# Can someone break down what exactly happens in this situation?

There is a circuit with a battery. Opposite the battery, we have a capacitor. I close the circuit. The current starts at the battery. Current is simply electrons passing through the wire, so the first electron in the line starts to move around until it reaches the 1st plate of the capacitor. What happens when that first electron hits the capacitor? I would say that the plate becomes negatively charged.

Then many electrons hit the plate and it begins to build up more and more negative charge. Now I have 2 questions - when does it stop building charge? And what is happening to the other side (the other plate of the capacitor)?

After whatever happens to the other plate, something must happen to the wire leading back to the negative side of the battery. Do electrons continue to travel along this wire?

• All circuit questions should have a circuit diagram. Otherwise, it's very hard for readers to know what the question is. – DanielSank Oct 22 '16 at 4:46

You can think of a battery as a pump for electrons. It pulls electrons in through one terminal and pushes electrons out through the other. A battery does not create electrons, it only moves them from one place to another.

When you connect a battery to a capacitor, the battery pulls electrons off of one plate (making it positively charged) and pushes them onto the other (making it negatively charged).

Now, electrons repel each other, so as more and more electrons get pushed onto the negatively charged plate, their combined electric field pushes on the electrons still in the wire, opposing the force due to the battery. Eventually, the force from the electrons on the capacitor plate becomes so large that it equals the battery force, stopping the current.

A similar effect happens at the positive plate. As electrons are removed, the plate becomes more positively charged, attracting the electrons in the wire. Eventually, the positive plate pulls so hard on the electrons that the battery can't pull any more off that plate.

You are basically correct as far as you go. The capacitance of the capacitor, how much charge it can store for a given potential difference, is given by:

$${C=q/V}$$

where q is the charge stored on the capacitor and V is the potential difference - in this case the voltage of your battery. So, as you say, electrons build up on the side of the capacitor connected to the -ve terminal of your battery and electrons are pulled off the side of the capacitor connected to the +ve terminal of your battery. A current flows, briefly. Pretty soon the potential difference across the capicitor becomes equal to that provided by the battery and you reach a point where the battery cannot provide enough energy to transfer any more charge and the current stops flowing. Now the capacitor is fully charged and no further changes occur.