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Spherical Capacitor, equation for voltage

  1. Feb 5, 2013 #1
    I am studying capacitors in an electromagnetism course and I am having trouble understanding/deriving the equation for voltage.

    We have a spherical capacitor with a positive charge on the surface on the center conductor (sphere radius R1), and negative charge on the outer conductor (sphere radius R2), and we have found the equation for the Electric Field using Gauss's law, and now we want to integrate to find the voltage.

    We want to find the equation for the voltage between the two conductors as a function of some arbitrary radius (r), so we integrate the electric field from R2 to r.

    My question is, would integrating from r to R1 be incorrect? My professor seems to think so but he isn't doing a good job on helping me understand why.

    In my years of working with voltage, you have always have a reference point when measuring it, and if your leads backwards, you just get a negative voltage.

    I tried both ways (integrating from R2 to r and integrating from r to R1), and then used these equations to find the total voltage needed to establish the field:

    Vtotal = V(R1)-V(R2)

    And this yielded the same result in either case....so what makes integrating from R2 to r correct and integrating from r to R1 incorrect?
  2. jcsd
  3. Feb 5, 2013 #2

    Simon Bridge

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    In physics we usually take the potential at infinity as the reference, unless someplace else is more convenient. How does that affect your judgement of which direction to do the integration?
  4. Feb 6, 2013 #3
    I guess the convenience comes from integrating from R1 to r since a radius of zero....well zero is where most charts/plots start

    If I went from infinity to R2, then the voltage would be zero since there does not exist an electric field outside the spherical capacitor, then the voltage would start to rise as r moved inside the spherical capacitor towards R1, and then it would peak and hold once inside of R1, since there doesn't exist any charge inside of a conductor (all at the surface), and thus no electric field inside the conductor.

    But when I reverse that, I start at r=0, there doesn't exist any charge inside the conductor so voltage is zero, once I cross over there is a sudden push from the positive charge since it is very close to the conductor, and it dies down as I move towards R2, and then is back to zero once outside of R2.

    Maybe it's unconventional to the usual way of doing it, but is it incorrect?
  5. Feb 6, 2013 #4

    Simon Bridge

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    The path does not matter to the potential difference - the endpoints will determine the sign though. You have to pick the approach that is appropriate to what you need to do with it.
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