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Electric potential of a charged shell- Self potential?

  1. Sep 5, 2014 #1
    1. The problem statement, all variables and given/known data

    I recently encountered a problem wherein I was asked to calculate the potential difference between two non conducting concentric spherical shells of radii R1 and R2.
    2. Relevant equations

    ΔV=-∫E.dr

    3. The attempt at a solution

    While I was able to calculate the potentials of individual spheres due to the other's electric field, I could not understand why, in the solution, an extra term in each sphere's potential was mentioned. The extra term was KQ/R (Where Q=charge of the sphere in consideration and R=its Radius). I later understood that the term can be indirectly called Self Potential (like we have self energy). But How can we calculate the self potential?
     
  2. jcsd
  3. Sep 5, 2014 #2

    BvU

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    Hello Adithyan,

    Somewhat incomplete, your problem statement: I suppose the shells were uniformly charged ?
    Never mind. Extra term had a sign, too ? What were the exact expressions for these potentials ?

    A potential is determined to an arbitrary constant, that generally is chosen in such a way that the potential is zero at infinite distance. There is only one constant, so with two shells there must be a common reference (again, generally V at infinity = 0). I suspect that's where these terms come from.
     
  4. Sep 5, 2014 #3
    Sorry I didn't mention it earlier- the shells were uniformly charged. The extra term had a sign equal to the sign of the shell's charge. The exact expression was: V1 (Potential of the smaller sphere) = KQ2/ R2 (Potential due to the bigger sphere) + KQ1/R1 (The 'extra' term I was talking about.) And yes, infinity was the reference point for this question.

    The extra term could be the potential of the sphere due to its charge. I tried deriving it this way:
    Let us consider a spherical shell of uniformly distributed charge Q and radius R. For a small element charge dq, potential due to the remaining charge (Q-dq) can be written as:
    dV =[itex]\frac{ K(q-dq)}{R}[/itex] (We can think of the distribution to be situated at the centre at a distance of R from dq)
    But, integrating this term does not give the desired answer.
     
    Last edited: Sep 5, 2014
  5. Sep 5, 2014 #4

    ehild

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    You seem to mix potential function with the potential energy of a system.

    There is an electric field around and in between the concentric charged spheres, which depends on the distance from the centre. That field has potential, U(r) and the negative gradient of the potential is equal to the electric field: E=-∇U.
    If you put a test charge q into the field it feels force equal to qE. When the test charge moves from r1 to r2 the field does q(U(r1)-U(r2)) work on it.

    The charged spheres have some potential energy. It is equal to the work done when bringing the charge onto the spheres.

    You can talk about the potential energy of a charged sphere, or the potential at a certain point in the field of that charged sphere. If that point is on the surface of a conducting sphere, one can speak of the "potential of the sphere", but then you need to specify the reference point.

    ehild
     
  6. Sep 5, 2014 #5

    I agree with your statements but I am afraid that does not answer my queries. The question clearly asks for potential difference between the spheres with infinity as the reference point. And I believe I have not mixed potential and potential energy in any of my steps/ statements.

    Forget my attempt at the solution and queries; can you derive and state here the potential difference between the spheres?
    Thanks in advance.
     
  7. Sep 5, 2014 #6

    ehild

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    What was the problem exactly? You wrote in the OP
    Was it a non-conducting shell with inner radius R1 and outer radius R2? And what about the charge distribution?

    ehild
     
  8. Sep 6, 2014 #7
    Yes, the inner and outer radii are R1 and R2 respectively. And they are uniformly charged.
     
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