Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Method of images for two charges and two planes

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider two charges, +q each, placed between two grounded plates at a distance d along the z-axis from each plate. thus, the potential V(x,y,-d) = 0 and V(x,y,d) = 0 and the charges are placed at (+R/2,0,0) and (-R/2,0,0) being a distance R apart. Draw the location and magnitude of the image charges needed to solve for the potential in the region between the plates.


    2. Relevant equations

    I've assumed two charges of -aq (a being a proportionality constant), one placed on the other side of plate a and the other on the opposite side of plate b at distances of h and -h.

    3. The attempt at a solution

    I'm assuming that one can calculate the potential (in the z-direction) at d and -d from all four charges and equate them to zero by the boundary conditions. At that point, I'm left with the equation a(d^2-(R/2)^2) = d^2 - h^2. If I only have one equation, how can I solve for both a and h? Or am I going about this the wrong way?
     
  2. jcsd
  3. Oct 11, 2009 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    I'm not sure I understand your problem description correctly...
    [​IMG]

    Is this the setup you are given (with the solid lines representing the grounded conductors)?

    If so, I don't think you can do this with just two images charges. In fact, I think you'll need an infinite number of image charges....just break the problem into pieces...

    [1] Where would you put an image charge, and what would its magnitude be in order to make the potential due to this image charge and the point charge at [itex]z=\frac{R}{2}[/itex] zero at [itex]z=d[/itex]?

    [2] Where would you put an image charge, and what would its magnitude be in order to make the potential due to this image charge and the point charge at [itex]z=-\frac{R}{2}[/itex] zero at [itex]z=d[/itex]?

    [3] Where would you put an image charge, and what would its magnitude be in order to make the potential due to this image charge and the point charge at [itex]z=\frac{R}{2}[/itex] zero at [itex]z=-d[/itex]?

    [4] Where would you put an image charge, and what would its magnitude be in order to make the potential due to this image charge and the point charge at [itex]z=-\frac{R}{2}[/itex] zero at [itex]z=-d[/itex]?

    These 4 image charges will cancel the potential due to your original charges at [itex]z=\pm d[/itex], but now there will be a non-zero potential at [itex]z=-d[/itex] due the first two image charges, and a non-zero potential at [itex]z=d[/itex] due to the second two image charges...

    [5&6] Where would you put two new image charges, and what would their magnitudes be in order to cancel the potential due to the first two image charges (image charges 1&2) at [itex]z=-d[/itex]?

    [7&8] Where would you put two new image charges, and what would their magnitudes be in order to cancel the potential due to the second two image charges (image charges 3&4) at [itex]z=d[/itex]?

    These 4 image charges will cancel the potential due to your first 4 image charges at [itex]z=\pm d[/itex], but now there will be a non-zero potential at [itex]z=d[/itex] due to image charges 5&6, and a non-zero potential at [itex]z=-d[/itex] due to image charges 7&8...

    ...Repeat Ad Nauseum until a clear pattern emerges...
     
    Last edited: Oct 11, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?