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Capacitance of two unequal spheres

  1. Feb 2, 2017 #1
    Assume that we have two charged spheres with unequal radii and which do not overlap. How cold we express their capacitance?
  2. jcsd
  3. Feb 2, 2017 #2


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    That's a pretty complex problem. What is the application? Is this a schoolwork problem? Do you have access to COMSOL or another modeling software package?
  4. Feb 2, 2017 #3
    Literary it's my curiosity at work. I know how to compute capacitance for system of two charged spheres with equal radius, but I never came across example of how two charged spheres with unequal radius.

    How could we, for example, find such capacitance? I do have COMSOL.
  5. Feb 2, 2017 #4


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    To estimate it roughly, you can just draw a picture like this (each end of each line is orthogonal to the sphere):
    MokWZ.png ,
    then assign to each line a value of electric field, roughly equal to the voltage divided by the line's length;
    then draw a rough diagram of the field on the big sphere, depending on the spherical angle,
    multiplied by the electric constant, it's the charge distribution;
    then, using some math, estimate the spherical integral and get the whole charge.
    I did such estimations for similar problems, there was acceptable agreement with experiment.
    Last edited: Feb 2, 2017
  6. Feb 3, 2017 #5


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    In COMSOL it is almost trivial, it is just an electrostatic model. I believe there is a somewhat similar example in the manual.
  7. Feb 6, 2017 #6
    If you have the charge stored in the spheres you can find and vectorial equation for the electric field, right?
    With this equation you can use numerical methods to evaluate the integral of electrical field through the line which is the smallest path between the spheres (a line) and evaluate the potential field difference (voltage).
    The capacitance will be the charge of one of the spheres divided by the voltage between they.
    Can someone point any mistake in my thoughts?
  8. Feb 9, 2017 #7
    This was very helpful. In my free time I did some extra experimentation. Thank you so much for your time and patience!
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