The question is as follows (taken from a previous exam):
Two balls (conductors w/ Radius $R_1$ for left ball ball, $R_2$ for right) are attached with a very long and thin conducting wire to two parallel conducting plates. (where $d^2 \lt \lt A$) The ball on the left has a charge of Q, and initially the switch is open.
Assume that there is no charge on $R_2$ initially.
I'm having difficulties understanding what exactly the capacitor does in this situation. If it wasn't there, after a long time, finding the charge on the left ball would be quite trivial; Using the fact that the potential will be the same after a long period of time we would compare the two, and and another equation would be made using the conservation of charge.
I know the capacitor will store charge on its plates, but what exactly does it do after a long period of time? Does it transfer charge to the right ball? Does it keep any charge at all?
I'd appreciate if someone could clear this up for me. I'm looking for more of an explanation of the situation rather than a solution to the problem.