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Capacitance of two tangential spheres

  1. Jun 7, 2012 #1
    Hi,

    Question:
    Consider two conducting spheres with radius R, which are tangential each other (i.e. they touch right at one point)
    If C = Q/V, where V is the potential at the surface, find the capacitance of this configuration.
    --------------------------------------------------------------------------------------------

    I've came along this question in some sort of old physics exam...
    I've been giving it some thought for a while but haven't really came up with anything..
    Anyone can shed some light on how to do this?
    (For instance what charge distribution should it have? +ve on one, and -ve on the other?
    And then integrate for all the charges on the surface?)

    Thanks.
     
  2. jcsd
  3. Jun 7, 2012 #2
    If they are touching each other then they do not form a capacitor.
     
  4. Jun 7, 2012 #3
    Thanks for the reply.

    True...I guess that part was confusing me..
    The question says that they are tangent to each other, so I guess we'll have to assume that they have infinitesimally close to each other but not touching?

    But in that case, wouldn't the potential blow up at the point where they are tangent to each other?
     
  5. Jun 8, 2012 #4
    Anyone, Please? :smile:
     
  6. Jun 8, 2012 #5
    No, it is meant that the combined surface of the two spheres forms one plate of a capacitor, the other being at infinity. Think of it as just one isolated metalic sphere having a capacitance [itex]C = 4 \, \pi \, \varepsilon_0 \, R[/itex].

    Your problem is really hard. I don't know if it has an answer in a closed form expression.
     
  7. Jun 9, 2012 #6
    Thanks a lot, I wasn't aware that a single sphere could have capacitance too.
    And yes, the question comes from an advanced level exam..

    For a single sphere, it seems that C = Q/V makes sense only because V is uniform over the surface...
    How about for a non-uniform potential in this case? Would you need to integrate over the surface?
    Any input is appreciated! :smile:
     
    Last edited: Jun 9, 2012
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