Question regarding Image Charges

[linda300]

Hey,

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

 

[anathema]

Thank you for your question about the use of image charges in problem solving. I can assure you that the final check you mentioned is indeed necessary in order to properly represent the physics of the system.

The purpose of using image charges is to simplify the problem by creating a mirror charge that cancels out the effect of the original charge, thus making the potential zero at certain points. However, it is important to note that this is an approximation and may not perfectly represent the physics of the system.

By not including the final check, you are essentially neglecting the contribution of the original charge and only considering the effect of the image charge. This may work in some cases, but it is not a reliable method for solving problems.

In order to accurately represent the physics of the system, it is important to also consider the effect of the original charge when you are very close to it. This will ensure that the potential is only due to the charge you are close to, as you mentioned in your post.

In some cases, the final check may not work out as you expected, but this does not mean that it should be ignored. It simply means that the problem may be more complex and require further analysis.

I hope this helps to clarify the use of image charges in problem solving. It is always important to thoroughly understand the approach and make sure that it accurately represents the physics of the system. Thank you for your interest in this topic.