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Electric potential, spherical conductor

  1. Aug 28, 2011 #1
    Hello,
    in this diagram, the shaded regions are spherical conductors.
    a1a6777d.png
    What's the potential at A=B?
    Ignoring the outer sphere, it should be kQ/R.
    When you add the outer sphere, potential at C=D=0 and electric field between B and C is kQ/x^2
    so i integrated (kQ/x^2) dx with interval [2R, R] and got -kQ/2r
    Is that right? I think the potential at A=B should be positive though, and the answer says so too :(
    Thanks!
     
  2. jcsd
  3. Aug 28, 2011 #2

    ehild

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    Do not forget that the electric field is a vector, and it points outward in this case. When you calculate the potential, you determine ΔU with the line integral -∫Edr. If you go inward E and dr point to opposite directions, the integrand is negative. You do not get confused when integrating from B to C and get U(B) knowing that U(C)=0.

    ehild
     
  4. Aug 28, 2011 #3

    ehild

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    The potential is the same at every point of a metal. So U(A)=U(B) and U(C) = U(D). I think you meant the electric field, kQ/r2.
    The potential is not kQ/r when the outer sphere is present, as this would give non-zero potential outside. You need an additive constant: U=kQ/r + C so as it give zero at r=2R.

    ehild
     
  5. Aug 28, 2011 #4

    I like Serena

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    Sorry, I've deleted my post to avoid confusion.
    I was mixing it up with another problem where the spheres were non-conducting.
     
  6. Aug 28, 2011 #5
    Thanks to all of you! :D
    just to be sure, so the potential at B is kQ/R without the outer sphere, and with the outer sphere its kQ/2R, and ehild's additive constant C is -kQ/2R to give U(B) = kQ/R - kQ/2R = kQ/2R?
    Thanks again!
     
  7. Aug 28, 2011 #6

    I like Serena

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  8. Aug 28, 2011 #7
    Thanks! :)
     
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