**Calculating the Capacitance**

Our task here is to calculate the capacitance of a capacitor once we know its ometry. Because we will consider a number of different geometries. it seems wise to develop a general plan to simplify the work. In brief our plan is as follows: (I) Assume a charge q on the plates; (2) calculate the electric field f between the plates in terms of this charge, using Gauss’ law; (3) knowing f. calculate the potential difference V between the plates from Eq. 25·18; (4) calculate C from Eq. 26·1.

Before we start. we can simplify the calculation of both the electric field and the potential difference by making certain. assumptions. We discuss each in turn.

**Calculating the Electric Field**

To relate the electric field f between the plates of a capacitor to the charge q on0 either plate, we shall use Gauss’ law:

Here q is the charge enclosed by a Gaussian surface, and p f .ciA is the net electric flux through that surface. In all cases that we shall consider, the Gaussian surface will be such that whenever electric flux passes through it, f will have a uniform magnitude £ and the vectors E and etA will be parallel. Equation 26·3 then will reduce to

Q=EA

in which A is the area of that part of the Gaussian surface through which flux passes. For convenience, we shall always draw the Gaussian surface in such a way that it completely encloses the charge on the positive plate; see for an example.

**Calculating the Potential Difference**

In the notation of Chapter 25 (Eq. 25·18), the potential difference between the plates of a capacitor is related to the field E by.

In which the integral is to be evaluated along any path that starts on one plate and ends on the other. We shall always choose a path that follows an electric field line, from the negative plate to the positive plate. For this path, the vectors f and d’S will have opposite directions, so the dot product E” d’S will be equal to – £ ds. Thus,

the right side of Eq. 26-5 will then be positive. Letting V represent the difference Vf – Vi’ we can then recast Eq, 26-5 as.in which the – and + remind us that our path of integration starts on the negative plate and ends on the positive plate. We are now ready to apply Eqs. 26-4 and 26-6 to some particular cases.

**A Parallel-Plate Capacitor**

We assume, as Fig. 26-5 suggests, that the plates of our parallel-plate capacitor are so large and so close together the.twe can neglect the fringing of the electric field at the edges of the plates, taking E to be .constant throughout the region between the plates. . We draw a Gaussian surface that encloses just the charge q on the positive plate.as in 26-. From Eq. 26-4 we can then write.

Thus, the capacitance does indeed depend only on geometrical factors-namely. the plate area A and the plate separation d. Note that C increases as we increase the plate area A or decrease the separation d.As an aside, we point out that Eq. 26-9 suggests one of our reasons for writing the electrostatic constant in Coulomb’s law in the form 1/41TEo. If we had not done so. Eq. 26-9-which is used more often in engineering practice than Coulomb’s law-would have been less simple in form. We note further that Eq. 26-9 permits us to express the permittivity constant EO in a unit more appropriate for use in problems involving capacitors: namely, EO = 8.85 X IO-I~ Film = 8.85 pF/m. (26-10) We have previously expressed this constant as EO = 8.85 X 10-12 c2/N . m2.