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Homework Help: Electric field of ring

  1. Apr 4, 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
    dE= k_edq/r^2costheta
    costheta= x/r

    3. The attempt at a solution
    i think i have to take the derivative of some forumla
    that looks like this:
    (kex)/(a^2 + x^2)^3/2 all multiplied by Q

    and set this to zero and find a max

    i really have no idea how to do this
  2. jcsd
  3. Apr 4, 2009 #2


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    That looks about right.

    Maybe take the differential using the quotient rule?

    (d(f(x)/g(x))/dx = ( f '(x)g(x) - f(x)g '(x) ) / g2(x)
    Last edited: Apr 4, 2009
  4. Apr 4, 2009 #3
    yes i applied the quotient rule
    but it gets extremely confusing..i end up with an answer like (a+x)(a+x) =0

    which im pretty sure is incorrect
  5. Apr 4, 2009 #4


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    I didn't work it out, but I noticed that the divisor goes away when you set it to 0, leaving you just

    ( f '(x)g(x) - f(x)g '(x) ) = 0

    or when f '(x)g(x) = f(x)g '(x)
  6. Apr 4, 2009 #5
    so that would leave us with

    3a+3x(a^2+x^2)^3/2(kex) - ke(a^2+x^2)^3/2 = 0

    wat abt the Q in the original equation
    im not very good at calculus so this is pretty tough for me
  7. Apr 4, 2009 #6


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    Going by the link:

    That gives you the equation to begin with and basically there's all the constants invariant in x, so Q just gets carried along through as part of that constant. Let's call it C.

    E = C*x/(x2 + a2)3/2

    Since it seems to be a common factor I don't think it affects Xmax as it looks to cancel out when you set it to 0.
    (Of course it does matter in evaluating |E|.)

    Using the quotient rule where f(x) = C * x, and g(x) = (x2 + a2)3/2

    So doesn't it look like (using the chain rule to determine g'(x)) and then setting to 0 that :

    C*(a2 + x2)3/2 = 3*C*x2(a2 + x2)1/2 ?

    If that's right then it looks like it simplifies to

    3x2 = (a2 + x2)


    x = (a*√2)/2 ?

    Not sure if it's right, but it feels right. No time to double check it. You certainly should.
    Last edited: Apr 5, 2009
  8. Apr 5, 2009 #7
    you are a genius!
    thanks for clearing that up

    but now using the x value how do i find Emax?
  9. Apr 5, 2009 #8


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    By substitution into your original equation, now that you know x in terms of a.
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