# Question Video: Analyzing a Series Combination of Capacitors

A 4.00-pF capacitor is connected in series with an 8.00-pF capacitor. A 400-V potential difference is applied across the capacitors. What is the charge on the 4.00-pF capacitor? What is the charge on the 8.00-pF capacitor? What is the voltage across the 4.00-pF capacitor? What is the voltage across the 8.00-pF capacitor?

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### Video Transcript

A 4.00-picofarad capacitor is connected in series with an 8.00-picofarad capacitor. A 400-volt potential difference is applied across the capacitors. What is the charge on the 4.00-picofarad capacitor? What is the charge on the 8.00-picofarad capacitor? What is the voltage across the 4.00-picofarad capacitor? What is the voltage across the 8.00-picofarad capacitor?

Lots of questions here, but they really boil down to two types: finding the charge and the voltage across these two capacitors. Let’s call the charges 𝑄 sub four and 𝑄 sub eight respectively. And then the voltages we’ll call 𝑉 sub four and 𝑉 sub eight. Since we’re told how these capacitors are connected, they’re connected in series, let’s draw out the circuit consisting of them and the 400-volt supply. Here we have a simple sketch of that circuit: the 4.00-picofarad and 8.00-picofarad capacitors in series and a 400-volt supply. Now onto our first question: what’s the charge that sits on either side of the 4.00-picofarad capacitor in this circuit?

One way we can start figuring this out is by recalling a relationship between charge capacitance and potential difference. In general, the capacitance of a capacitor is equal to the charge on its plates divided by the potential across them. If we rearrange this relationship to suit our scenario, we can say that we do know the capacitance of this capacitor, that’s given, but we don’t know the potential difference across it. We know it will be somewhere between zero and 400 volts. But what it is we don’t quite know yet. It seems then that we might not be able to solve for the charge across this capacitor. Well if we can’t solve for the charge across one, what if we combine the two capacitors in our circuit by adding them in series? That way we would have a total capacitance and we also have a total potential difference, so that would give us a total charge across that total capacitor plate.

Well it’s an idea! Let’s try it out and see where it leads. To calculate 𝑄 sub 𝑇, the total charge across the total capacitance of the circuit, we’ll need to know that total capacitance. We’ve called it 𝐶 sub 𝑇. That will involve adding together the capacitances of these two capacitors arranged in series. Since they are arranged in series, we’ll have to recall the special rule for adding together capacitances arranged this way. In particular, if we have two capacitors set up in series, then the total capacitance of those two 𝐶 sub 𝑇 is equal to their product divided by their sum. Let’s write that in then for 𝐶 sub 𝑇, the total capacitance of our circuit. As we look at the expression then, where 𝐶 sub four and 𝐶 sub eight represent the 4.00- and 8.00-picofarad capacitors respectively, we see that everything on the right-hand side of this expression is known.

Recall that 𝑉 sub 𝑇, the total potential difference in the circuit, is 400 volts. This means then that we can calculate 𝑄 sub 𝑇, the total charge across the total capacitance of the circuit. Speaking of the circuit, let’s take another look at it. When we think about the circuit turning on or current starting to flow, we know that that current will be the same all throughout this entire series loop. Since the current is the same everywhere at a given point and time, that means that the charge that builds up on each of our two capacitors will also be the same. After all, it’s supplied by the current. Even though these two capacitors have different values, the charge that builds up on each one, because they’re arranged this way in this kind of circuit, will be the same. And in fact, that charge will match up with the total charge that would accumulate on an equivalent capacitor in this circuit. In other words, the charge 𝑄 four, the charge across the 4.00-picofarad capacitor, is equal to 𝑄 sub 𝑇. And not only that, but so is 𝑄 sub eight, the charge across the other capacitors plates. And again, the reason for this is that the same current at any given instant flows throughout the circuit and, therefore, charge builds up equally on the plates.

That’s good news then. If we solve for 𝑄 sub 𝑇, which we know how to do since everything in the right-hand side is known, we’ll be solving simultaneously for 𝑄 sub four and 𝑄 sub eight. So let’s go ahead and do that now. We insert the values for our four and 8.00 times 10 to the negative 12 farads. And we sum them in our denominator to give 12 times 10 to the negative 12 farads and multiply all this by the potential difference 400 volts. This calculates out to a value of 1.07 nanocoulombs. That’s the charge not only across the 4.00-picofarad capacitor, but also that across the 8.00-picofarad capacitor.

Now that we know that, let’s move on to the next part. where we want to solve for the voltages, the potential differences across these two capacitors. And we see that our relationship from earlier, 𝐶 equals 𝑄 over 𝑉, will be helpful to us at this stage. For example, we can say that 𝑉 sub four, the potential difference across the smaller capacitor, is equal to the charge on that capacitor divided by its value; that is, its capacitance. We can use our result from earlier for the charge on this capacitor. And we know its capacitance is 4.00 times 10 to the negative 12 farads. When we compute this fraction, we find a result to three significant figures of 267 volts. That’s the voltage drop or potential difference across the 4.00-picofarad capacitor. Finally, we want to do the same thing with the 8.00-picofarad capacitor: find the voltage across it. We could use the same mathematical relationship as we did to solve for 𝑉 sub four or, alternatively, we could also subtract 𝑉 sub four from the total potential difference across the circuit, which we know to be 400 volts. Either way will work. We can try this second suggested method: 400 volts minus 267 volts is 133 volts. That’s the voltage across the larger of the two capacitors.