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Charged circle; electrostatics

  1. Jan 2, 2006 #1
    Electrostatics problem

    The problem I'm trying to get help on is the following:
    You have an evenly charged circle with the total charge Q and radius R. There is also a point-charge q located a (thats a symbol, not an article) meters from the center of the circle (the point charge is still inside the circle, but not excactly in the middle of it). Find the force on the point-charge. (Notice, it's in 2D, it is a circle, not a sphere)
    Now, that's how I thought to approach: I picked the centre of the circle at O(0;0) and put q to A(a;0), so i had a circle (x-a)^2+y^2=R^2
    First of all, I exerted the circle's density of charge, so I could later replace dQ in Coulomb's law.
    Now i put down the Coulomb's law for my case. Because of y-symmetry, dF_y-s obviously compensate each other and the sum of all forces would be the sum of F_x-s. So i derive dF_x by putting down the general case and then multiply it with the ratio of sides of the new-formed triangle (dF,dFy,dFx).So i finally get dF_x.
    Now I should replace dl with the function of x and dx, but that is going to be very uncomfortable to do (i tried this at first, but look, how the angle of dl varies around the circle, when you look at dl through q-s eyes). So polar coordinates would help a little:
    By integrating over the angle of 2 pi, I get
    I bet, you can see the problem now. The integrand is impossible (beyond my abilities) to solve it generally. When i do it numerally, it shows sensible results, but the general solution is what bothers me.
    I imagine, there is a niftier general approach. So all the advice is welcome.
    I'll hope you are able to help.
    Alright, the thread didn't come out to be a hit, but could you at least read the problem and guess something about how you think it might be done?
    Last edited: Jan 3, 2006
  2. jcsd
  3. Jan 4, 2006 #2
    i mayb e wrong on this ... but shouldnt the answer be zero??

    your limits of integration should be the same, no? Thus giving you a zero integral?
  4. Jan 4, 2006 #3
    If the charge is at the center obviously forces will cancel. But since the charge q is off centered, there should be a net force on q..
  5. Jan 5, 2006 #4
    Is the "circle" a conductor?
  6. Jan 5, 2006 #5
    Thank you for your answers (and questions)!
    If the point charge would be in the center of the circle, then yes, the answer would be zero. It also would be a zero, if there would be a sphere instead of the circle.
    The limits are not the same, they are form [tex]-\pi[/tex] to [tex]\pi[/tex], or they could also be from [tex]0[/tex] to [tex]2\pi[/tex].
    The integral won't be zero, you can substitute R and a with numbers and then do it numerically with the help of some fancy math app, like MathCad, for example. You'll see that the integral differes form zero.
    Also, the function under the integral sign makes sense: [tex]dF_x[/tex] is the biggest, when [tex]\alpha=0[/tex] and the smallest, when [tex]\alpha=\pi[/tex].
    Makes sense, though doesn't help me much.
    No! The circle itself is evenly charged, thus the voltage between any two points on the circle is always 0, so there is no way it can be interpreted as a conductor.
    Last edited: Jan 5, 2006
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