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FInding point where E=0

  1. Aug 29, 2015 #1
    1. The problem statement, all variables and given/known data
    I'm given an equilateral triangle of side length b with charges q at each corner and I have to find the y coordinate for point P on the y-axis where the electric field is equal to zero (E=0 other than the center of the triangle)
    1VmMG4D.png
    2. The attempt at a solution
    I figured I'd start by finding the formula for the electric field and started here:
    %7D%7Br_%7B2p%7D%5E%7B2%7D%7D+%5Cfrac%7B%5Chat%20r_%7B3P%7D%7D%7Br_%7B3p%7D%5E%7B2%7D%7D%29.gif
    Then I tried to put the distances in terms of b and y:
    g_white%20%5Cfn_phv%20r_%7B1P%7D%5E%7B2%7D%3D%20%5Cfrac%7Bb%5E%7B2%7D%7D%7B4%7D+y%5E%7B2%7D.gif

    g_white%20%5Cfn_phv%20r_%7B2P%7D%5E%7B2%7D%3D%20%5Cfrac%7Bb%5E%7B2%7D%7D%7B4%7D+y%5E%7B2%7D.gif

    gif.gif
    Because of symmetry I figured that I could just take the y compponent of r1P and r2P by multipling it by sin of theta:
    gif.gif
    From there I tried to set the formula equal to zero, but it got immensly complicated.
    D+%5Cfrac%7B1%7D%7By%5E%7B2%7D-by%5Csqrt3%20+%20%5Cfrac%7B3%7D%7B4%7Db%5E%7B2%7D%7D%29.gif
     
  2. jcsd
  3. Aug 29, 2015 #2
    P is inside the triangle. The field from the top charge is in the -y direction. It has to be for the three fields to add to zero.
     
  4. Aug 29, 2015 #3
    I didn't even think about that, if they are all q then P has to be the center. But the problem explicitly states that the point can't be the geometric center. Would there be any points in 3 dimensions that could satisfy this or when y = gif.gif ?
     
  5. Aug 30, 2015 #4

    gneill

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    Staff: Mentor

    I think you'll find that there are two locations on the y-axis within the triangle where the field goes to zero. Can you imagine why this might be?

    A suggestion: Since q and b are not specified they can be treated as scaling factors along with k. So without loss of generality take your equilateral triangle to have a side length of 2, making the base vertices (-1,0) and (+1,0), and the top vertex (0, ##\sqrt{3}##). Your equation of interest should then look simpler. I'd suggest tackling this with a numerical solver for y between 0 and ##\sqrt{3}##. Multiply the results by "b/2" to regain the original scale.
     
    Last edited: Aug 30, 2015
  6. Aug 31, 2015 #5
    Would the second point be very close to point 3 on the triangle because it would be very close to the test charge and thereby able to counter the force contributions in the y direction of both 1 and 2?

    Am I on the right track with that last equation I wrote or do I need to rethink my approach?
     
  7. Aug 31, 2015 #6

    gneill

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    Staff: Mentor

    Very close to any point charge the field strength increases without bound, so it's unlikely to be the case. Instead look nearby the triangle center.
    It's the right track, but you're on the scenic route :smile:
    If you follow my suggestion you should have a sleeker equation that can be attacked numerically.
     
  8. Aug 31, 2015 #7
    The only thing I could think of is how ovals have a center and a focus, but I don't think triangles have a focus. Sorry, I'm kind of slow.
    I think this is the condensed version, but when I plug in the triangle's center point if.latex?%5Cdpi%7B80%7D%20%5Cbg_white%20%5Cfn_jvn%20%280%2C%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B3%7D%29.gif , I get 1.5 intead of 0. (Though if it were subtraction rather than addition between the 2 terms it does equal zero)
    29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D+%5Cfrac%7B1%7D%7B%28y-%5Csqrt%7B3%7D%29%5E%7B2%7D%7D%29.gif
     
  9. Aug 31, 2015 #8

    gneill

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    Staff: Mentor

    The b/2 term applies to the value of y that makes the expression zero, not to the value of the field; it's a scaling factor pertaining to the size of the triangle.

    What was your reasoning for choosing addition versus subtraction in the equation?
     
  10. Sep 1, 2015 #9
    Oh I see now, in the middle of the triangle the y component of the field from 3 is in the opposite direction from 1 and 2, so it should be a minus sign. Although I'm not really sure how to simplify it from here. I tried expanding everything out, but nothing cancels. Is this were I would use something like Mathematica?
     
  11. Sep 1, 2015 #10

    gneill

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    Staff: Mentor

    Right.

    As I suggested, you might want to turn to a numerical solver for the roots. Wolfram or Mathematica should be up to the task, but you'll have to select roots that are real and lie within the range of applicability of your equation.
     
  12. Sep 1, 2015 #11
    I plugged "Solve[(2 x/((1 + x^2)^(3/2))) - ((x - sqrt[3])^(-2)) == 0, x]" into Mathematica, but I am new to the software and don't understand the result:

    {{x -> Root[-1 + 3 #1^6 - 16 #1^5 sqrt[3] - 16 #1^3 sqrt[3]^3 + #1^4 (-3 + 24 sqrt[3]^2) + #1^2 (-3 + 4 sqrt[3]^4) &, 1]},
    {x -> Root[-1 + 3 #1^6 - 16 #1^5 sqrt[3] - 16 #1^3 sqrt[3]^3 + #1^4 (-3 + 24 sqrt[3]^2) + #1^2 (-3 + 4 sqrt[3]^4) &, 2]},
    {x -> Root[-1 + 3 #1^6 - 16 #1^5 sqrt[3] - 16 #1^3 sqrt[3]^3 + #1^4 (-3 + 24 sqrt[3]^2) + #1^2 (-3 + 4 sqrt[3]^4) &, 3]},
    {x -> Root[-1 + 3 #1^6 - 16 #1^5 sqrt[3] - 16 #1^3 sqrt[3]^3 + #1^4 (-3 + 24 sqrt[3]^2) + #1^2 (-3 + 4 sqrt[3]^4) &, 4]},
    {x -> Root[-1 + 3 #1^6 - 16 #1^5 sqrt[3] - 16 #1^3 sqrt[3]^3 + #1^4 (-3 + 24 sqrt[3]^2) + #1^2 (-3 + 4 sqrt[3]^4) &, 5]},
    {x -> Root[-1 + 3 #1^6 - 16 #1^5 sqrt[3] - 16 #1^3 sqrt[3]^3 + #1^4 (-3 + 24 sqrt[3]^2) + #1^2 (-3 + 4 sqrt[3]^4) &, 6]}}

    Do you know what these mean?
     
  13. Sep 1, 2015 #12

    gneill

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    Staff: Mentor

    No, sorry. I'm not familiar with the details of Mathmatica either. Perhaps there's another method other than "Solve" that will give real numerical results? FindRoot perhaps?
     
  14. Sep 1, 2015 #13
    I'll just toy around with it and see where I get, thanks for your help!
     
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