Differences
This shows you the differences between two versions of the page.
Both sides previous revision Previous revision Next revision | Previous revision Next revisionBoth sides next revision | ||
184_notes:cap_charging [2017/10/05 19:34] – dmcpadden | 184_notes:cap_charging [2021/03/11 22:13] – bartonmo | ||
---|---|---|---|
Line 1: | Line 1: | ||
Section 19.1 in Matter and Interactions (4th edition) | Section 19.1 in Matter and Interactions (4th edition) | ||
+ | |||
+ | / | ||
+ | |||
+ | [[184_notes: | ||
+ | |||
===== Charging and Discharging Capacitors ===== | ===== Charging and Discharging Capacitors ===== | ||
- | Over the last two weeks we have been building a fairly robust model of what happens to the charges both on the surface of the wires and those moving through the wire. Now we are going to introduce | + | Over the last two weeks we have been building a fairly robust model of what happens to the charges both on the surface of the wires and those moving through the wire (through resistors). Now we are going to introduce a new circuit element called a capacitor and see what changes about the electron current, the electric field and the surface charges. [[184_notes: |
{{youtube> | {{youtube> | ||
- | === Role of the Resistor === | + | ==== Discharging Capacitors ==== |
- | When charging or discharging a capacitor, there is usually a resistor placed in the circuit (like a lightbulb or some other kind of resistor) because the resistor helps control both the maximum current possible and the time it takes to charge/ | + | [{{ 184_notes: |
- | + | Let's start by considering a set of parallel plates that have a charge of +Q on one plate and charge of -Q on the other plate. In circuits, we call this set up a **capacitor**. Then we connect the plates | |
- | ==== Discharging Capacitors | + | |
- | {{ 184_notes: | + | |
- | Let's start by considering a set of parallel plates that have a charge of +Q on one plate and charge of -Q on the other plate. Then we connect the plates with a conducting wire. As we said before, the excess electrons from the negative plate are going to feel a strong repulsion from the other electrons and will be pushed through the wire toward the positive plate, which also has a strong attraction. If we put a light bulb in the middle of the circuit, what would we expect to happen to the lightbulb in this case? Because a lot of electrons are moving through the wire (or rather there is large electron current), the lightbulb would light up much in the same way that it would if it were attached to a battery. | + | |
When the lightbulb is lit up brightly (only a short time after the wires have been connected), the electric field and surface charges in the wires are also very similar to if there were a battery in the place of the capacitor. The electric field points away from the positive plate and toward the negative plate and is constant along the wire. The electric field in the bulb filament would be larger since it is a resistor and would have surface charge piling up on either side. The surface charges would have a small gradient along the wires from the capacitor plates to the light bulb. | When the lightbulb is lit up brightly (only a short time after the wires have been connected), the electric field and surface charges in the wires are also very similar to if there were a battery in the place of the capacitor. The electric field points away from the positive plate and toward the negative plate and is constant along the wire. The electric field in the bulb filament would be larger since it is a resistor and would have surface charge piling up on either side. The surface charges would have a small gradient along the wires from the capacitor plates to the light bulb. | ||
- | {{184_notes: | + | [{{184_notes: |
+ | [{{ 184_notes: | ||
After a short time has passed, the electron current moves away from the negative plate, meaning the plate has lost some of its negative charge. Since the electron current moves toward the positive plate, that plate becomes less positively charged. This means that not only does the electron current decrease, but that the electric field between the capacitor plates and in the wires also decreases. With a smaller electron current, there is a smaller build up of charges across the lightbulb, making the electric field inside the bulb filament also smaller - so the lightbulb becomes dimmer after some time. | After a short time has passed, the electron current moves away from the negative plate, meaning the plate has lost some of its negative charge. Since the electron current moves toward the positive plate, that plate becomes less positively charged. This means that not only does the electron current decrease, but that the electric field between the capacitor plates and in the wires also decreases. With a smaller electron current, there is a smaller build up of charges across the lightbulb, making the electric field inside the bulb filament also smaller - so the lightbulb becomes dimmer after some time. | ||
- | {{ 184_notes: | ||
As this process continues, the capacitor plates lose more and more of their charge, so the electric field gets smaller and smaller causing the bulb to get dimmer and dimmer. Eventually, the capacitor plates lose all of their charge (both become neutral plates), so the electron current stops completely because there is no longer an electric field around the wire. | As this process continues, the capacitor plates lose more and more of their charge, so the electric field gets smaller and smaller causing the bulb to get dimmer and dimmer. Eventually, the capacitor plates lose all of their charge (both become neutral plates), so the electron current stops completely because there is no longer an electric field around the wire. | ||
