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Location of maximum electric field due to a ring of charge?

  1. Jan 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Hi,
    Having some trouble with answering this question:
    A thin nonconducting rod with a uniform distribution of +'ve charge 'Q' is bent into a circle of radius R. There is an axis, 'z' which originates in the center of this ring.

    In terms of 'R', at what +'ve value of z is that magnitude maximum?

    I'm not precisely sure what this question is asking (slightly ambiguous), however i'm assuming it's asking where the electric field due to this ring is at a maximum. Any help is appreciated!

    2. Relevant equations

    E = (q*z*K)/(Z^2 + R^2)^(3/2)
    E = F/Q
    Where K = 1/(4*Pi*E(naught))
    3. The attempt at a solution

    I have determined z in terms of R to be
    z = R/Tan(Pi/2 - Theta)

    Where 'Theta is the angle of elevation between the 'point' on z and the edge of the ring.

    Thanks!
     
  2. jcsd
  3. Jan 15, 2012 #2
    Looks as though the problem wants you to find the electric field on the axis of the ring. So you will want the charge distribution (hint: make the circle into a line to get charge per length), and you will want to employ symmetry. Is this a class that uses calculus? If so set up the integral and I, or someone else, will tell you if it's right.
     
  4. Jan 15, 2012 #3
    It kind of does, however the magnitude of the z components is zero when it lies between the ring.
     
  5. Jan 15, 2012 #4

    SammyS

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    So, the ring of charge lies in the xy coordinate plane, and is centered at the origin.

    I assume you have determined the E field at any point along the z-axis.

    I general, how do you find the maximum of a function?
     
  6. Jan 15, 2012 #5
    No, you have to find the field for all points along the z-axis and then maximize the function you get for the E-field. Maybe it's zero, maybe it's not. :p What did you get, and how did you get it?
     
  7. Jan 16, 2012 #6
    Yes, but I can't seem to simplify the equation I get.
     
  8. Jan 16, 2012 #7
    So, take the equation
    [tex]dE=\frac{\lambda dl}{r^2}[/tex]
    and use symmetry/geometry to tell me what this equation (differential form of Coulomb's law) becomes in terms of your parameters and coordinates.
     
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