Capacitance and voltage drops

Let's say that Switch $S$ is closed. I have a conceptual question:

1. If a stream of particles acquire a potential difference upon exiting the battery, why will they lose some of that potential across the capacitor?

2. How does voltage get reduced in a capacitor?

Voltage drops occur across resistors because (I guess this is an axiom), electrons flowing across resistors encounter resistance (and thus force).

1. What understanding frames the voltage drop across capacitors?

There are two aspects here, the steady state aspect (what happens long after the switch closes or long after the switch opens) and the transient aspect, what happens just when the switch closes.

In your circuit, you would measure the full battery potential across the capacitor in the steady state with the switch closed. No current flows, and hence there is no U=I*R drop across the resistor.

Also, since no current flows, it is not particularly intuitive to talk about "voltage drop" across the capacitor. There will be a potential, but you can simply remove the capacitor completely and you will have the same potential (if you measure at the nodes where the capacitor was once attached).

For the transient state exactly when the switch closes, you will have a brief surge of charge flowing "through" the capacitor (as one side gains electrons and the other loses some).

If you reformat your circuit a bit, you will see that it is an "RC" filter where the capacitor is charged through the resistor as the switch closes. During that brief time, current flows through the resistor and into the capacitor until it's charged fully. The bigger the resistor is, and the bigger the capacitance is, the longer this takes, but eventually it reaches the steady state.

You can see it like the capacitor delays the equalization of potential in your circuit, but when it's in equilibrium, it doesn't do anything.

Just comment if something's still unclear..

• "In your circuit, you would measure the full battery potential across the capacitor in all steady states." This isn't really right. With the switch open, the model doesn't define what you'd measure across the capacitor. You'd need to know the leakage conductance through the capacitor and through the switch (and through the measuring device) to know what the circuit would do in real life. – The Photon Jan 26 '15 at 18:46
• @BjornW "For the transient state exactly when the switch closes, you will have a brief surge of charge flowing "through" the capacitor." So, during transient states, voltage drops across the capacitor because of "surge." Ah yes, your explanation about voltage "drops" during the steady state makes sense. – Muno Jan 27 '15 at 0:04
• @ThePhoton you are of course correct, I wrote too fast :) – BjornW Jan 27 '15 at 0:10
• @user21945 the cap is a dynamic component and its difficult to find good words for it, but I would say that with the cap empty and the switch just closing, I guess you can imagine the cap starting as a resistor with 0 resistance, and as the cap fills up, it approaches infinite resistance and the current stops. – BjornW Jan 27 '15 at 0:13

Voltage differences occur across resistors because electrons flowing across resistors are pushed by a force. The force is drag force type of force exerted by the ion lattice.

Voltage differences occur across capacitors because electrons flowing across the insulator between the capacitor plates are pushed by a force. The force is drag force kind of force exerted by the molecules of the insulator.

It seems to be exactly the same thing with capacitors and resistors.

There's a just a little complication: Total drag force exerted to the electrons by the insulator depends on how many slowly moving electrons there are inside the insulator.

If the number of moving electrons in the insulator is small, and the number of moving electrons in the conductor is large, the electrons in the conductor near the insulator will feel a force that tends to prevent them from going into the insulator.

That force is the force that prevents free electrons leaving metal objects. It works like this: Metal ion lattice pulls quite hard on any free electron, usually it pulls equally into all directions. But at the boundary of the ion lattice there is a net pull, which tends to keep the "free" electrons in the ion lattice.