and it were perfect, we shouldn't be expected to measure figures in the textbook. I'd still be nervous about assuming that it were centered.
     
  8. Feb 2, 2008 #7
    Just a note tony, you may want try it and see what you get if they were the same distances and see if it matches the solution. I don't know if that's a Walker physics text, but mine is. . and sometimes I wonder about their "diamonds of difficulty" system.

    Sometimes the "no diamond" problems make me want to eat razor blades while a three diamond is a walk in the park.

    Caesy

    EDIT: Sorry, just noticed it's an even-numbered problem=no solution to check. . . and I spelled my name wrong!!
     
  9. Feb 2, 2008 #8

    tony873004

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    and I'm taking Physics advice from you?! :rofl:

    I'm not using Walker. I'm using "Physics, the Nature of Things" by Lea. The 3-diamond questions do tend to be tough in Lea. I like Walker better. Lea over-explains stuff, and Walker gets straight to the point. But Lea tends to be meticulous, which is why I'm doubting that the z-axis is centered in the illustration. It looks off-center, it measures off-center, and it's not labelled properly, missing some <-a-> s that a properly-labeled, z-axis in the middle drawing should have.
     
  10. Feb 2, 2008 #9
    well, I'm not sure, but if it isn't in the center, and all data is given in figure, how is suppose to you find out the distance of q, 6q and -3q to the center?? use center square.
     
  11. Feb 4, 2008 #10

    tony873004

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    You were all correct. We were told to assume the z-axis is centered.
     
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