Can I charge a capacitor using 2 batteries?

1 capacitor, 2 separate batteries (Battery A and Battery B). Connect A+ to one side of the capacitor and B- to the other side of the capacitor. A and B are not connected, there is no closed circuit. looking like:

-A+__________CAPACITOR_______-B+


Can the capacitor be charged this way or not? if not, why?

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No. Batteries supply potential difference.

The positive terminal of A(I'll call it A+) is at a higher potential than the negative terminal of A(A-). The same goes for B. However, we don't know if A- and B+ are at the same potential, so we can't conclude that A+ is at a higher potential than B-.

In fact, A+ and B- are at the same potential, as it is the lowest energy configuration of this system.

For a capacitor to work, there needs to be a potential difference across its ends. Here, there isn't.

Besides, a battery only works when charge is being drawn/added from/to both terminals. Electrostatic repulsion will not let the battery supply charge from just one terminal. Don't look at a battery as a producer of charge. Look at it as a separator of charge. For every positive charge A shoots out of its positive terminal, there will be a negative charge that gets stuck on its negative terminal; which will work to prevent more negative charges from accumulating on A-. If negative charge can't accumulate on A-, then A+ will stop shooting out charges. This happens very quickly -- you won't be able to measure the amount of charge that A+ released.

However, if you connect A- and B+, then A- and B+ will be at the same potential, and A+ will be at a higher potential that B-, and the capacitor will charge.

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What if I connected a negative ion generator to one side of the capacitor, and touched the other side with my finger, rubbing my foot against the carpet? would that charge the capacitor? –  Moha Apr 7 '13 at 15:03
@moha well, yes, though "generating" just one type of charge is hard (requires a massive amount of energy --you could do it by tearing a charged capacitor apart, but the force is huge). –  Manishearth Apr 7 '13 at 15:20

No, the capacitor won't work..!

1) The capacitance of a capacitor is always a constant. And, you differentiate two capacitors only by using this parameter. Thus in order for the capacitor to function, the voltage between either of its plates should be the same. I haven't seen a capacitor with a voltage difference between its plates. Make sure that $V_A=V_B$, so that $C=q/V$ is maintained a constant.

2) After another reflection into your question, I saw the phrase "there is no closed circuit". How can there be a current flow and hence a voltage be established, when there's no closed circuit..?

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Can the capacitor be charged this way or not?

To "charge" a parallel plate capacitor requires transporting electrons from one plate to the other via an external circuit.

The capacitor itself is not electrically charged by this process, but rather, energy charged. Work is done by the external circuit "charging" the capacitor.

The work done charging the capacitor is equal to the energy stored by the capacitor. When the capacitor can do work on an external circuit, we say the capacitor is "charged".

Since there is no circuit in your diagram, no way for electrons to move from one plate to the other of the capacitor, the capacitor cannot be "charged" in this way.

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Do you mean that electrons move from one plate to the other on the capacitor using the circuit? Because as far as I know, the medium between the two plates does NOT let electrons to pass through it. How exactly does charge build up on the plates? and how true is the statement of "current passing THROUGH the capacitor"? –  Moha Apr 7 '13 at 15:01
Yes, as I wrote above, the electrons are transported via an external circuit. As electrons are removed from one plate, that plate becomes more positively charged while, as electrons are added to the other plate, that plate becomes more negatively charged. While it is true that electrons do not flow through the dielectric, it is also true that there is a current through the capacitor, a displacement current. See: en.wikipedia.org/wiki/Displacement_current#Necessity –  Alfred Centauri Apr 7 '13 at 17:33