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Method of images and charges

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1. Homework Statement
To conducting planes are placed at an angle 60 degrees to each other. A point charge q is placed on the line bisecting this angle at a distance a/2 fro each plane. Find the image charges and the resulting potnetial

2. Homework Equations
[tex] V(0,0) = 0 [/tex]
[tex] V(x,0) = 0 [/tex]
[tex] V(x,2y/\sqrt{3}) = 0 [/tex]

3. The Attempt at a Solution
Since our charge is located at (a,a) the image charge to make the bottom plane grounded is loacted at a point (a,-a)

Now for hte plane at 60 degrees. the image charge has to be located a distance a/2 from the plate but at an angle. Finding the coordiantes of this point is what is troubling me.

Is it safe to assume that this point is going to be on the Y axis?? such that it turns into the placement like in the diagram??

i hope that is right... so far.

i think to neutralize the potnetial at the origin im goig to need three more charge located diametrically opoosite location. So in the end we will have 6 charges?? I am currently working on it.

thanks for your help!
 

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Answers and Replies

Since our charge is located at (a,a) the image charge to make the bottom plane
arounded is loacted at a point (a,-a)
WRONG... the charge is located at (a/2, sqrt(3)/2) as in the figure... and the image charge
is (a/2, sqrt(3)/2)...

Now for hte plane at 60 degrees. the image charge has to be located a distance a/2 from the plate but at an angle.
Yup.. you are correct...

Is it safe to assume that this point is going to be on the Y axis?? such that it turns into the placement like in the diagram??
WRONG... the image charge is not on the Y axis....

Finding the coordiantes of this point is what is troubling me.
Then hard work on it... :biggrin:

i think to neutralize the potnetial at the origin im goig to need three more charge located diametrically opoosite location. So in the end we will have 6 charges?? I am currently working on it.
Yes, you will end up with 6 charges (including the original one)...

GOOD LUCK
 

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