5.3 Procedure for calculating capacitance from Office of Academic Technologies on Vimeo.

**5.03 Procedure for calculating capacitance**

In order to calculate the capacitance of a capacitor, we will follow a certain procedure. Let’s say calculating the capacitance. As a first step, we will assume that our capacitor is charged to a certain value. In other words, we take our capacitor, which consists of two parallel conducting plates separated by an insulating medium, and we connect them to terminals of the power supply, let’s say, to a battery such that its positive terminal is here and the negative terminal is here. In doing so, we store some certain amount of charge to the capacitor so we end up with some certain amount of positive charge stored on the upper plate and a negative amount of charge stored on the lower plate.

The first step is we will assume that a charge q is on the plates. The second step, we will calculate the electric field between the plates in terms of the charge stored on the capacitor by using Gauss’s law. As you recall, it was expressed as the closed surface integral of E dot dA over some hypothetical Gaussian surface that we choose is equal to net charge enclosed in the volume surrounded by this closed surface s divided by Epsilon zero.

Once we obtain the electric field between the plates, as our third step, we will calculate the potential difference between the plates directly from the definition of potential difference that we obtained earlier, which was V sub f minus V sub i was equal to minus path integral of E dot dl from an initial point i to a final point f.

In this case, let’s draw back our plates one more time that they’re charged to some certain q value. From step two, we will know the electric field between the plates by applying Gauss’s law. If the upper plate is charge positively and the lower plate is charged negatively, to some plus q and minus q charges, respectively, and the electric field is therefore filling the region originating from positive plate and entering into the negative plate.

Let’s say we are going to do this integral along a straight line of path. That is the easiest path pointing from positive to negative plate, and therefore the dl is an incremental displacement vector along this path. And let’s call the potential associated with the positive plate as V plus and the potential associated with the negative plate as V minus. Therefore along this path, the potential difference between these two plates, our final potential will be V minus and the initial potential will be V plus, and we’re going to be integrating from positive to negative plate of E dot dl in order to get this potential difference.

Here, since we’re dealing with the potential difference, the origin is completely irrelevant. We can define the potential associated with the negatively charged plate as zero and call the potential associated with the positively charged plate as V. So in terms of this new notation system, we can say that V minus is equal to zero and V plus is equal to V. Then our equation takes the form of zero minus V is equal to minus E dot dl integrated from positive to negative plate. In this case, these negative signs will cancel on both sides, giving us an expression for the potential difference between the plates of a capacitor is equal to integral from positive to plate to negative plate of E dot dl.

Now, once we determine the electric field from step two by applying Gauss’s law, therefore we can take this integral along a path in order to get the potential difference between the plates, calculating the potential from the electric field. The last step of the procedure is basically calculating the capacitance from its definition. In other words, calculate C from its definition, which is C is equal to q over V. Knowing the charge stored on the plates, as well as the potential difference between the plates, then we are able to determine the capacitance of that specific capacitor.