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Finding potential of conducting sphere

  1. Nov 26, 2015 #1

    Titan97

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    1. The problem statement, all variables and given/known data
    Find the potential at the centre and at P of a hollow uncharged conducting spherical shell if a charge q is placed at a distance of 2r from centre of shell.
    (r is the radius)
    IMG_20151126_190834_538.JPG

    2. Relevant equations
    All the charge induced on the sphere has to reside on the sphere's outer surface.

    Mirror charge is given by ##\frac{-qR}{a}##
    At a distance ##\frac{R^2}{a}## from centre.
    (R is radius and 'a' is distance of charge from centre)

    3. The attempt at a solution
    Using method of images, mirror charge is -q/2 and its distance from C is r/2
    Potential at centre is then $$\frac{-k\frac{q}{2}}{2.(r/2)}+\frac{kq}{2r}=0$$
    Answer given is ##\frac{kq}{r}##

    Similarly at Point P, I got ##0## as the potential. But answer given is ##\frac{-kq}{6r}## at P.
     
    Last edited: Nov 26, 2015
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  3. Nov 26, 2015 #2

    TSny

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    Note that the problem does not state that the spherical conductor is grounded.

    Is point P on the surface of the conductor?

    The given answers do not appear to be correct to me.

    Putting an image charge inside the sphere is useful for finding the potential outside the sphere. What can you say about the potential everywhere inside the sphere?
     
  4. Nov 26, 2015 #3

    Titan97

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    P is a point on the surface of the sphere.
    Actually, the problem specifies "charged conducting spherical shell". But charge of the shell is not given. So I thought it was a printing mistake. Is there any difference?

    Potential at every point inside the shell is constant.

    I would like to add this point. Let's say the induced charge density on the outer surface of the shell is σ (a function in terms of the point on the sphere).

    Then, potential is centre $$V_C=\frac{kq}{r}+\frac{k\int\sigma da}{r}$$
    Since total charge on sphere is zero,
    $$V_C=\frac{kq}{r}$$
     
    Last edited: Nov 26, 2015
  5. Nov 26, 2015 #4

    TSny

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    Would you please state the problem word for word as it was given to you.

    I agree that the potential is constant throughout the hollow region inside the shell. So, I don't see how the potential at C can be different than at P.
     
  6. Nov 26, 2015 #5

    SammyS

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    The solution to this problem can be done with superposition together with the method of images.

    The image charge together with the charge, q, produce an equipotential surface with potential of zero co-incident with the surface of the sphere.

    However, that's the solution if the sphere has a charge of -qr/(2r) = -q/2 .

    Superimpose upon this, the solution for an isolated conducting sphere with the opposite charge (+q/2). That is, replace the sphere with a charge of q/2 at the sphere's center.

    I.e., combine the image charge and this charge at the center.
     
  7. Nov 26, 2015 #6

    Titan97

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  8. Nov 26, 2015 #7

    TSny

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    I'm not sure of the interpretation of the problem.You might be right that the problem meant to state that the sphere is uncharged rather than charged.

    Also, it could be that the problem wants you to find the contribution to the potential at P due to just the induced charges on the sphere. When I work the problem with this interpretation, I get kq/(6r) which differs in sign from the given answer. So, I'm not sure what's going on.

    I don't see where the problem asks for the potential at the center.
     
  9. Nov 26, 2015 #8

    Titan97

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    I added that part. The minus was a typo. Did you solve using method of images?
     
  10. Nov 26, 2015 #9

    SammyS

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    The potential at the center is the same at the potential at P, as is the potential anywhere on the surface or in the interior.

    The answer key is almost correct.
     
  11. Nov 26, 2015 #10

    TSny

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    Yes, I used the method of images. (See SammyS post #5 for more details.)
     
  12. Nov 26, 2015 #11

    Titan97

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    Why?
     
  13. Nov 26, 2015 #12

    SammyS

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    As I understand the problem, the net charge on the sphere is zero.
     
  14. Nov 26, 2015 #13

    Titan97

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    In which case does the sphere get -q/2 charge? Is it when the sphere is grounded? But in that case as well, initial charge on he sphere is zero.
     
  15. Nov 26, 2015 #14

    SammyS

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    For there to be an equipotential surface at the location of the surface of the sphere, you replace the sphere with the image charge ( -q/2 ). That as it turns out gives a potential of zero for the sphere. However, Mr. Gauss tells us that this solution is equivalent to the sphere having a net charge of -q/2 .

    So we need another charge. One with charge of +q/2. Where can we place it?
     
  16. Nov 26, 2015 #15

    Titan97

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  17. Nov 26, 2015 #16

    SammyS

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  18. Nov 26, 2015 #17

    Titan97

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    It will only be due to the charge inside the sphere.
     
  19. Nov 26, 2015 #18

    SammyS

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    The sphere is neutral. There is no charge inside the sphere. Therefore the flux is ____ .
     
  20. Nov 26, 2015 #19

    Titan97

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    Zero.
     
  21. Nov 26, 2015 #20

    SammyS

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    Right.

    Replacing the sphere with individual charges must give the same result (at locations external to the sphere) as the sphere gives.
     
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