# Can one capacitor charge another connected by a single lead only?

My instructor presented the following problem on capacitors:

Plates 1, 2, 3, and 4 form two parallel plate capacitors as shown. Plates 1 and 2 are charged, plates 3 and 4 are not. Both pairs of plates are separated by distance d.

Describe what you think would happen if you connected plate 3 to plate 2 with a wire. How many positive charges will plate 3 have on its right side? What does that mean about the electric field between plates 3 and 4 (bigger, smaller, or the same as the electric field between 1 and 2)?

I was unsure so I performed a quick experiment with two capacitors.

I charged one capacitor (representing plates 1 and 2) with a battery and measured it with a multimeter to confirm it was holding a charge. Then, I touched one lead of that capacitor to another, uncharged capacitor (representing plates 3 and 4). Afterwards, I measured the voltage of each capacitor with a multimeter and found that the originally charged capacitor was still charged but the uncharged one was not. None of the charge transferred to the second capacitor by touching a single lead to the charged capacitor.

So I answered that plates 3 and 4 would not become charged at all.

Unfortunately, that was incorrect. The instructor says that plate 3 will have the same amount of positive charges as plate 2 has negative charges and since plates 3 and 4 have a smaller area, the electric field between them will be bigger.

I don't see how the second set of plates can become charged without forming a circuit. This question looks similar, and in that case one capacitor would not charge the other until the switch is flipped completing the circuit.

What am I missing here? Is my experiment with the capacitors not representative of the problem?

• The pictures are as given by your instructor, or are they your depiction of your understanding of the question? May 13, 2020 at 0:11

But if dimension of the plates of capacitor 1-2 is much larger than their distance $$d$$, this charge transfer will be much smaller than charge on the plate 2 and can be neglected. This is so in your experiment, where the capacitor has plates so close the electric field outside is so small that it is negligible and does not lead to any appreciable charge transfer.