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Electric Potential due to conducting sphere and conducting shell

  1. Jun 13, 2014 #1
    1. The problem statement, all variables and given/known data
    A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of – 3Q, the new potential difference between the same two surfaces is????
    Answer is V.Let R be the radius of the outer surface of shell and r of inner surface and let the radius of sphere be X.
    Consider epsilon not as 'ε'

    2. Relevant equations
    potential at a point =Q/4∏ε.


    3. The attempt at a solution
    If Q charge is on the surface of sphere then it will induce -Q charge on the inner surface of the shell and Q on the outer surface of shell.Potential at the surface of sphere due to charge inside sphere is Q/4∏εX but I am unable to calculate NET potential at surface of sphere(potential due to charge outside and inside sphere).
    I know that potential on the outer surface of shell is Q/4∏εR.
     
  2. jcsd
  3. Jun 13, 2014 #2

    ehild

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    Yes, it is true. And you also know the electric field between the sphere and the inner wall of the shell (think of Gauss' Law). How do you get the potential difference then?

    ehild
     
  4. Jun 13, 2014 #3
    E at any point between sphere and shell is kQ/a^2 where a is the distance of a point from the center of sphere.
    since,dV/dr=-E. so v=kQ/a.I do not know whether it is right or wrong.
     
    Last edited: Jun 13, 2014
  5. Jun 14, 2014 #4

    ehild

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    You get V, the potential difference between the sphere (a=X) and the inner surface of the shell (a=r) by integrating -E between these boundaries. You also know the potential at the outer surface of the shell is Q/(4∏εR), and that a conductor has the same potential everywhere. Your formula misses the integration constant that you need to match to the potential values at a=X and a=r.

    ehild
     
  6. Jun 14, 2014 #5

    SammyS

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    BTW: It's epsilon naught (ε0) .

    (Homonyms)
     
  7. Jun 14, 2014 #6
    So potential on the outer surface of sphere is kQ/R+kQ/r-kQ/X.so,potential difference between the two surfaces is kQ/r-kQ/X.Now if shell is given the charge -3Q then net charge on it will be -2Q(as initially it has Q charge).The shell will induce -3Q charge on the surface of sphere.So,net charge on the sphere will be -2Q. Now if we calculate potential difference we will get P.D=k(-2Q)/r-k(-2Q)/X.This value is not the same as previous one but the answer says that it should be same as previous one.
     
  8. Jun 14, 2014 #7

    ehild

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    .
    No, the inner and outer surface of the conducting shell are at the same potential.

    That is true, but what is the sign? What is V(sphere)-V(shell)?

    There was zero net charge on the shell, -Q inside and Q outside. Adding the shell -3Q charge, the net charge will be -3Q. How is it shared between the inner and outer surfaces?

    No, the charge on the shell will not induce any charge inside. Does the electric field change between the sphere and the shell if you add charge to the outer shell? Think of Gauss' Law.


    ehild
     
  9. Jun 15, 2014 #8
    In reply to ehild
    No, the inner and outer surface of the conducting shell are at the same potential.

    As the electric field inside conducting shell is 0.so,the inner and outer surface of the conducting shell are at the same potential.


    That is true, but what is the sign? What is V(sphere)-V(shell)?

    I do not understand what is meant by sign in this case.


    There was zero net charge on the shell, -Q inside and Q outside. Adding the shell -3Q charge, the net charge will be -3Q. How is it shared between the inner and outer surfaces?

    Please explain the sharing of charge between inner and outer surface.As initially inner surface of shell has charge -Q so as to get net charge -3Q outer surface of charge should have charge -2Q. I do not know if it is right or wrong.So please explain.


    No, the charge on the shell will not induce any charge inside. Does the electric field change between the sphere and the shell if you add charge to the outer shell? Think of Gauss' Law.

    That was my mistake.Yes,while calculating E using gauss law we do not consider charge outside the gaussian surface.So E between shell and sphere will not change.
    I was not able to quote so I replied in this way.
     
    Last edited: Jun 15, 2014
  10. Jun 15, 2014 #9

    ehild

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    Use the quote button under the post. If you want to quote only some sentences, write quote in front in [], and end the quotation with /quote also between [].


    ehild
     
  11. Jun 17, 2014 #10
    I got it .Thank you for your help.
     
    Last edited: Jun 17, 2014
  12. Jun 17, 2014 #11
    As the electric field inside conducting shell is 0.so,the inner and outer surface of the conducting shell are at the same potential.


    kQ/R+kQ/r-kQ/X- kQ/R.


    Please explain the sharing of charge between inner and outer surface.As initially inner surface of shell has charge -Q so as to get net charge -3Q outer surface of charge should have charge -2Q. I do not know if it is right or wrong.So please explain.


    That was my mistake.Yes,while calculating E using gauss law we do not consider charge outside the gaussian surface.So E between shell and sphere will not change.
     
  13. Jun 17, 2014 #12

    ehild

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    What is the potential at the outer surface of the shell?

    ehild
     
  14. Jun 17, 2014 #13
    As I have mentioned above,potential at the outer surface of shell is kQ/R.In second case I think that it is k(-2Q)/R as the inner surface of shell has charge -Q as as to get net charge -3Q on shell,outer surface of shell should have charge -2Q.I do not know if it is right or not.
     
    Last edited: Jun 17, 2014
  15. Jun 17, 2014 #14

    ehild

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    The outer surface of the shell is at distance R from the centre. The electric potential is the same as the whole charge of the system was in the centre. In the first case, it is Q, and the potential both at the outer and the inner surface of the shell is kQ/R. In the second case, the net charge is -2Q. The potential of the shell - on both sides- is -2kQ/R.

    At the same time, the electric field between the sphere and the shell does not change. Nor does the potential difference.

    ehild
     
  16. Jun 17, 2014 #15
    So the charge on the outer surface of shell in second case is -2Q and the potential difference in two cases will remain same . THANK YOU sir for answering.Now I understood it completely.
     
  17. Nov 12, 2015 #16
    ∏ is it π?
     
  18. Nov 12, 2015 #17

    ehild

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    Yes, Π is π, but the potential at a point is not Q/4∏ε.
     
  19. Nov 12, 2015 #18
    I have found the answer of this question in a google book as
    ANSWER:Because the potential difference between solid sphere and hollow shell depends on the radii of two spheres and charge on the inner sphere.Since the two values have not changed therefore potential difference does not change.
    I want to ask

    that the answer mentions the potential difference between solid sphere and hollow shell
    but the question asks potential difference between
    surface of the solid sphere and that of the outer surface of the hollow shell
    Are both same?
     
  20. Nov 12, 2015 #19
    And the answer also mentions
    Which two spheres?

    I think there is shell and sphere in the question but the answer mentions two spheres?
     
  21. Nov 12, 2015 #20
    Does it mean charge on solid conducting sphere ?
     
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