Energy stored in a capacitor question

[magu1re]

Hey.

I am having major trouble understanding why the energy under the V against Q graph formed when a capacitor is charging up is equal to the energy stored in the capacitor.

I think that I do not understand this because I do not truly appreciate what the existence of a potential difference across the capacitor means. What does it mean? (Because electrons are unable to flow across this electric field due to the presence of the dielectric).

The problem I have is that the electrons do not flow through the capacitor like they do in most components and so I cannot explain the fact that W = Integral of V with respect to q using the, perhaps basic, definition "V is energy lost per unit charge as it flows through the component."

What does it mean to say there is a difference in electrical potential between two points?

Please help me.

 

[Drakkith]

Think of how a capacitor is built. You have 2 plates separated by a dielectric. When charged, you have an excess of electrons on one plate and a lack of on the other. The dielectric keeps these electrons from flowing to the other plate. However, let's look at what happens once you complete the circuit.

Lets just imagine a single wire from one terminal to the other. Normally, the electrons on one plate don't have anywhere to go and are attracted to the positive plate. Because of the dielectric, they can't go anywhere. On the other plate you have a lack of electrons, but because of the dielectric and because there is nothing connected to the terminal the positive plate has no way to gain any electrons and keeps its positive charge.

Now, once you connect the wire, what happens? Suddenly you have a material connected to the positive plate that has electrons in it. Because of the lack of electrons on the plate the wire experiences a positive force pulling its electrons toward the plate. The negative plate has the reverse of this since it has excess electrons. They push the other electrons with a certain amount of force. Because the wire is connected to the positive plate, there is somewhere for the electrons in the wire to go to when they experience a force. So the electrons in the negative plate flow out into the wire, and the electrons in the wire flow into the positive plate. The greater the difference in charge between the 2 plates, the more force you have. IE Voltage.

 

[Curl]

I think that I do not understand this because I do not truly appreciate what the existence of a potential difference across the capacitor means. What does it mean? (Because electrons are unable to flow across this electric field due to the presence of the dielectric).

It turns out that it doesn't matter if the electrons travel straight through from one plate to another or if they go around the dielectric via a wire. The end result is the same so the path doesn't actually matter.

This is explained by the concept of conservative fields, so if you studied that type of Math, that's where it comes from.

 

[Penn.6-5000]

Let's say you charge a capacitor up to five volts. A volt is one joule per coulomb. Now, imagine that a hypothetical electron were to manage to jump across the capacitor, from the negative plate up to the positive plate. It feels the repulsion of the negative plate and the attraction of the positive plate, so as it's flying it gets faster, that is, it picks up kinetic energy. It hits the other side with more energy than it started. How much more energy? The electron was 1.6*10^-19 columbs, and it fell across five volts, so the energy gained is qV = 8*10^-19 Joules.

Now let's talk about something completely unrelated. Instead of talking about the electron that *did* jump across the gap, let's talk about the electron that didn't jump. I have my capacitor here and I try to force some current through it. So an electron comes through the wire and up into the negative plate. For whatever reason, it doesn't jump the gap. It stays on the plate. But it does "repel" an electron on the opposite-facing plate and that repulsed electron then goes out the other wire. Viewed from the outside, an electron went into one end of the capacitor and came out the other side. But we know that what *really* happened was that one electron went in and a different electron came out. So now the negative plate has a surplus negative charge on it and the positive plate has a surplus positive charge. Where there are charges there's an electric field, and we can measure the strength of the electric field by a voltage. So what's the voltage you get across a capacitor when an electron passes through it? It turns out to have a lot to do with the design of the capacitor. If you put a sheet of dielectric material inside the capacitor, for example, the resulting voltage is lower than if the dielectric weren't present. If you can't afford a dielectric, moving the plates further apart will have the same effect. So the relationship between the charge accumulated on the plates and the resulting voltage between the plates varies from capacitor to capacitor... But those two values are always proportional, so we call the constant of proportionality the "capacitance" of that particular capacitor. Then if we have a 1 farad capacitor (1 farad = 1 Coulomb per volt) and a 1.6*10^-19 coulomb electron, we expect the voltage across the capacitor to now be 1.6*10^-19 Volts.

Notice that we solved two entirely different problems:
Problem A: When electrons *don't* jump across the capacitor, a current flows but a voltage builds up. How much voltage? (Answer: voltage = charge / capacitance).
Problem B: When electrons *do* jump across the capacitor, they get a lot of energy out of it. How much energy? (Answer: Energy = charge * voltage).
Recognizing this distinction should help keep you out of several types of trouble.

To answer your actual question: "What does it mean to say there is a difference in electrical potential between two points?" An electric potential difference is measured in volts, so I'm going to refer to this concept as a voltage between two points. The voltage between two points tells you how much energy a charged particle can get by moving from one point to the other. It doesn't matter whether a particle actually *does* undergo that motion; it's sufficient to imagine a hypothetical particle. The ultimate destination of that energy depends on the type of object:
1. When a 1 Coulomb particle moves across the terminals of a 5-Volt battery, 5 Joules of chemical energy inside the battery are converted into 5 joules of electric potential energy in the particle.
2. When a 1C particle moves across the terminals of a 120-Volt light bulb, 120 Joules of electric potential energy in the particle are transformed into light (and a lot of heat).
3. When a 1C particle moves through a 15-Volt motor, 15 Joules of electric potential energy in the particle are transformed into motion (and a little bit of noise, waste heat, etc.)
4. When a 1C particle jumps across a 30-Volt gap between the plates of a capacitor, 30 joules of electric potential energy are transformed into the particle's kinetic energy... but when it hits the other plate a lot of that kinetic energy is going to turn into heat.

