How does charge build up and dissipate in a capacitor when it is hooked to a...

How does charge build up and dissipate in a capacitor when it is hooked to a battery and then unplugged?

Expert Solution

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Initially, the capacitor can be thought of as neutral. Meaning, it has no net charge. A mix of positive and negative charges on both plates (well call them upper and lower plates to identify which is which).

When you connect a battery, you have current flow. It doesn't matter if the circuit is open or close (from the perspective of the source, it does not know what is 1nm, 1mm, or 1m ahead of it). So you have these positive charges that start to loiter around the upper plate. At first, first, its 1 charge, 2 charge, 10 charges and it keeps doing that until the entire surface area of the plate is the same charge. At this point, there is no current flow on the capacitor because there is no longer a change in the electric field.

But as the feild is changing, (such as when the charges are accumulating), the positive charges on the upper plate, attract negative charges on the lower plate. But in order of a negative charge to accumulate on the lower plate, a positive charge on the lower plate needs to be....displaced...from the lower plate. Essentially being kicked out from that spot.The plates have finite space, so there is only enough room for so many charges. This repeats until every positive charge has attracted a negative charge.

This is how current flows in a capacitor. Through what is called, displacement current.

This is why current in an uncharged capacitor is instantaneous, and exponentially tapers off, as the charge builds up, there is less and less negative charges needed and so less and less positive charges get displaced.

Suppose that with the capacitor fully charged, the switch is now closed this time consisting of a resistor and capacitor. Electrons will now flow around the circuit via the resistor as the charge on capacitor acts as the source of current

The charge on the capacitor will be depleted as the current flows. The rate at which the capacitor voltage reduces towards zero will depend on the amount of current flowing..

genius_generous answered 1 year ago

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