# Two balls are attached with a very long and thin conducting wire to two conducting plates. How does the capacitor react in this situation?

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.

The switch is now closed, Find the total charge on the right ball after a very long period of time.

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.

• Is there any charge on $R_2$ or the capacitor plates initially? – EL_DON Aug 10 '16 at 15:40
• None, the only charge is Q on the left-most ball. – RonaldB Aug 10 '16 at 15:42

The charge on $R_1$ repels itself so it redistributes onto $R_1$ and the left capacitor plate when the switch is closed. The new charge on the left plate attracts the opposite charge, which flows from $R_2$ to the right plate. Now $R_2$ is charged with the same sign as $R_1$, $R_1$ has only some of its initial charge, the left plate is charged with the same sign as $R_1$, and the right plate is charged with the opposite sign. You could think of this as some charge indirectly being transferred to $R_2$, but actually, no charges can cross the capacitor gap.

• What exactly do you mean by "repels" itself? – RonaldB Aug 10 '16 at 16:01
• It means that the charge on $R_1$ will spread out so that some of it is transferred to the LH plate. – sammy gerbil Aug 10 '16 at 18:02
• @RolandB the total charge on $R_1$ is composed of many many like charges. One electric charge will exert a force that tends to push another charge of the sign away from itself. Thus, a group of positive charges will all be pushing each other away, or repelling each other. In this case, the only way to get away is to transfer along the wire to the plate of the capacitor, so this is what happens. – EL_DON Aug 10 '16 at 19:37

This reminds me of the problem of connecting two capacitors, one or both of which are charged. There is (usually) a redistribution of charge, and half the stored energy magically disappears. The answer is that the system oscillates, and unless there is some damping (ie some loss through resistance in the connecting wires) it never reaches the equilibrium which you have assumed in the answer.

This question shares some similar features, viz. three capacitors connected in series. (The balls also have capacitance.) "After a very long period of time" refers to the fact that the charges will oscillate when the switch is closed but equilibrium (a steady state) will eventually be reached, so energy will be lost through damping. (I don't know if this fact will be useful in a solution.)

The wires not only allow a transfer of charge to the parallel-plate capacitor and $R_2$, but also ensure that $R_1$ and the LH plate are at the same potential, and $R_2$ and the RH plate are at the same potential, when equilibrium is reached.

As in the 2-capacitor problem, you only need to think about the final distribution of charge, and how it can satisfy the constraints which you have already identified on charge conservation and potentials - also the application of $Q=CV$ to each capacitor.

Once you start working through the problem your understanding of it will improve. I recommend that you post your solution, or as much as you can do if you get stuck.

• $R_1$ and $R_2$ are definitely not at the same potential at the end of the problem unless maybe for a certain special setup. – EL_DON Aug 10 '16 at 19:43
• @EL_DON : In the 2-capacitor problem exactly half the stored energy is lost, regardless of the resistance of the wires. The time constant $CR$ determines how long it takes to reach equilibrium. I did not write that $R_1$ and $R_2$ will be at the same potential. I have edited to make that clear. – sammy gerbil Aug 10 '16 at 20:50