# Why isn't a capacitor an open circuit?

Alex Hughes
I understand the basics of how capacitors work. If you hook one up to a power source, negative charge will build up on one plate, and positive charge will build up on the other plate. As the charges build up, an electric field forms between the plates and a potential difference is established. This potential difference stops once it reaches that of the battery. This is because eventually, the charge builds up to a point where there is no electric field between one of the terminals of the battery and one of the plates.

However, what I'm confused on is how the charges build up on the plates to begin with. If you look at a capacitor, it consists of two plates separated by a certain distance. To me this is an open circuit? My book says that originally an electric field exists between the two terminals of the battery which drives the electrons to one plate, and pulls them away from the other. But this would mean that electrons still flow even if the circuit isn't complete.

If that was the case why couldn't you just take one wire and attach it to the negative terminal, take another wire and attach it to the positive terminal, and bring them near each other (without touching) and have a current? Do electrons still flow up to the point of a break in a circuit? If so, that makes zero sense to me. It would mean you would have current even when a circuit isn't complete. Would love some help. Thanks.

Last edited by a moderator:

Homework Helper
Two real wires running parallel and near to each other do indeed have some capacitance. Not much, but it is enough to notice.

In my younger days I worked with video terminals (VT-100's). Unlike today's computers, these were simply dumb terminals. ASCII coded data was sent down the wire and displayed on the screen. In the same way, ASCII coded data was sent from the keyboard to the mainframe (or mini-computer). We had a quite a number of these terminals scattered in the building at distances between 100 and 200 feet from the main computer.

The wires running from computer to terminal were a number of twisted pairs. There was strand carrying transmit data to the terminal, a strand carrying receive data back to the computer. one for signal ground, one for protective ground and two more for status signals -- DSR and DTR. This was pretty vanilla RS232-C.

While troubleshooting one particular terminal, I found that what I typed on the terminal was displayed on the screen, even when the far end was not plugged into anything. But if one plugged the far end in, the behavior went away.

This was my first introduction to stray capacitance. At 9600 bits per second, there was enough capacitance between the transmit data wire and the receive data wire that the signals sent down the one wire resulted in levels that could be received by the other.

When the far end was plugged in, the transmitter at the far end with its direct connection was able to hold the signal level steady in spite of the capacitance.

[There was probably some inductive coupling as well. We crimped our own DB25 end connectors and I was a rookie just out of college who didn't know enough to put transmit and receive data on separate twisted pairs]

Mentor
If that was the case why couldn't you just take one wire and attach it to the negative terminal, take another wire and attach it to the positive terminal, and bring them near each other (without touching) and have a current?
In fact, that is indeed possible. However, the capacitance is so small that the amount of current that you get this way is basically negligible. This kind of parasitic capacitance happens between any pair of conductors.

The physical basis of this current is the displacement current in Amperes law.

vanhees71, cnh1995, NFuller and 1 other person
Homework Helper
Gold Member
But this would mean that electrons still flow even if the circuit isn't complete.

Indeed. Electrons flow into one plate of the capacitor and electrons flow out of the other plate. However they are not the same electrons.

Alex Hughes and davenn
Homework Helper
Gold Member
Do electrons still flow up to the point of a break in a circuit? If so, that makes zero sense to me. It would mean you would have current even when a circuit isn't complete.

You do have a current but not for very long...

When you connect a constant voltage (eg DC) source like a battery to a capacitor electrons flow onto one plate and are expelled from the other. Charge builds up on one plate which causes a voltage to appear across the gap in the capacitor. The polarity of this voltage is opposite to the applied voltage so quite rapidly the flow of electrons stops.

If the applied voltage was constantly changing (for example AC) then electrons would continue to flow in and out of the capacitor as they try to make the voltage on the capacitor "keep up" with the changing applied voltage. These electrons flow in and out of the capacitor but not through it.

Alex Hughes, davenn, Dale and 1 other person
Homework Helper
Gold Member
The large area face-to-face, not touching plates of capacitors allow a lot of electrons to flow into one plate before they saturate it beyond the input voltage. After that, those excess electrons can only be drawn back out of that plate. In and alternating current, they will start to a accumulate in the other plate and the situation is reversed. The faster the frequency, the further from saturation either plate gets and the lower the resistance to the AC. For very fast frequencies, the circuit appears almost as though the circuit is complete. For very slow frequencies, the circuit appears almost as though the circuit is open.

The larger the plates are, the more electrons they can take before saturating. That is where a capacitor differs from your two-wire example. Two wires are a tiny, tiny capacitor, unless they are very long.

Last edited:
Alex Hughes
Alex Hughes

davenn and Dale
$$I=C \dot{U}=\frac{U_{\text{bat}}-U}{R}.$$
$$U(t)=U_{\text{bat}} \left [1-\exp \left (-\frac{t}{RC} \right) \right]+U_0 \exp \left (-\frac{t}{RC} \right),$$