I have been trying to find an equation (or some solution) of how to calculate the electric field strength (in N/C) of a conducting rectangular (nearly flat) plate which has non-zero potential to it, say the plate has a potential of 5V.
I have searched over the internet (and read Griffiths book on the method of images section), and it seems one would need to apply the method of images to solve the problem. However, since almost all examples regarding method of images is about "grounded" plates at 0V, it is very difficult for me to understand how to use method of images to solve for the electric field of a plate at 5V?
Also, I would like to mention that within the electric field of the 5V conducting plate, there is a small freely-moving positively charged particle in the vicinity (I assume because the charge on this small particle is the same as the charge on an electron/proton– it can be regraded as a test charge and therefore, modification to the combined electric field between the conducting plate and the charge is not necessary?-please correct me here if I am wrong.)
Best Answer
It sounds a bit like you're missing something from the problem description? 5V potential relative to what? It would make sense if you have the field between two conducting plates of different potential for example.
The "0V" they are talking about in the method of mirrors is really a way of saying that any inherent excess or deficit charge in the conductor vanishes (since you assume the plate has a connection to a 0V potential reservoir of vanishing resistance) hence you have a neutral overall charge distribution on the plate.
This allows you to consider the field-distribution when for example a point-charge with charge q is placed in the vicinity of the grounded plate; by the method of images the solution is equivalent to no plate but a charge of charge -q an equivalent distance behind the plate. By differentiating the potential field you can find the equivalent distribution of charge on the plate. Without reading the actual problem, the test charge if it has a specified charge really sounds like it has to be considered in the solution.
Simply saying that a plate has a 5V potential doesn't give you the information needed to calculate a charge distribution on it or a potential field strength. It doesn't even mean that the plate is non-neutral, because Volt is a relative unit. It would have made sense if there was an equivalent plate of, say, 0V in parallel. It might also be a red herring to compound the exercise, which might be a "standard" pointcharge vs. plate problem.