Line 23: | Line 25: | ||
==== Charging a Capacitor ==== | ==== Charging a Capacitor ==== | ||
+ | [{{ 184_notes: | ||
We could also consider the reverse situation: what would happen if you started with neutral plates in your capacitor and connected it to a battery and a lightbulb? In this case, rather than discharging the capacitor, you would be charging the capacitor. | We could also consider the reverse situation: what would happen if you started with neutral plates in your capacitor and connected it to a battery and a lightbulb? In this case, rather than discharging the capacitor, you would be charging the capacitor. | ||
- | {{184_notes: | ||
If we think about what happens to the charges immediately after the circuit is connected, the surface charges arrange themselves in the first few nanoseconds and set up an electric field in the wires just like before. Because the capacitor is initially neutral, the capacitor doesn' | If we think about what happens to the charges immediately after the circuit is connected, the surface charges arrange themselves in the first few nanoseconds and set up an electric field in the wires just like before. Because the capacitor is initially neutral, the capacitor doesn' | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
As each electron from the electron current hits the capacitor plate, it will become stuck on the plate - it cannot jump the gap between the two capacitor plates. So how does the current continue and not just get stuck on the plate? When an electron stops on one of the plates, it causes an area of excess negative charge. This will then repel one of the electrons in the opposite plate, leaving a net positive charge on the opposite plate, while the repelled electron continues to move in the wire beyond the capacitor. | As each electron from the electron current hits the capacitor plate, it will become stuck on the plate - it cannot jump the gap between the two capacitor plates. So how does the current continue and not just get stuck on the plate? When an electron stops on one of the plates, it causes an area of excess negative charge. This will then repel one of the electrons in the opposite plate, leaving a net positive charge on the opposite plate, while the repelled electron continues to move in the wire beyond the capacitor. | ||
- | {{184_notes: | ||
After some time has passed, the electrons from the electron current pile up on one plate in the capacitor, leaving a net positive charge on the other plate. When this happens, there is now an electric field from the charges on the capacitor that points // | After some time has passed, the electrons from the electron current pile up on one plate in the capacitor, leaving a net positive charge on the other plate. When this happens, there is now an electric field from the charges on the capacitor that points // | ||
This process is also a **quasi-static state** - the electron current is decreasing over time, while the charge on the capacitor and potential difference across the capacitor is increasing. The total process from uncharged to fully charged takes between fractions of a second and a few minutes depending on the type of capacitor and resistor involved. | This process is also a **quasi-static state** - the electron current is decreasing over time, while the charge on the capacitor and potential difference across the capacitor is increasing. The total process from uncharged to fully charged takes between fractions of a second and a few minutes depending on the type of capacitor and resistor involved. | ||
+ | [{{184_notes: | ||
+ | |||
+ | ==== Role of the Resistor ==== | ||
+ | When charging or discharging a capacitor, there is usually a resistor placed in the circuit (like a lightbulb or some other kind of resistor) because the resistor helps control both the maximum current possible and the time it takes to charge/ | ||
- | === Video Demo === | + | ==== Video Demo ==== |
Here is a video demonstration of a lightbulb and capacitor circuit, charging and discharging. | Here is a video demonstration of a lightbulb and capacitor circuit, charging and discharging. | ||
Line 45: | Line 50: | ||
==== Dielectrics ==== | ==== Dielectrics ==== | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
If you have a fully charged capacitor, one way to add more charge to the plates (making your capacitor more efficient) would be to add a small insulator between the plates of the capacitor. Since it is an insulator, the charges on the plates would polarize the molecules in the insulator but those charges would not be able to move freely. This means that inside the capacitor plates, there would be a strong electric field from the plates pointing in one direction, with a smaller electric field from the dielectric material pointing in the opposite direction. This means that the net electric field between the two plates with the dielectric would be smaller than the field between the two plates by themselves. | If you have a fully charged capacitor, one way to add more charge to the plates (making your capacitor more efficient) would be to add a small insulator between the plates of the capacitor. Since it is an insulator, the charges on the plates would polarize the molecules in the insulator but those charges would not be able to move freely. This means that inside the capacitor plates, there would be a strong electric field from the plates pointing in one direction, with a smaller electric field from the dielectric material pointing in the opposite direction. This means that the net electric field between the two plates with the dielectric would be smaller than the field between the two plates by themselves. | ||
Line 52: | Line 57: | ||
==== Examples ==== | ==== Examples ==== | ||
- | {{[[: | + | [[: |