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Point charge outside a grounded conducting sphere

  1. Dec 23, 2017 #1
    1. The problem statement, all variables and given/known data

    upload_2017-12-23_12-42-35.png
    2. Relevant equations


    3. The attempt at a solution

    Option (a), because the potential is 0 on the surface, and potential due to point charge is positive so there must be negative charge on the surface. Since opposite charges attract each other, there is an attractive force between the sphere and the point charge.

    (b) As the potential due to point charge is different at different points of surface, the surface charge density should not be same everywhere. So, this is correct.

    (c) In case of electrostatics, electric field inside a conductor is 0. So, this is correct.

    (d) By memory, I feel that this option is correct. Is there anyway to find out this information without solving the question fully?

    (e) For large d, the point charge and the sphere will act as a dipole and so the potential falls off as ## \frac { 1} {d^2} ## .


    So, the required option is (e).

    Is this correct?
     
  2. jcsd
  3. Dec 23, 2017 #2

    Orodruin

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    Your argumentation for (a-c) are correct. You may want to think a bit more about (d) and (e). "Feeling" that an option is correct without being able to argue for it is not a very strong argument.

    Hint: When you deal with a solution to Poisson's equation with one charge and an equipotential sphere (which this is for the exterior of the sphere), the solution can be written as a solution in the entire space with a single mirror charge inside the removed sphere such that the potential of the charge and mirror charge exactly cancel on the surface. What would happen if the two charges were the same?

    Edit: Clearly the two options (d) and (e) are in contradiction to each other. If (d) is false, then there is a net monopole moment and the potential falls off as 1/d and (e) must be true. If (e) is false then there monopole moment cannot be non-zero, which means (d) much be true.
     
  4. Dec 23, 2017 #3

    Delta²

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    I THINK that d) is false so as @Orodruin argues in his last few lines this means that e) must be true.

    However I myself cant argue exactly why I believe d) is false, I just think that the induced charge depends on the geometry setup of the problem too and not only on the point charge that initiates the electrostatic induction.
    In other words I would expect the induced charge to depend on the radius of the sphere a, as well as the distance of the point charge from the center of the sphere, and since the problem doesn't give specific values for those two factors, I just very doubt the induced charged would be exactly -q but I might be wrong.
     
    Last edited: Dec 23, 2017
  5. Dec 23, 2017 #4
    Let's consider another image problem. There is a point charge q along z- axis at a distance d above the infinite conducting grounded plane.

    I know that its image charge is –q and –q is the charge induced on the plane.

    Is it so that the charge induced on the conducting material equal to the image charge, always?
    If the image charge is equal and opposite to the charge, then on the spherical surface, the magnitude of potential due to image charge at some point can be greater than that of original charge and hence the total potential is not 0 at that point.

    So, the magnitude of image charge could not be equal to original charge.

    If the charge induced on the spherical surface is equal to the image charge, then the magnitude of induced charge is not equal to that of original charge.

    So, option (d) is false.
     
  6. Dec 23, 2017 #5

    Orodruin

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    The fact that you get zero potential on a plane if you introduce an equal but opposite charge is a hint, so yes (d) is indeed false, which makes (e) true.
     
  7. Dec 23, 2017 #6
    How does the following answer the above question? I couldn't follow it.
    The image charge is equal and opposite in case of plane. This doesn't mean that the induced charge is equal to the image charge.
    After calculating I get to know that induced charge is equal to image charge in plane case.
    But, how to generalize it? How to say that induced charge is always equal to image charge?
     
  8. Dec 23, 2017 #7

    Orodruin

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    Gauss law.
     
  9. Dec 23, 2017 #8
    I construct a Gaussian sphere of radius bigger than that of the original sphere but smaller enough to not consists of the original point charge and concentric with the original sphere.
    Now, flux through Gaussian surface is same in both cases i.e. image charge or original induced charged sphere. And so the charge enclosed by the Gaussian surface is same in both cases. Hence, the induced charge is equal to the image charge.


    This method will be applicable if I replace the sphere by any finite conducting object. So, in this case the induced charge is always equal to the image charge.


    But how to prove it in the case of infinite object, say in case of infinite conducting plane?

    I construct a Gaussian cubical surface, whose one face will be just above the conducting plane and all other sources will be infinitely away from the image charge, and the surfaces perpendicular to the conducting plane is very small in height.

    Then, in both cases i.e image charge or conducting plane, the only surface which will contribute to the total flux is the surface which is just above the conducting plane. And hence the total flux through the Gaussian surface will be same in both cases. Consequently, the image charge will be equal to induced charge on the plane.


    Does this mean that for any kind of conducting object or image problem, it could be shown that image charge is equal to induced charge?

    So, in general, induced charge is always equal to the image charge.
     
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