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Method of Images in Electrostatics

  1. Dec 30, 2009 #1
    1. The problem statement, all variables and given/known data

    A point charge q is held a distance d > a below the centre of a large spherical conductor of radius a. Show that, if the spherical conductor is earthed, the potential outside the sphere may be found by placing an image charge at a distance c=a2/d from the centre of the sphere, and determine the charge stored on the spherical conductor.

    Instead of being earthed, the net charge on the spherical conductor is now set to zero. Explain why, in order to represent the electric field outside the conductor an additional image charge must now be included inside the sphere, and determine the potential of the conductor.

    2. Relevant equations
    I found the image charge to have a charge -aq/d

    3. The attempt at a solution

    I can do the first part of the question fine. But I'm getting confused when we set the conductor to zero charge.

    My reasoning for why we need an additional image charge is that ( I Think) the overall charge of all the image charges inside the conductor should now also be zero. Hence, we would need an image charge of +aq/d inside the sphere...

    My intuition was to put it at a distance a^2/d in the opposite direction to the other image charge so it makes a dipole with it centred at the origin of the sphere... But I don't really no if this is right or if its justified.

    I don't know how I can then calculate the potential easily without doing alot of trig and maths?

    Thanks very much
  2. jcsd
  3. Dec 31, 2009 #2


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    Homework Helper

    hmmm... I had a nice long writeup and then I realized that I was completely misunderstanding the problem :rofl:

    Anyway, you had a good idea, thinking about the total of all the image charges. To justify it (I don't know if you did this), I would use Gauss's law with a very large Gaussian sphere. Since the two configurations (point charge with uncharged conductor vs. point charge with image charges) need to generate the same electric field, they had better have the same total charge, q.

    To find exactly where you need to place the other image charge, I would try using the fact that the electric field at the surface of the conductor needs to be perpendicular to the conductor. (I haven't done the problem myself, so I can't guarantee this will work, but I suspect it'll get you somewhere) Once you figure out where that other charge goes, you can use the fact that the potential is constant throughout a conductor: in order to find the potential of the conductor, you only need to compute it at one point. And I'm sure you know how to compute the potential produced by a system of point charges at one point...
  4. Jan 1, 2010 #3
    I'm having trouble calculating the position of the second image charge. I tried putting it at an arbitrary distance from the centre on line with the other image particle.

    Then I try to calculate the field due to all three particles at the radius of the conductor. But I think I'm making a mistake as I get a quartic equation which I had to get my computer to solve.

    Whats the best way to do this bit? Is there a trick to simplify the mathematics of working out the field on radius at an arbitrary angle?

  5. Jan 1, 2010 #4


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    Don't calculate it at an arbitrary angle - calculate it at whatever angle is most convenient, and then check it at an arbitrary angle.

    But here's my line of thought: if you know that there is exactly one additional image charge, there's really only one place you can put it that will make the net electric field perpendicular to the surface of the sphere at all points.
  6. Jan 3, 2010 #5
    ah got it now

    thanks very much
  7. Jan 11, 2010 #6
    2005 Paper 1 Question B7? :D
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