# Video: Analyzing the Effect of Changing the Plate Separation of a Parallel Plate Capacitor

An air-filled (empty) parallel-plate capacitor is made from two square plates that are 25 cm on each side and 1.0 mm apart. The capacitor is connected to a 50-V battery and fully charged. It is then disconnected from the battery and its plates are pulled apart to a separation of 2.0 mm. What is the capacitance of this new capacitor? What is the charge on each plate? What is the electrical field between the plates?

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### Video Transcript

An air-filled empty parallel-plate capacitor is made from two square plates that are 25 centimeters on each side and 1.0 millimeters apart. The capacitor is connected to a 50-volt battery and fully charged. It is then disconnected from the battery and its plates are pulled apart to a separation of 2.0 millimeters. What is the capacitance of this new capacitor? What is the charge on each plate? What is the electrical field between the plates?

Let’s give labels to each of these three quantities being asked for. Let’s call the capacitance of the new capacitor 𝐶 sub two, we’ll call the charge on each plate 𝑄, and we’ll call the electrical field between the plates capital 𝐸.

In this example, we have a capacitor in an initial setup and then the setup is changed. Let’s sketch out what happens. At the start, we have an air-filled or empty capacitor, which is made of two square plates 25 centimeters on a side, and the plates are 1.0 millimeters apart. These plates are then connected to a 50-volt power supply, and this connection is held until the plates are fully charged.

Once the plates are fully charged, they’re disconnected from the battery supply. Then with the charges still on the plates, so their distance apart is now 2.0 millimeters instead of 1.0 millimeters. It’s this capacitor that we’ve called the new capacitor, and we want to solve for its value, 𝐶 sub two.

To do that, we can recall that the capacitance of a capacitor is equal to the permittivity of free space, 𝜖 naught, times the area of one of the capacitor plates divided by the distance separating the two plates. 𝜖 naught is a constant whose value we can look up. And knowing this, we now have all the information needed to solve for 𝐶 sub two, the capacitance of the new capacitor.

We plug in the value for 𝜖 naught. 𝐴 is equal to 25 centimeters all squared, and 𝑑, the distance separating the plates, is 2.0 millimeters. Before we calculate 𝐶 two though, we’ll want all of our distances to be in the same set of units.

Right now, we have an area in square centimeters and a distance in millimeters. Let’s convert those to agree with the length units of 𝜖 naught, which are meters. Looking at the area, 25 centimeters is equal to 0.25 meters. And for the separation distance, 2.0 millimeters is equal to 2.0 times 10 to the negative third meters.

Now that our units all agree, we can notice that all of these units of meters will actually cancel out, and we’ll be left only with units of farads, the units of capacitance. To two significant figures, the capacitance of this new capacitor is 0.28 times 10 to the negative ninth farads, or 0.28 nanofarads. That’s the capacitance of these plates now that they’re 2.0 millimeters apart.

Next, we’re going to solve for the magnitude of the charge, 𝑄, that’s on each one of these opposite parallel plates. Here’s how we’ll go about doing that. It turns out that capacitance is not only equal to 𝜖 naught 𝐴 over 𝑑 but that it’s also equal to the charge on each plate divided by the voltage setup across them. If we combine these two equations for capacitance, we see that 𝑄 over 𝑉 is equal to 𝜖 naught 𝐴 over 𝑑.

Now there’s something very important to realize about this equation. See that it refers to a potential difference, 𝑉, of voltage. That voltage, as we see from our sketch, is no longer being applied to this capacitor. It was only applied when the capacitor separation distance was 1.0 millimeters. This tells us that if we’re going to use 50 volts of potential difference for the voltage in this equation, we can’t use 2.0 millimeters for the separation distance. We have to go back to the original configuration where that separation distance is 1.0 millimeter.

With that caveat in place, we’re ready to continue solving for 𝑄, and we’ll do it by multiplying both sides of this equation by the potential difference 𝑉. There we have it, an equation for 𝑄 in terms of known values. We’ll just need to be careful as we plug in to use the correct ones.

𝜖 naught is a given constant. 𝐴, the area of each plate, is also constant, but we’ll be sure to plug in in units of square meters. The potential difference is 50 volts; that’s due to the battery. And 𝑑 is the original separation distance between the plates, 1.0 millimeters.

Let’s plug all these values in in SI standard units to solve for 𝑄. When we do plug in for all these values, notice that, once again, the units of meters cancel out from numerator and denominator, and we’re left with units of farad-volts, in other words coulombs. To two significant figures, 𝑄 is 28 times 10 to the negative ninth coulombs or 28 nanocoulombs. That’s the charge magnitude on each of our capacitor plates.

Finally, let’s solve for the electric field magnitude that exists between these two plates. Since these plates separate positive and negative charge, we know an electric field will be created between them, pointing from the positive to the negative plate. This field between the plates will be uniform. And to solve for it, we can recall a mathematical relationship between the field and the charge density on the plates.

The electric field between oppositely charged parallel plates is equal to the charge density on the plates divided by 𝜖 naught. This is the same thing as saying it’s equal to the charge on either one of the plates divided by 𝜖 naught times the plate area.

We know the area of these plates, we’ve solved for charge in an earlier step, and 𝜖 naught is a constant value that we’re ready to plug in. When we do insert these values, we’re careful to write our area in units of square meters rather than square centimeters. And when we calculate all this out, we find an electric field magnitude of 50 kilovolts per meter. That means that, for every meter of separation between these plates, the potential changes over that distance by 50000 volts.