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Homework Help: Intermediate E&M (Should Be Easy To Help)

  1. Sep 1, 2010 #1
    1. The problem statement, all variables and given/known data

    We have a line charge of density λ and length 2L along the x axis. What is the electric field at the point (0,1,1)m for L=1cm and λ=200μC/m? (Hint: no new integral is required.)


    2. Relevant equations

    E=(kq)r^2


    3. The attempt at a solution

    I drew a figure and noticed that the equation for E should be multiplied by 2 since the charged line at both sides of the x-axis are affecting the point. I think I remember how to do this in 2D, but for some reason 3D is confusing me. Thanks in advance.
     
  2. jcsd
  3. Sep 1, 2010 #2

    diazona

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    Are you able to post the figure you drew?

    How would you do it in 2D?
     
  4. Sep 1, 2010 #3
    I currently don't have a scanner, so unfortunately I can't right now. I know if the problem were something similar, but in 2D, the X components would cancel and the Y components would double.
     
  5. Sep 1, 2010 #4
    I also think that the components of E in the Y and Z direction will both be doubled, but I am having a hard time going from there.
     
  6. Sep 1, 2010 #5

    diazona

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    I was suggesting that you actually go through the entire process of solving the equivalent problem in 2D. In other words, do this one first:
    Once you manage to do that, it won't be hard to apply your reasoning to the 3D case.
     
  7. Sep 1, 2010 #6
    Don't you just do...

    dE=2k*(lamda*dx/r^2)cos(theta) where cos(theta)= z/r, r=sqrt(z^2+x^2)

    E= k* integral of [2*lamda*z/(z^2 + x^2)^3/2] dx

    the bounds of the integral are from 0 to .02m


    Sorry about how messy it is, I don't know how to do it neater.
     
  8. Sep 1, 2010 #7

    diazona

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    That's the idea, but a couple of questions: where exactly are the endpoints of the line of charge? And where did the factor of 2 come from?

    Note that in the original problem, it said "Hint: no new integral is required." That suggests that you should already have the result of that integral somewhere in your references. So go ahead and use that result.
     
  9. Sep 1, 2010 #8
    I believe the endpoints are from x=-1 to x=1. The factor of two comes in from where the components in the y direction add up.
    And unfortunately, I looked through my references earlier and couldn't find anything I thought would help.

    I am off to bed for the night... early class in the AM. Thanks for your responses and I will get back on here tomorrow morning immediately after class so you can continue helping me (if you would have the time). Thanks again.
     
  10. Sep 2, 2010 #9

    diazona

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    If the endpoints are from x=-1 to x=1 (what units, by the way?), why are you integrating from 0 to 0.02?

    You don't have to include the extra factor of 2 in this case, it will be automatically taken care of by the integral.
     
  11. Sep 2, 2010 #10
    The units are in cm, so x=.01m to x=-.01m.

    I believe these are also the bounds of the integral.

    I also see what you are saying about the extra factor of 2.

    I am still a bit confused on how to apply this to the other problem though.
     
  12. Sep 2, 2010 #11

    diazona

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    Doing the problem in the 3D case is the same except that instead of just [tex]z[/tex], you have [tex]\sqrt{z^2 + y^2}[/tex].
     
  13. Sep 2, 2010 #12
    So would it be
    E= k* integral of [lamda*sqrt(z^2+y^2)/(z^2 + x^2)^3/2] dx
    and the bounds from -.01m to .01m?
     
  14. Sep 2, 2010 #13

    diazona

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    You have to change the z in the denominator as well.

    Although now that I think about it, there's something misleading about the expression as you wrote it. In the 2D case, you used z to refer to the z-coordinate of the point P, but you used x to refer to the x-coordinate of the charge element, [itex]\lambda \mathrm{d}x[/itex]. You might consider using a different variable instead of z, since that's really just a number, not a coordinate that varies the way x does. Maybe d.

    If you do that, when you move to the 3D case, nothing changes. Well, one thing changes: at the end, after you're done with the integral, when you calculate what number to plug in for d, you'll use [itex]d = \sqrt{z_P^2+y_P^2}[/itex] instead of [itex]d = z_P[/itex] as you had in the 2D case. (The subscript P's indicate that those are the coordinates of the point P)
     
    Last edited: Sep 2, 2010
  15. Sep 2, 2010 #14
    I went and asked my professor what he meant by the last statement of the problem. He said that we just had to use symmetry to solve this without integrating. I will try it and let you know what I get.
     
  16. Sep 2, 2010 #15
    Ok I didn't make progress. My teacher told me that you just have to use sqrt(2) as the magnitude of the vector from the origin to (0,1,1) and somehow use that information to get the answer. He even gave me the answer (if I am remember i correctly) but I still need to understand how he did it. He didn't even work out an integral.
     
  17. Sep 2, 2010 #16

    diazona

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    Maybe you're supposed to approximate the line as a point charge?
     
  18. Sep 2, 2010 #17
    No that wasn't it either. I remember he was saying to draw the y-z plane and use that to look at how the vectors are related. He told me about a value of 10 on both the y and z axes, but I don't remember what those values were for. Those were used to calculate the answer. If I remember correctly the answer was supposed to be E=10/srqr2 in both the y and z directions. I was in a rush to catch my next class and he took longer than expected, so I didn't pay as much attention as I should have.
     
  19. Sep 2, 2010 #18

    vela

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    The hint refers to a "new integral," so it sounds to me like you had already solved the 2D case earlier, perhaps in class or in another problem. This problem is the same problem but in a 3D setting, so I think your professor just wants you to figure out how to apply the previous solution to this configuration.
     
  20. Sep 2, 2010 #19
    Maybe, but when I visited him he didn't mention anything like that. He just kept stressing to think about it in terms of the y-z plane.
     
  21. Sep 2, 2010 #20

    vela

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    He probably did. You just didn't realize that's what he meant. The symmetry he's talking about is a rotation about the x-axis. Think about how you can turn the 3D problem into the 2D problem by rotating the system about the x-axis. If you understand how to do that, I think diazona's earlier comment about how you get the answer will make sense to you.
     
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