- #1

linda300

- 61

- 3

I have a question about the use of image charges to solve problems,

Once you have determined the potential of the problem by including image charges you have to check if the potential describes the physics properly,

so if you have an infinite earthed conducting sheet with chat q at distance a above the sheet, the image charge would be -q at distance -a under the sheet.

So first you must check that the potential is zero when you are located at the sheet,

So say the x and y plane are a sheet, when z=0, V=0,

Then you have to check that as you move very far away the potential is zero,

So as z -> infinity V-> 0,

And the final thing to check is that when you are very close to the charge, the potential is only the potential due to the charge you are close to.

Is this final check 100% necessary?

I have seen a few examples were people have just ignored this check because if they would apply it, it wouldn't work out.

By not working out i mean, you let the distance vector r = (x +dx, y+ dy, z+ dz)

where x,y,z are the location of the charge.

then you would expect to have a V= q k /sqrt(dx^2 +dy^2 +dz^2) after substituting that r into your potential, but in these examples I saw, if you substitute this in you get the required V with the addition or subtraction (depending on image charges) of other terms, so that would not really properly represent the physics of the system would it?

Or is this final check not really necessary?

Just wanted to more thoroughly understand this approach to problem solving,

Thanks