A parallel plate capacitor has plates of area 1.5 meters squared. The plates are separated by 0.020 millimeters of neoprene rubber that has a dielectric constant of 6.7. What is the capacitance of the capacitor? What charge is stored in the capacitor when a 9.0- volt potential difference is applied across it?
In this question, we see we want to solve for two values: first, the capacitance of the capacitor mentioned and then what charge is stored in the capacitor. Let’s call these values 𝐶 for capacitance and 𝑄 for charge, and we will start off with a sketch of our parallel plates. We’re told these plates are separated by a distance we can call 𝑑, and each of them has an area we will name 𝐴. 𝐴 and 𝑑 of course are both given to us in the problem statement.
We also know that the space separating these two plates is filled with a dielectric material, neoprene rubber. This material has a dielectric constant we’ll refer to as 𝜅, and that value also is given to us. Knowing all this, let’s start off on the first part of our question: what is the capacitance 𝐶 of this capacitor? There’s a mathematical relationship we can recall which connects the capacitance of our capacitor with its geometry as well as its dielectric constant. That relationship says that the capacitance 𝐶, what we want to solve for, is equal to 𝜅 times 𝜖 naught, the permittivity of free space, multiplied by the plate area all divided by the distance separating the plates.
𝜖 naught, the permittivity of free space, is a constant whose value we can look up; it’s 8.85 times 10 to the negative twelfth farads per meter. Since we know all the values on the right-hand side of this expression for capacitance, we’re ready to plug those in. When we do though, we’ll make sure to convert units so that all the units are consistent with one another. For example, when we plug in the distance 𝑑, which is given in units of millimeters, we will change that to units of meters. When we do insert these values converted to consistent units, notice that the units of meters cancel out completely from this expression. we’re left only with units of farads, a good sign.
Calculating this expression, we find the capacitance of this capacitor to be 4.4 microfarads. That’s its capacitance with this dielectric in place. Now we move on to part two. In this part, we want to solve for the charge, we’ve called it 𝑄, which is stored in the capacitor when a 9.0-volt potential difference is set up across it. If we call this potential difference 𝑉, we can add this to our library of information about the scenario. To solve for the charge 𝑄 stored on the capacitor when a potential difference is setup across it, we’ll recall a second relationship defining capacitance. We saw that capacitance is equal to 𝜅 times 𝜖 naught times the area of the plates divided by the distance between them.
And it is also equal, generally, to the charge on the plates divided by the potential difference across them. This is a very helpful relationship because as we remember from part one, we’ve already solved for the capacitance 𝐶. If we rearrange this equation to solve for 𝑄, we see it’s equal to capacitance times potential difference. Capacitance, we recall from part one, is 4.4 times 10 to the negative sixth farads, and the voltage is 9.0 volts. Multiplying these two terms, we find a charge of 4.0 times 10 to the negative fifth coulombs. That is the amount of charge stored in the capacitor when this potential difference is applied across it.