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Potential energy in concentric shells

  1. Jan 16, 2016 #1
    1. The problem statement, all variables and given/known data
    There are two concentric shells of radii a and b respectively ,inner shell has charge Q and outer shell is neutral what will be potential energy of outer shell due to inner shell

    2. Relevant equations
    ##V##=##\frac{KQ}{R}##

    PE=Charge .V

    3. The attempt at a solution
    I know what will be potential of outer shell due to inner shell it will be ##\frac{KQ}{b}## but as outer shell is neutral it does not have any charge and hence potential energy of outer shell should be zero,because we know potential energy is equal to charge multiplied by potential. charge is zero on outer shell hence it's PE would also be zero.Right?
     
    Last edited: Jan 16, 2016
  2. jcsd
  3. Jan 16, 2016 #2

    haruspex

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    Sounds good to me.
     
  4. Jan 16, 2016 #3

    BvU

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    Hello Gracy,

    Right !
    It looks to me as if you sense a contradiction somewhere ?
    The electric field ( in 'all of space' ) with the outer shell in place is identical to the electric field without, so putting it in place shouldn't require energy.
    I wouldn't say
    but

    potential at r = radius of outer shell due to inner shell will be ##\frac{KQ}{b}##

    [edit] this time haru was faster !
     
  5. Jan 16, 2016 #4
    Ok.There is a question.

    A solid conducting sphere of radius a having a charge Q is surrounded by a conducting shell of inner radius 2a and outer radius 3a as shown.Find the amount of heat produced when switch is closed.
    Heat produced=Initial energy -final energy
    calculation.png

    calc.png

    I have taken interaction energy to be zero on the basis of the answer of OP.And I have considered formula of self energy of conducting and and non conducting spheres to be same.
    I know formula of self energy of conducting sphere is ##\frac{KQ^2}{2R}##

    Heat produced=##\frac{KQ^2}{4a}## - ##\frac{KQ^2}{6a}##

    = ##\frac{KQ^2}{12a}##

    But it is wrong!
     
  6. Jan 16, 2016 #5

    TSny

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    In your original question, the outer shell was assumed to have negligible thickness so that the charge on the inner surface of that shell was essentially at the same location as the charge on the outer surface of the shell.

    But now the outer shell has a significant thickness. You will need to reassess the initial potential energy.
     
  7. Jan 16, 2016 #6
    But I wrote the outer shell is neutral.So no charge in inner or outer surface of the outer shell.
     
  8. Jan 16, 2016 #7

    TSny

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    The net charge is zero on the outer shell (initially). But, the inner and outer surfaces each have charge.
     
  9. Jan 16, 2016 #8
    Are you hinting at induced charges?
     
  10. Jan 16, 2016 #9

    TSny

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    Yes.
     
  11. Jan 16, 2016 #10
    Why do you think that ?There is nothing such mentioned in the problem.
     
  12. Jan 16, 2016 #11

    TSny

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    According to the statement of the problem, the outer shell has an inner radius of 2a and an outer radius of 3a.
     
  13. Jan 16, 2016 #12
    What kind of changes shall I make in my calculations now?I don't know what is self energy of shells having significant thickness and I also don't know how to calculate interaction energy in such situation.
     
  14. Jan 16, 2016 #13

    TSny

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    If you have a system of static charges where an amount of charge Q1 is located where the potential is V1, an amount of charge Q2 is located where the potential is V2, and an amount of charge Q3 is located where the potential is V3, then the total potential energy of the system is

    U = (1/2) (Q1V1 + Q2V2 + Q3V3)

    This is a generalization of the formula for the potential energy of a capacitor: U = (1/2)QV.

    Try to apply this to your system where you have three charges: charge on the sphere, charge on the inner surface of the shell, and charge on the outer surface of the shell.

    U = (1/2) (QsphereVsphere + QinnerVinner + QouterVouter)

    You should find that the last two terms simplify nicely. The work will be in getting the first term.
     
  15. Jan 17, 2016 #14
    I solved it as follows
    last.png


    But it is not correct where I went wrong.
     
  16. Jan 17, 2016 #15

    ehild

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    Why are the potentials at the surfaces of the outer shell different ?
     
  17. Jan 17, 2016 #16
    I calculated it. Just the sign is opposite.
     
  18. Jan 17, 2016 #17

    ehild

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    You did not show your calculation. The outer shell is a conductor. What do you know about the distribution of potential on a conductor?
     
  19. Jan 17, 2016 #18
    That it is same everywhere inside the conductor.
     
  20. Jan 17, 2016 #19
    CA.png
     
  21. Jan 17, 2016 #20

    ehild

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    Is it different on the surfaces of the conductor?
     
    Last edited: Jan 17, 2016
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