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Method of images for two charges and two planes

by wakko101
Tags: charges, images, method, planes
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wakko101
#1
Oct11-09, 03:28 PM
P: 68
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?
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gabbagabbahey
#2
Oct11-09, 11:23 PM
HW Helper
gabbagabbahey's Avatar
P: 5,003
I'm not sure I understand your problem description correctly...


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...


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