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Charge distribution on spheres

  1. Jul 7, 2017 #1
    1. The problem statement, all variables and given/known data

    Two conducting spheres having same charge density badadd759c20b0351498736507.png and with radius “R” & “2R” are brought in contact and separated by large distance. What are their final surface charge densities ?
    2. Relevant equations
    No equation in question.

    3. The attempt at a solution
    Tried using the fact that distribution of charges on spheres are proportional to their radii but end up with the wrong answer.

    Also how to calculate charge distribution for same question but on different shapes.
     
  2. jcsd
  3. Jul 7, 2017 #2

    gneill

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    Hi Suyash Singh,

    Welcome to Physics Forums.

    According to the forum rules you need to show some details in your attempt at solution. What equations did you use and how did you apply them? What result did you get (even if you believe it to be incorrect)?
     
  4. Jul 9, 2017 #3

    haruspex

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    The distribution of charge on conducting spheres in direct contact is a very hard problem. It is quite different from two well-separated spheres connected by a wire.
    See e.g. http://www.physics.princeton.edu/~mcdonald/examples/twospheres.pdf.
     
  5. Jul 9, 2017 #4

    Delta²

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    Where the following reasoning goes wrong?
    "During contact the spheres exchange electric charge so that the final equilibrium is when both have the same potential. The potential of each sphere is given by [tex]V=K\frac{Q}{R}[/tex]" so the equilibrium is when [tex]\frac{Q}{R}=\frac{Q'}{R'}[/tex]"
     
  6. Jul 9, 2017 #5

    haruspex

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    That's for a sphere in isolation. Each sphere's potential is affected by the charge on the other sphere, and the distributions will not be uniform.
     
  7. Jul 9, 2017 #6

    Delta²

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    Ok I see but , as also the paper that you linked says, this reasoning is correct for the zero order approximation. This problem probably is given in high school, so for high school physics this approximation can be taken as correct?
     
  8. Jul 9, 2017 #7

    haruspex

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    According to Suyash, that got the wrong answer.
     
  9. Jul 9, 2017 #8
    well the answer is 5/3sigma and 5/6 sigma

    i did this in my exam too but it doesnt give the correct answer.
     
  10. Jul 9, 2017 #9
    by the way its school physics so only basic formulas would be required and everything would be ideal.
    I also did this through "charging by contact method" but that also doesnt seem to work:cry:
     
  11. Jul 9, 2017 #10

    haruspex

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    I believe it does give the answer you quote. Please post your working.
     
  12. Jul 9, 2017 #11
    here's what i did
    let $ be sigma(surface charge density)
    k(q)/r=k(q')/2r
    whch can be written as
    $ pi r r/r=$ pi 4 r r/2r
    -$=0
    which doesnt make sense
     
  13. Jul 10, 2017 #12

    haruspex

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    They have the same surface charge density and different potentials before they are brought into contact.
    When they come into contact, both densities change and the potentials become equal. Only then can you apply k(q)/r=k(q')/2r.
     
  14. Jul 10, 2017 #13

    rude man

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  15. Jul 10, 2017 #14

    rude man

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    Your approach is OK as far as it goes but you also need to relate charges on the two spheres before and after contact. There is Q and Q' before contact and there is different Q and Q' after contact, but the sum of before & after is constant.
     
  16. Jul 10, 2017 #15

    haruspex

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    The charges become redistributed while in contact. Once they lose contact, those charges are fixed. Only the way each charge is distributed changes after that.
    To make the problem work, it should have specified that they are kept well apart but connected by a thin wire . That is a very standard formulation.
    .... as I wrote, and Suyash acknowledged, in post #12.
     
  17. Jul 10, 2017 #16

    rude man

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    The wire is not needed; why should the potentials change relatively once they have made contact?
    EDITed again.
    @haruspex I will send you a private note, see what you think.
     
    Last edited: Jul 10, 2017
  18. Jul 10, 2017 #17

    haruspex

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    While in contact, they share the charge, distributed so as to have the same potential on both surfaces. But what is that potential, and how much charge is on each sphere? This is a tough problem, and as I understand it there is no exact solution.
    Did you follow the link I posted?
     
  19. Jul 10, 2017 #18

    rude man

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    ??
     
  20. Jul 10, 2017 #19

    rude man

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    I wind up with the assumption that the potential, upon contact, whatever it may be, is unchanged for both spheres after they're separated. I still can't rigorously defend that assumption but I'm pretty certain it's assumed in the problem statement, especially since it's supposed to be below college engineering level physics. Connecting the two spheres by a wire as haruspex describes would remove the need for that assumption.
     
  21. Jul 10, 2017 #20

    haruspex

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    Very unlikely. While in contact, the potential at each is the result of the charge distribution over both spheres.
    Have you read the paper I linked to yet?
    Take a look at eqn 4.1 at http://rspa.royalsocietypublishing.org/content/royprsa/468/2145/2829.full.pdf.

    Edit: using that equation I get that for spheres radius R, 2R and total charge Q the potential while in contact is ##\frac Q{2R\ln 3}##. The individual charges are ##\frac Q2\left(1-\frac{\pi\sqrt 3}{9\ln 3}\right)## and ##\frac Q2\left(1+\frac{\pi\sqrt 3}{9\ln 3}\right)##
     
    Last edited: Jul 10, 2017
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