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Capacitance of an isolated spherical conductor

  1. May 15, 2015 #1
    So it says here that a conducting sphere of radius R with a charge Q uniformly distributed over its surface has V = Q/4πεR , using infinity as the reference point having zero potential,,V (∞) = 0. This gives C = Q/|ΔV| = Q/(Q/4πεR)=4πεR. Does ,V (∞) mean that you are taking the potential of a sphere of infinitely large radius and compressing it into a sphere of radius R to find V? Sorry if my understanding is completely wrong, haha. But why is the potential difference equated to ,V (∞) - V? Also, assuming that the sphere is charged to a potential V1. A spark then occurs which discharges the sphere to a potential V2. Would the energy of the spark be E=c(V1-v2)^2/2 or E=c((V1)^2-(V2)^2)/2 ? I'm confused about the potential difference part. Thank's for your time!
  2. jcsd
  3. May 15, 2015 #2


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    If it is truly isolated, there is no capacitance. There is also no way to define the voltage. Voltage is always measured as a difference between two points. Similarly, capacitance is caused by opposite charges on two or more nearby objects. That excludes isolated objects.

    So your puzzle about the Q and V goes away is there is no V.
  4. May 15, 2015 #3
    I thought that the potential of the object was the work done in bringing a unit charge from infinity.
  5. May 15, 2015 #4


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    Thst is the potential difference between the object and the reference value at infinity. You may assume the voltage at infinity is zero, or any other reference value. There is no such thing as absolute voltage.

    In capacitance, it is the proximity of two objects which is the origin of capacitance, so the proximity is central to the concept, not incidental,
  6. May 15, 2015 #5
    Oops, that was a typo sorry. What I meant was insulated spherical conductor.
  7. May 15, 2015 #6
    I see your argument, but the following link describes the capacitance of an isolated sphere: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capsph.html#c2
    Another example is the Hertzian Dipole, where the end plates have much more capacitance than calculated by the parallel plate formula. They seem to act as isolated plates having quite large self capacitance.
  8. May 15, 2015 #7


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    I looked at the link you provided. It appears that they implicitly assume that the sphere sits above a ground plane. The voltages in the formulas are with respect to ground. In my definition, that's not isolated. Otherwise every charged particle free in empty space would have capacitance.

    Re The Herzian Dipole: I really can't speak about RF frequencies, antennas, or the impedance of free space. I'll bow to your knowledge on that.
  9. May 16, 2015 #8
    I see no such thing over there.
    No they are with respect to infinity. To find them you need to integrate the electric field from the surface of the sphere to infinity.
    I really don't see the problem with this. It's true that the self-capacitance can be ignored in most cases.
  10. May 17, 2015 #9


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    The sphere is surrounded by a conducting sphere at infinite distance. We're not talking about large values of capacitance, the self-capacitance of the Earth is just 700uF.

    Scroll down to Self-Capacitance here: http://en.m.wikipedia.org/wiki/Capacitance
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