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Electric field over a ring

  1. Jul 9, 2007 #1
    1. The problem statement, all variables and given/known data

    Why is it that if you want to find an electric field over a ring, you take the derivative of the E equation (kq/r^2)? Like, here's an example.


    "A 0.5 microcoloumb charge is distributed ona ring of radius 3cm. Find the electric field on the xis of the ring at 1cm, 2 cm, 3cm, etc."

    Why is it that you have to take the derivative of the electric field equation instead of, like, the integral or something?
     
  2. jcsd
  3. Jul 9, 2007 #2

    Dick

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    You would need to integrate to get the field. What makes you think that phrase implies taking a derivative?
     
  4. Jul 9, 2007 #3
    Well, taking the derivative was what got me the answer. Let me do the problem for 1 cm

    E=kq/r^2, right, and since it's a ring, we'd have to change r into (x^2+r^2)^(1/2) to get the hypotenuse between the radius r of the ring and the distance which we want to find the electrical field.

    So derivatating the dquation ke*q/(x^2+r^2)^(1/2) would get me -ke*q*x/(x^2+r^2)^(3/2).

    Plugging in the numbers would be (8.99x10^9)(.5mC)(1cm)/(1cm^2+3cm^2)^(3/2), which would ultimately get me the answer (at least the one that I saw in my book).

    Of course, I still don't know why I derivatate the equation.
     
  5. Jul 9, 2007 #4

    Dick

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    dE=k*dq/(r^2+x^2), ok, that's the length of dE vector. But you only want the component perpendicular to the plane of the ring since the horizontal will cancel when integrated. So you need to multiply that by a trig function. What is it? See, no differentiations. What you did above is turn r^2 into (r^2+x^2)^(1/2) (how?????) and then differentiate incorrectly.
     
  6. Jul 9, 2007 #5
    Hmm...it's a bit unclear without a picture, sorry.

    The reason I turned r^2 into (r^2+x^2)^1/2 is because the radius of the circle and the x would make two sides of a triangle. The two lines woudl meet at the center of the circle, and I just drew the hypotenuse joining the two lines at their opposite ends. Hence, the hypotenuse would be (x^2+r^2)^(1/2).

    Thus, it was that line that I was using in place of the r. You know...the slope along the radius.
     
  7. Jul 9, 2007 #6
    [​IMG]

    Here's a kind of good picture I found. Mine is like this one, except without the blue part at the end. That's how I was finding my r.

    EDIT: This isn't my problem btw...just a pic I found on Google search that looks kind of like my own
     
  8. Jul 9, 2007 #7

    Dick

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    Yes, that's the length of the hypotenuse. But E field is proportional to 1/length^2 - not 1/length.
     
  9. Jul 9, 2007 #8

    Dick

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    Exactly the right picture. Notice that when you integrate dE that the dE_perp parts cancel, so you only want to integrate dE_x.
     
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