Consider a setup, a point charge q is brought in front of infinite conducting plate, to calculate the net force on the q we use the image method, however let the induced charges devloped on the outer face be -q, then the field due to the charges developed on the outer surface is given as plane, is given by $ \sigma/2A\epsilon$ where $\sigma$ denotes the surface charge denisty on the outer face, and A the area of the plate, and while the opposite surface would have $+q$ charge devloped, as a result of which electric field due to this is surface is $ -\sigma/2A\epsilon$, hence cancelling the net electric field at the point charge . So shouldnt the charge experience zero force?
2
-
$\begingroup$ I think you are getting many things wrong with the system, so it is hard to know what you are really asking about. First, the induced charge density will not be uniform. Second, the typical example is a thin conducting sheet. Are you wanting to consider one with thickness? Have you actually tried to determine the image charge configuration you would need for this case? $\endgroup$– BioPhysicistCommented May 2, 2021 at 19:11
-
$\begingroup$ @BioPhysicist, even if we consider a thin conducting sheet, there's always a positive and negative charges induced so won't that cancel each other ? And can you please explain why the induced charges wont be uniform in the outer surface after a long time? I have tried the image method, I understood it how it works, but I'm not able to figure out what I have done incorrectly $\endgroup$– green_32Commented May 2, 2021 at 19:19
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
5
The issue here is that you are giving to much "power" to the image charge here. The only thing the image charge does is give the same potential at the boundary (the sheet) so that we can use the uniqueness of solutions to Poisson's equation to determine the potential everywhere on the side of the sheet that has the real charge. It is a trick to find the potential. That is it.
The image charge, therefore, doesn't induce an equal but opposite charge distribution on the sheet. Only the real charge does that.
answered May 2, 2021 at 19:38
-
$\begingroup$ Yes, I agree that real charge does it, but since the conducting sheet is electrically neutral, for the negative charges induced due to real charge there must be a positive charge induced as well on the opposite side, So , for every positive charge induced there is a negative charge induced on the opposite side, hence for every field produced due to negative induced charges ,there exists an equal and opposite field due to positive induced charges, making the net field zero. I think I am not able to visualise clearly what is happening in 3-D, please help I'm very confused. $\endgroup$– green_32Commented May 2, 2021 at 20:41
-
$\begingroup$ @green_32 Oh, I thought you were doing the typical example of a grounded sheet. $\endgroup$ Commented May 2, 2021 at 20:44
-
$\begingroup$ @green_32 In the case of a grounded sheet it won't be electrically neutral. Grounded $\neq$ neutral $\endgroup$ Commented May 2, 2021 at 21:00
-
$\begingroup$ Thanks a lot for clearing, so this means that if a body is neutral then the charge a will experience zero force? $\endgroup$– green_32Commented May 2, 2021 at 21:02
-
$\begingroup$ @green_32 No. But I'm order to use method of images you need to know the potential at the boundaries. If you don't ground the conductor then how will you know what potential it is at? $\endgroup$ Commented May 2, 2021 at 22:25