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Electric Field of a ring?

  1. Aug 30, 2009 #1
    1. The problem statement, all variables and given/known data

    Determine both the location and the maximum magnitude Emax of the electric field along the axis of a uniformly charged ring. (Use epsilon_0 for ε0, Q, and a as necessary.)

    2. Relevant equations
    1) dE= (ke)dq/r^2cos(theta)
    cos(theta)= x/r

    2) Ex=Q*((ke*(x))/(a^2 + x^2)^3/2)
    (a is the radius of the ring)


    3. The attempt at a solution

    I took the derivative of Equation 2 and set it equal to 0 to find the maximum x value
    x=(a*sqrt(2))/2 and I know that to be the correct answer for where the maximum E occurs; but upon attempting to plug this back into the original equation to find what the maximum E would be, I get the wrong answer every time. Any advice as to what I'm doing wrong?

    Any help would be greatly appreciated.
     
    Last edited: Aug 30, 2009
  2. jcsd
  3. Aug 30, 2009 #2

    kuruman

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    Homework Helper
    Gold Member

    Hi Stang289GT. Welcome to PF.

    If you do not show what you have done, I cannot figure out where you went wrong. Also, please note that sometimes when you write something up for someone else to see, you find your own mistakes.
     
  4. Aug 30, 2009 #3
    Hmmm...okay.
    (there's a figure that goes along with this, but I can't upload it...)

    Plugging in the maximum x=(a*sqrt(2))/2 value into equation 2 you get:

    Emax=(k_e*Q*((a*sqrt(2))/2))/(((a*sqrt(2))/2)^2 + a^2)^(3/2)
    simplifying this, I think it's supposed to be:
    Emax=(k_e*Q*sqrt(2))/6a
    but this answer is not correct. Is there an equation other than this one that I should be using?
     
  5. Aug 30, 2009 #4

    kuruman

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    Homework Helper
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    Assuming that your expression for the E-field is correct, your problem is probably in your simplification. It would help if you wrote

    [tex]\frac{a \sqrt{2}}{2}=\frac{a}{\sqrt{2}}[/tex]

    then the denominator becomes

    [tex](\frac{a^{2}}{2}+a^{2})^{3/2}=(\frac{3 a^{2}}{2})^{3/2}[/tex]

    and simplify from there.
     
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