# Question Video: Capacitance of a Parallel Plate Capacitor

An 8.0 pF vacuum capacitor has a plate area of 0.070 m². What is the separation between its plates?

03:04

### Video Transcript

An 8.0 picofarad vacuum capacitor has a plate area of 0.070 meter squared. What is the separation between its plates?

As we work towards solving for the separation, let’s first consider what a vacuum capacitor is. We knew that, in general, a capacitor consists of two conducting plates that face one another. When the capacitor is connected to a circuit and charge begins to flow, if this flowing charge is conventional current, then positive charge will begin to accumulate on the left plate here. And negative charge will accumulate on the right plate.

As we’ve drawn this capacitor, there’s nothing in between the two capacitor plates. They’re separated by air. Air, after all, is a good insulator. And that keeps the positive charges and negative charges separated on their opposite sides. That is, up to a point. As current continues to flow, charge continues to build up and build up on these plates. And since opposite charges attract one another, the attractive force between these positive and negative charges, as they accumulate, grows.

With enough charge on these plates, the charges will jump the gap. That’s what’s known as dielectric breakdown. The dielectric, the air, is no longer able to separate the charges. But this is where a vacuum capacitor comes in. In a vacuum capacitor, the whole gap between these two plates is enclosed in a sealed chamber. This chamber is then evacuated, brought to very low pressure, so that effectively, it’s a vacuum. A vacuum, it turns out, is an even better insulator than air. And it does an even better job of keeping these charges separated even as they continue to build up on the capacitor plates.

That’s just a bit of background on what kind of capacitor we’re working with. Now, let’s consider just how far apart these plates must be. And we can recall the mathematical relationship: capacitance, 𝐶, is equal to the permittivity of free space, 𝜖 naught, times the area of a capacitor plate all divided by the distance separating the two parallel plates.

In our scenario, we’ve been given the capacitance 𝐶 is 8.0 picofarads as well as the area in square meters of each of our plates. And we know that 𝜖 naught, the constant, is 8.85 times 10 to the negative 12th farads per meter. This tells us that we have everything we need in order to solve for the distance 𝑑 between this two parallel plates. To find it, we’ll first rearrange to isolate 𝑑 on one side of this equation. 𝑑 equals 𝜖 naught times 𝐴 divided by 𝐶. And as we plug in our values, we make sure to write our capacitance as 8.0 times 10 to the negative 12th farads. That’s what 8.0 picofarads is.

When we look at the units in this expression we’re about to calculate, notice that one factor of meters cancels out as well as the factor of farads. We’re gonna be left with an answer in units of meters. If we calculate this fraction and then convert to units of millimeters, we find an answer of 77. That’s the separation distance between the parallel plates of this capacitor.