1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Capacitance of ring

  1. May 20, 2013 #1
    I was reading the following thread at stackexachange.

    The first answer calculates the radial force using the capacitance of ring. Any ideas how the poster derived the formula for capacitance? Is the formula different for MKS units? I am interested in deriving it if it doesn't involves too much of mathematics.

    Any help is appreciated. Thanks!
  2. jcsd
  3. May 20, 2013 #2


    User Avatar

    Staff: Mentor

    Last edited: May 20, 2013
  4. May 20, 2013 #3
    Nope, I definitely cannot understand that. Does the formula remains same in MKS unit?

    Btw, I am much more interested in calculating tension in the ring. I tried calculating force on a single particle which subtends a very small angle of ##2\alpha## at the center. The charge on this small part is ##\lambda R (2\alpha)## where ##\lambda## is the linear charge density and R is the radius of ring. It is obvious that that the force on this small particle is horizontal in direction (along the radius R). Consider another particle subtending angle ##d\theta##. Charge on this is ##\lambda (Rd\theta)##. The force due to second particle on first is
    [tex]dF=\frac{k\lambda^2 R^2 2\alpha d\theta}{r^2}[/tex]
    where ##r^2=4R^2\sin^2(\theta/2)## (calculated from the law of cosines).
    Integrating the component along the horizontal direction i.e integrating ##dF\sin(\theta/2)##
    [tex]F=\int_0^{2\pi} \frac{k\lambda^2 R^2 2\alpha d\theta}{4R^2\sin (\theta/2)}[/tex]

    But wolframalpha says that its not possible to calculate the above integral. :confused:

    Thank you!

    Attached Files:

  5. May 20, 2013 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It clearly diverges near θ=0. This is not uncommon when treating charge as distributed on an infinitely thin wire. In reality, wires have thickness, and this becomes important at close quarters between the two regions of charge being considered in the integral. A possible way around it is to use a different approximation for that part of the integration where the two elements are within a distance equal to the wire's thickness, or thereabouts.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Capacitance of ring
  1. Capacitance Problem (Replies: 5)

  2. Capacitance Problem (Replies: 4)

  3. Equivalent Capacitance (Replies: 7)