So when a charged particle moves across a voltage, its electric potential energy either increases or decreases... Electric potential energy increases in gadgets that "generate" electricity and decreases in gadgets that consume electricity. When the electric potential energy increases, that energy had to be taken from somewhere: the chemical energy in a battery, the rotational motion of a turbine generator, light hitting a solar cell, etc. When the electric potential energy decreases, that energy goes somewhere: into a light bulb, a motor, a microwave oven, or into the kinetic energy of the particle itself.

 

[magu1re]

It turns out that it doesn't matter if the electrons travel straight through from one plate to another or if they go around the dielectric via a wire. The end result is the same so the path doesn't actually matter.

This is explained by the concept of conservative fields, so if you studied that type of Math, that's where it comes from.

Okay. Well if that's the case then I could explain the area under the graph being equal to the energy stored if it was an electron from the negative plate moving to the positive plate when the capacitor discharges but in reality this isn't the case, is it?

So when sites talk of the energy transferred in moving a charge from one point to another do they mean the same charge (i.e. the same particle)? Because if not, then I could argue that removing a tiny amount of charge from the negative plate to the positive plate requires qV J of energy. Then I could repeat this process until the voltage across the plates was 0V and this would mean all of the energy within the capacitor would have been transferred and this would be the integral in question.

 

[Zacku]

Okay. Well if that's the case then I could explain the area under the graph being equal to the energy stored if it was an electron from the negative plate moving to the positive plate when the capacitor discharges but in reality this isn't the case, is it?

Hi,

The area under the graph is something like
\int_0^Q V(q) dq Now if you decompose V(q) in something like U_1 - U_2 and express the integral as a virtual charging procedure in N steps. You will get that it is roughly the sum

\sum_{i=1}^N qU_1(i)-qU_2(i) which is basically the total work that has to perfom an external operator (a generator in general) to move all the charge bunches of charge q=Q/N from plate 1 to plate 2.

 

[whocouldshebe]

Good question, I was wondering the same thing, why sparks won't jump in both directions at the same time? At first I just explained it away as negative is no charge and + is charge, but then both negative and positive lightning bolts wouldn't exist. Positive lightning is typically 10 times stronger than negative lightning so I thought maybe the negative transference is there just we can't see it as well as the positive jump in electrons because it's visible frequency of light is outside of the visible spectrum, it is too small, and/or the negative jump is lost too easily to heat?

I have a question though. I want to charge up my Leyden jars on low voltage rather than pouring tons of little sparks into a jar. I have 2 antennas on my roof that put out between 300mV and 500mV each according to my multimeter.

A spark gap transmitter has a tuning coil between the Leyden jars and the antenna, I don't know why that is there, but if I place a high voltage single direction diode (say 15kV) to let the charge of 600mV to 1V into my capacitors but not let it return back out to my antenna, would that alone let the charge build in my capacitor up to 15kV? I don't need anything between the Earth ground and my capacitor ground do I to force a charge to build in my capacitor?

I also noticed that my 3 liter bottle capacitor shows 25 Megaohms with water only and 15 Megaohms with a 50/50 water/salt-vinegar solution? How does that play into things?

I also have a hydrogen peroxide bottle and aluminum can capacitor. This evening I will make the dual layer wax paper aluminum foil rolled capacitor. Another question on that too...

Most say leave wax protruding out on all sides, others say leave the wax in the middle and lay one foil on left the the other on the right, but don't let them touch. What difference does it make as long as the 2 different layers of foil don't touch? It seams like it would be easier to lay the the wax paper in the middle of the 2 aluminum sheets, then you would have aluminum foil terminals, positive and negative protruding well out both ends of the roll at the end that can be capped off with a round plate of aluminum? Is the spark more likely to jump to the foil with the less used design of wax paper in the middle?

(Eventually I want to drop the entire roll inside a 3 liter bottle filled with oil or topped off with wax, the 2 big pie plate terminals would make better connection points, with a hole in the bottom and one on top for the wire, where positive and negative won't matter, that's why I'm asking now before I waste a roll of paper and foil.)

 

[rcgldr]

Perhaps a simplfied answer. Voltage is potential per unit charge. The area under the V versus Q curve is voltage x charge, which is potential energy, and the work done is the change in potential energy.

 

[whocouldshebe]

Perhaps a simplfied answer. Voltage is potential per unit charge. The area under the V versus Q curve is voltage x charge, which is potential energy, and the work done is the change in potential energy.

So basically the charge would jump the gap in both directions if the charge for the negative pole were an exact inverse of the positive pole. Most multimeter make it look as if this is true, but the charge given by the meter is in fact only relational to that particular ground pole. A different ground would show a different reading and likewise have a different bolt strength using the same positive terminal, the same rule should also go for negative terminals linked to a different positive terminal, and why I can get different volts by moving my negative terminal ground around the yard.

But I'm still not sure how to make my capacitor charge with a trickle charge...?