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Electric potential inside a spherical shell

  1. Dec 9, 2011 #1
    Say we have a spherical shell of outer radius b and inner radius a. The shell has a total charge +3q and at it's center is a point charge of charge -q. I know that the E field for r>b would simply be: E = (3q-q)/(4πr^2ε0) and thus the electric potential inside the shell must be the same as the electric potential on the outer shell since there is no E field inside the shell. This brings me to the conclusion that the electric potential inside the shell is simply 2q/(4π(b^2)ε0).


    This confuses me however as can't it be said that if we take the electric potential at A with respect to the point charge, we get -q/(4π(a^2)ε0)? How can this be the case at the inner surface of the shell if the shell itself has an electric potential of 2q/(4π(b^2)ε0)?


    Am I looking at this wrong as in one case I'm comparing to infinity and in the other I'm comparing with the point charge? If this is the case does this mean the electric potential at the center of the sphere would simply be 2q/(4π(b^2)ε0) + -q/(4π(a^2)ε0)?
     
  2. jcsd
  3. Dec 9, 2011 #2

    Delphi51

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    Homework Helper

    A CONDUCTING shell, right? Spherical, so Gauss' Law applies and the E field outside depends only on the total charge inside the sphere so I think your first thought,
    E = (3q-q)/(4πr^2ε0) is correct. Integrating E*dr in from infinity gives V = 2q/(4πr)ε0 (oops, you had r²) and that should be good to radius b. No change going through the conductor. A charge of +q is induced on the inner surface which kills the E field due to the -q at the center (leaving a charge of +2q on the outer surface of the shell). Once r < a, E = (-q)/(4πr^2ε0), the potential change is the integral of this new E going the opposite way so V will get smaller as r gets smaller.
     
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