To specify what I am asking: How do we even get a current for a moment when the circuit technically never is completed? And what is it that does this? Here I am thinking of a very simple circuit where there is a DC power source and a capacitor (if we want to be accurate, I'm sure we'd need a resistor as well).

So, when we connect the DC power supply to this circuit, what happens inside? I understand that plates connected to a power supply will be charged according to the charge coming into it, so if the negative terminal of the battery were connected to a plate, it would cause that plate to be negatively charged as well. However, why does this happen and how?

(Just realizing this is only indirectly correlated to a capacitor, but whatever.)

First, I thought maybe the charge density in the two plates of the capacitor were changed according to the voltage applied, so that there would be a voltage across, but that also confused me and didn't make sense, because (unless I'm wrong) for there to be higher charge density in one part of the wire, wouldn't the charge density have to decrease at other places in the wire? And wouldn't this have created a local electric field through this wire?

I tried to draw a picture to visualize what I'm thinking (the stronger the color, the greater the charge density):

enter image description here

This just didn't feel right, so what is it that actually happens? Also, how does this give us a current for a second before it flattens out?

  • \$\begingroup\$ The idea that a circuit must be "closed" before anything happens is a simplification and doesn't account for transients. The usual circuit analysis works around that by considering a capacitor to be a sort connection between its terminals, at least a good enough one to make the circuit "closed". \$\endgroup\$
    – Ben Voigt
    Commented Jan 23 at 22:39

2 Answers 2


Your reasoning of a change in charge density is about right.

If you push an electron onto one plate of the capacitor, the charge density there will necessarily rise, but in doing so, the increased negative potential of that plate will repel an electron out of the other plate, thereby reducing the charge density on the other side of the dielectric.

Said another way, if an electron enters one side of a capacitor, an electron must leave the other side. Repeatedly forcing an electron onto one plate leaves you with an accumulation of electrons on one plate (the plate became negatively charged), and a deficit of the same number of electrons on the other plate (positively charged). To an external observer, Kirchhoff's current law (KCL) is not violated; current in both capacitor terminals is equal.

That's how a current can flow "through" a capacitor. There appears to be a current flow, equal on both sides in accordance with KCL, but in reality no electron ever crosses the dielectric barrier.

Also to an external observer, the net electric charge of the capacitor is still zero. There are still equal numbers of protons and electrons throughout the material of the entire capacitor but locally to the plates there is a charge imbalance, immediately either side of the dielectric, an imbalance which gives rise to the potential difference (voltage) across the capacitor.

Current stops flowing altogether when the potential difference of the voltage source (battery) equals the difference in potential of charges on the capacitor plates. That's logical, considering that to push an extra electron onto a plate, that electron must have greater potential energy than those already there. This means that current flow will only be momentary, until the source and capacitor potential differences equalise.

As far as the wires are concerned, it's true that for there to be a current flow along a wire, there must be a potential difference between the two ends, suggesting that there is a place of lower potential energy that a charge can/will move towards, and this would imply a charge density gradient along the wire. Initially, immediately after the capacitor is connected to the source, this will be the case.

If current were unconstrained by resistance in the loop, it would be infinite, but that's never the case. Everything in the system has resistance, including the battery and wiring, so current will be capped. Initially the voltage source has some potential difference, and the capacitor has zero potential difference, and by connecting them together you produce a situation in which charges must move from one to the other to make them equal. The imbalance must be resolved. Charges will move through whatever resistances are present in the loop, and it is those resistances which initially have the greatest potential difference across them, bridging the gap between source and capacitor voltages, but limiting the current that will flow according to Ohm's law. If those resistances are small, with a large potential difference across them current will be huge.

So, yes, there will initially be a potential gradient along the wires, and along resistances inside the voltage source itself, which is what is meant by an "electric field", and that gradient is what defines the forces on charges that find themselves there, and the potential energy they possess. That's what propels the charges, always to a place of lower potential energy.

Explicitly placing a resistor in the loop is a measure to control initial current, and the greater that resistance, the greater its initial share of the potential difference, meaning that the source and wiring resistances share a much smaller fraction of the overall potential discrepancy.

Eventually though, capacitor voltage will have risen enough to equal the voltage source, and there will be a state of equilibrium, in which there's no current because no conductive element in the system has any potential difference across it; all charges at all points along any conductive part of the loop have the same potential. The two wires each have the same electrical potential along their length, so there's uniform electron distribution (thermal and quantum fluctuations notwithstanding) along them. The charge imbalance is constrained to the vicinity of the capacitor plates.

All this is a macroscopic understanding of the system, in classical terms. Depending on how deep you want to go down this rabbit-hole, what really happens is less prosaic. Here's another answer I wrote addressing these mechanisms in terms of potential waves and, to some extent, quantum principles.


Your diagram is more or less what happens. After connecting the battery, you end up with a net positive charge on the top (red) plate and a net negative charge on the bottom (blue) plate. There's an electric field between the plates, and the potential difference across that electric field is equal to the battery voltage.

There is also an electric field between the wires, which we can represent as a "parasitic" or "stray" capacitance between the wires. But this capacitance is small (picofarads), so at low frequencies we can ignore it.

The brief current when the battery is connected comes from the battery pushing charge onto one plate and pulling it off of the other. No current flows between the plates. But from the outside, we see charge moving into one side of the capacitor and moving out of the other side. So in circuit theory we pretend that current is flowing "through" the capacitor.

  • \$\begingroup\$ I actually wouldn't "ignore" the wires. It might be key to understand that every conductor has some capacitance that must be "filled" with charge while the same amount of charge is taken from the conductor on the opposite side of the battery. The capacitor is merely a special conductor arrangement (large and rather close). \$\endgroup\$
    – tobalt
    Commented Jan 24 at 5:16

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