Hello everyone. In this video series, we are going to review the physics of capacitors and inductors, the voltage-current relationship, and how to combine them in a circuit. In this video, we start with the capacitors, study their physics, and voltage-current relationship.

So, what is a capacitor? A capacitor is a device that we use to store energy in the form of electric charge. So, it is made of two parallel plates separated by some distance D, and the plate surface area we denote it as A. Between these two plates, we have a dielectric. It can be air or any dielectric insulator. The ability of the capacitor to store or save charges is measured in farad, and it is known as the capacitance. So, the capacitance is what defines the capacitor's ability to save a charge.

The capacitance of a capacitor depends on its physical construction, as given by this equation. You can see that the capacitance is proportional to the surface area of the two plates. So, if I have larger plates, this means that I have a higher capacity to save charges. But the capacitance is inversely proportional to the distance between the two plates. Also, the capacitance depends on the medium permeability of the dielectric between the two plates.

So, how does it work? Here, I have my capacitor connected in a circuit. I have a resistor, and I have a battery. Let's assume that at the beginning, the capacitor doesn't have any charge. So, when placed in the circuit, the positive terminal of the battery starts to send positive charges, and the negative terminal will start to release negative charges. Those positive charges will now go and accumulate on the surface of the capacitor. So, on one side, I have a positive charge, and on another side, I have a negative charge. This process repeats, so I start to accumulate charge on both sides of the capacitor.

Okay, but after some time, I have here enough charge—too many volts of charge. The battery is sending another positive charge, but now we know that like charges will repel each other. So, those positive charges on the side of the plate will oppose the flow of this positive charge coming from the battery. Likewise, those negative charges on the other side of the plate will oppose those negative charges coming from the negative terminal of the battery.

This opposition would result in a reduced rate of the flow of the charges in the circuit. And as we know, current is equal to the rate of charge flow over time. So, as a result, after some time, the current flow from the battery to the capacitor will be reduced until we reach saturation, in which there is no current because the capacitor is fully charged with no more electrons or positrons moving through the circuit.

Okay, so as we develop a charge here on the two plates, we actually develop a potential or a voltage difference between the two plates because here I have a positive charge on one plate and a negative charge on the other plate. At steady state, the voltage difference on the capacitor equals the voltage difference from the battery. And as a result, there will be no current flowing into that circuit because now we have too many charges on the plate opposing any other charge coming from the battery.

There is a relationship between the amount of charge on the capacitor and the voltage drop across the capacitor. It is linearly proportional. So, Q, the charge, is equal to C * V. So, you can see that the voltage is proportional to the amount of the charge on that capacitor. You can also compute the capacitance by dividing the total number of charges over the voltage drop across that capacitor.

Now, given this relationship Q = C * V, mathematically, we know that any change in Q would have a change in V. We cannot change C because C is a property of the capacitor itself. Okay, so now if I take the derivative with respect to time, I get dQ/dt equal to dV/dt times C. But what is dQ/dt? As we studied earlier, current is the rate of flow of charge, so dQ/dt is actually the current going through this capacitor.

So, the current in the capacitor is equal to the capacitance times the rate of change of the voltage across that capacitor. If you integrate this equation, you can see that the voltage across the capacitor is equal to 1/C times the integral of the current going through the capacitor. These two equations define what we call the voltage-current relationship across the capacitor.

Now, if you'd like to compute the power in the capacitor, we know that the power is equal to voltage times current. So, if I have the voltage V, the current of the capacitor is equal to C * dV/dt. So, we can have this equation for the power.

Now, I mentioned that by accumulation of the charge, you're actually accumulating energy there. And if you disconnect the battery from the capacitor, the capacitor would still hold the charge. So, what is the energy stored in that capacitor?

We say, okay, if I have the power, I know that the power is the rate of change of the energy. So, the rate of change of the energy would equal to C * V * dV. Integrating that to get the energy would end up with this equation here. So, the total energy stored in a capacitor is equal to 1/2 * C * V², where V here is the voltage across the capacitor.

Let's practice those definitions with a simple example. Here, I have a network of capacitors, and I know that the charge on capacitor 1 and on capacitor 2 equals 20 microcoulombs. And I'd like to determine the energy stored on the three capacitors and also the total energy in all three capacitors.

Okay, I know Q, I know the value of the capacitance, and I need to compute the energy E. I know that the energy is equal to 1/2 * C * V². So, we need to compute the voltage across those capacitors. But the good news is that knowing C and Q, we can obtain V because we know that Q is equal to C * V. C is known, Q is known, so we can solve for V.

For capacitor number one, we know the capacitance is 4.67 microfarads, and the charge is 20 microcoulombs. (This should be "coulombs" here.) Okay, so we can compute the voltage across the first capacitor as Q1 / C1, which is 4.3 volts. Once I have the voltage, we can compute the energy on capacitor 1 as 1/2 * C1 * V1², which gives us 43.2 microjoules.

Now, we can do the same for the second capacitor. We have the capacitance, and we have the charge, so we can compute the voltage of the second capacitor as Q2 / C2, which will give us 5.1 volts. Now, we are ready to compute the energy as 1/2 * C * V².

For the third capacitor, we don't have any information about it, but we still can compute the energy because the third capacitor was in parallel to the second capacitor. And if they are in parallel, this means that they have the same voltage. So, the energy stored in capacitor 3 would be 1/2 * C3 * V3². We know C3, and V3 is equal to V2, so we substitute it here and we can obtain the capacitance.

So, in conclusion, in this video, we learned about the characteristics of a capacitor. The charge in the capacitor is proportional to the voltage across the capacitor, and the proportionality constant C is the capacitance, measured in farads. We also learned about the current-voltage relationship—the current in the capacitor is equal to C * dV/dt, where V is the voltage drop across the capacitor. Likewise, the voltage would equal 1/C times the integral of I dt.

Finally, the energy stored in a capacitor is related to the voltage across the capacitor and is equal to 1/2 * C * V². Okay, thank you so much for listening, and I'll see you in the next video.