not sure if this question's been asked before, though I couldn't see one in my brief search on here… Well, the problem I am trying to solve is that I want to determine the electric field strength between two parallel rectangular-shaped plates with different voltages on each plate. For example, rectangular plate 1 has a voltage of +180V and rectangular plate 2 (parallel to plate 1) has a voltage of -5V. What equation(s) could we use to determine the field strength at any point between these two plates? The distance of separation of the plates is finite, say 8.6mm.
[Physics] Electric field between two conducting plates of different potential
electrostatics
Related Solutions
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.
The exact formula to calculate the electric field at a distance $z$ from the centre of a disk of radius $R$ is given at
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html
As you can see for $R\gg z$ the magnitude of electric field is constant and given by $E=\frac{\sigma }{2\varepsilon_{0}}$.
The $V=Ed$ formula can be applied to the case where two parallel plates kept at voltage $V$ (external) and separated by distance $d$. See the animation below from
http://www.regentsprep.org/Regents/physics/phys03/aparplate/
As you can see if the potential is constant as the distance gets smaller the electric field increases. If you want to apply the $E=\frac{\sigma }{2\varepsilon_{0}}$ formula here you need to calculate a new $\sigma$ for each $d$ because in this case $\sigma$ is not constant, it increases as the plates come closer as illustrated in the animation by more $+$ and $-$ charges on the plates.
Edit: Answers to the questions in the comments.
Question: What is $\sigma$ and why it increases as the two plates come together under a constant external potential $V$?
Answer: $\sigma$ is a measure of charge density. It can be calculated as total charge divided by total area. $\sigma$ increases as the plates come closer because the charges on each plate can attract more of the opposite charge to the other plate.
Question: How would one calculate the new electric field if the distance between the plates is reduced but there is no external voltage, that is the plates has constant $\sigma$?
Answer: There are two ways. If the $R\gg z$ case is valid then irrespective of the distance between the plates the electric field is constant and given by $E=\frac{\sigma }{\varepsilon_{0}}$.
If $R\gg z$ is not valid then one needs to use
$$E_{z}=\frac{\sigma}{2\varepsilon_{0}}\left ( 1-\frac{z}{\sqrt{z^{2}+R^{2}}} \right )$$
to calculate the electric field.
Best Answer
If you are not worried about fringing effects near the edges of the plates (which you shouldn't, if the plate dimensions are much larger than the gap distance), then finding the field is simple: V/d, where V is the voltage between the plates and d is the gap distance between the plates.
So, in the example you gave, there would be a field of 185V/(0.0086m) = about 21512 V/m, between the two plates. The voltage 1mm from the -5V plate would be -5 + (0.001m)(21512 V/M) = 21.512V-5V = 17.512V