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the circuitIt might sound absurd but is it possible to charge a capacitor solely by connecting the positive terminal of a battery to one plate of the capacitor and the negative terminal of another identical battery with the other plate of the same capacitor?

I understand (partially) the following things:

  1. Earthing may be required.
  2. If earthing is not done then the circuit might have to be closed with a wire to allow current to flow.

I don't understand that (treat these as my questions):

  1. Why is earthing required? If earthing is not done is a potential difference not present between the positive terminal of first battery and negative terminal of the second battery?

  2. If the answer to 1 is true and a potential difference exists without the need of earthing (or closing the circuit) then the capacitor will acquire a charge because there exists a potential difference between the plates of the capacitor. For this purpose Where does the current come from in the open circuit and where does it go?

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  • $\begingroup$ Voltage between two unconnected parts of a "circuit" is undefined. $\endgroup$
    – Solomon Slow
    Commented Jul 14 at 19:05
  • $\begingroup$ That's what I fail to understand $\endgroup$
    – Nightwing
    Commented Jul 14 at 19:05
  • $\begingroup$ The batteries are just a complication. Get rid of them. Suppose you only have a capacitor that is not connected to anything. What is its voltage? Adding back either one or both of the two batteries doesn't change the answer. $\endgroup$
    – Solomon Slow
    Commented Jul 14 at 20:09

2 Answers 2

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In order for a battery to charge a capacitor its positive terminal needs to pull electrons from one plate making it positively charged while the negative terminal has to push an equal number of electrons onto the other plate making it negatively charged. That takes work (energy) which comes from the chemicals between the battery terminals. If only one terminal of a battery is connected, the battery can’t supply the necessary energy to move electrons from one plate to the other.

Bottom line: a complete circuit involving both terminals of a battery is needed to charge the capacitor. The use of the two batteries, with no connection between them by earth or other means to form complete circuit, cannot supply the necessary energy.

Hope this helps.

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  • $\begingroup$ the plate that is attached to one of the terminals will develop a net surface charge and the unconnected opposite plate while staying charge neutral will, per force, develop via electrostatic induction a nonuniform surface distribution. If the plates are of different metal from the battery electrode then the distribution will get even more complicated because of the Volta effect. $\endgroup$
    – hyportnex
    Commented Jul 14 at 20:40
  • $\begingroup$ @hyportnex when you say develop a net surface charge I assume you are referring to the migration of the existing charge to the surface, but that the net charge on each plate remains zero, correct? $\endgroup$
    – Bob D
    Commented Jul 14 at 22:27
  • $\begingroup$ just like the negative electrode of the battery has excess negative surface charge (the positive electrode has equal excess positive surface charge), when you connect another metal to the (-) electrode without closing it on the other end will have excess negative charge on its surface. The total charge is zero but not locally on the electrode surfaces. The voltage between the electrodes in my car battery is 12V and are apart by, say 30cm, then there is a field between them about $E=12V/0.3m = 40V/m$ corresponding to an average surface charge density $\approx \epsilon_0 40V/m=....$ $\endgroup$
    – hyportnex
    Commented Jul 14 at 22:56
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You can't charge a capacitor to any meaningful extent this way. The reason is those unconnected nodes at the end.

Assume everything in the circuit starts out neutral. To put a charge on the capacitor, each battery would have to put an equal and opposite charge on the end nodes. That is the only way to conserve charge.

So how much charge can the battery shove onto the unconnected nodes? This is also a question about capacitance: how much charge $Q$ can a voltage $V$ put on the unconnected node? The answer is $CV$ where $C$ is the capacitance of the unconnected node and the wire leading, towards it. Essentially the wire and node act as one plate of an ad-hoc capacitor, wile other conductive objects (the ground, and/or the other terminal) act as the second plate. [1]

Remember that to make a capacitor with a high capacitance, the plates need to be close together and have a large area. Wires tend to have a small area, and let's assume the battery terminal is not especially close to any of these other conductive objects. So the ad-hoc capacitor scores badly on both counts. It is going to have a very small capacitance. The charges on the "real" capacitor and the ad-hoc capacitor are always equal, so will be limited by the charge on you can get onto the ad-hoc capacitor. That will prevent the real capacitor from fully charging... at least, assuming the "real" capacitor has a greater capacitance than the ad-hoc one. For example, if the real capacitor is 1000 times bigger than the ad-hoc capacitor, it will only get charged to 0.1% of full.

In many circuits, especially ones where the conductors are close together, like integrated circuits or multilayer PCBs, it is necessary to take this kind of capacitance into account. But if you are building the circuit out of human-scale wires and alligator clips, it isn't going to be significant, and the capacitor can be assumed not to charge until you actually connect those two end nodes.

[1] An object also has an intrinsic capacitance even if nothing is anywhere nearby. In practice, there's probably something nearby.

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