5.5 Cylindrical Capacitor from Office of Academic Technologies on Vimeo.

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**5.05 Cylindrical Capacitor**

Now we will calculate the capacitance of a cylindrical capacitor. As the name implies, now we’re dealing with a capacitor, which consists of two concentric conducting cylindrical surfaces, let’s say these are, this is the larger surface, or outside surface, and the smaller concentric inner surface. All right. Therefore, our cylindrical capacitor is something like this. And let’s give some dimensions to our capacitor also.

Let’s assume that the length of the capacitor is h and the inner radius is a, other radius is b. And we charge our capacitor such that we connect the inner surface to the positive terminal of a power supply, and the outer surface to the negative terminal of the power supply of our battery, let’s say. In doing so, therefore, the inner plate will be charged positively everywhere, and the outer plate is going to be charged negatively. And let’s say the magnitude of this charge is plus q, and the inner plate, minus q, is along the outer plate.

Therefore, by charging the capacitor, we have already done the first step of calculating the capacitance, and we say that the capacitor is charged to some q Coulombs. And as a second step, we will calculate the electric field between the plates of this capacitor by applying Gauss’ law, which was integral of E dot dA over a closed surface is equal to net charge inside of the volume surrounded by this closed surface, q enclosed over Epsilon 0.

Again, the insulating medium between these two conducting plates is air, so this is a conductor, and this is a conductor, like aluminum plates, for example. All right. So now we are interested with the electric field between the plates. If we look at the geometry of the plates, we see that the electric field is going to originate from the inner positively charged plate, and it will enter into the negatively charged outer plate, in this case, in radially outward direction. Therefore, the electric field is going to be filling the space from positive plate to negative plate, in radially outward direction.

The whole region between the plates will be filled by the electric field generated from these two charged plates. Well, since we are dealing with a cylindrical geometry, we’re going to choose our Gaussian surface in the form of a cylinder, such that its side surface is passing through the point of interest. Our point of interest, in this case, is going to be any point between the plates. Let’s choose that point somehow away from the ends of this cylinder, somewhere around the middle at this point, p. This is the region between the plates, and we choose our Gaussian cylinder, Gaussian surface in the form of a cylinder, such that its side surface is passing through the point of interest. Therefore, it is going to be a surface something like this, a cylindrical surface.

Now as we apply this Gauss’s law, in order to calculate the electric field at this point between the plates, we’re going to make an assumption that we’re far away from the end points, and we’re going to neglect the end effects. Neglect end effects means that, of course, when we go to this upper boundary, the electric field over here is not going to be radially outward direction because we will have electric field pointing in upward direction and the net field will be the vector sum of those two. But we will neglect those end effects. Otherwise, we cannot apply the Gauss’s law, because this cylindrical surface in that case will not satisfy the conditions in order to apply Gauss’s law. So we have to make this assumption, and we will say that neglect end effects, and assume that the electric field is between these two plates in radially outward direction.

And so, considering this, then again, we can divide this closed surface integral into the sum of the open surfaces, which gradually makes this whole cylinder. And as you recall from the cylindrical geometry, when we apply the Gauss’s law to the infinite straight route problem, since the open form of this cylinder consists of a rectangular sized surface, and a circular upper and lower surfaces which this rectangular surface wraps around, we can say that this integral can be expressed as integral over the top surface of the cylinder, which is this one. And then the, that is E magnitude dA magnitude.

And if we look at the angle between E and dA vector for the top surface, and where the electric field is in this region right here, since for the top surface dA is perpendicular to the surface like this, and the electric field is radially outward, the angle between them for the top surface is 90 degrees.

So we’ll have cosine of 90 here. And since cosine of 90 is zero, this integral will not make any contribution. And plus same for the bottom surface; integral over the bottom surface, again, dA is perpendicular to the surface, pointing in downward direction, and electric field is radially out, filling this whole space. So for that part, it’s going to be pointing like this, and again the angle between them will be 90 degrees. So that, too, will not contribute. We will have E magnitude dA magnitude times cosine of 90, which is going to give us zero.

And the only surface left is the side surface integral, for the side surface, if we look at our point of interest, electric field is radially out here, and the dA is again perpendicular to the surface. Therefore, for that point, and for the whole side surface, the angle between electric field vector and the incremental surface area vector will be just zero degree. So for the side surface integral, we will have EdA times cosine of 0.

When we add all these open surface integrals, then it will be an integral over the whole closed surface of the cylinder. And on the right-hand side we will have q enclosed over Epsilon 0. Cosine of 0 is, again, 1. And as long as we are on the side surface of this Gaussian cylinder, we’re going to be same distance away from the charge that it is enclosing, which is the inner cylinder here. So the electric field will be constant over the side surface. Then we can take it outside of the integral.

Well, moving on, we will have E times integral over the side surface of the cylinder, of dA is equal to q enclosed over Epsilon 0. Well, if we look at our open surface form of this Gaussian cylinder here, it has the radius of r; that is the location of the point to, relative to the center. And so this side therefore will be equal to the circumference of either the top circle or the bottom circle, which is then will be equal to 2 Pi r, since it wraps around those circles. And we gave the dimension of h for the height or the length of the cylinder. So integral of dA over the side surface, which will give us the side surface area, and that will be equal to e times the side surface area, which is 2 Pi r h. And on the right-hand side we have q enclosed over Epsilon 0.

Again, q encloses the net charge inside of the region surrounded by Gaussian surface, in this case this purple cylinder. When we look at the inside of that region, we see that it wraps around the whole inner surface, and therefore it encloses whatever the charge distributed along that inner surface of this capacitor, and that is equal to the total charge along that surface, which is q.

So, solving for the electric field, the magnitude of the electric field becomes equal to q over 2 Pi Epsilon 0 h times r. When we look at this expression as different from the parallel plate capacitor, we see that it is not a constant quantity; it varies with 1 over r. In other words, as we go from inner surface to the outer surface, from inner surface to the outer surface, the electric field strength decreases with 1 over r, with 1 over the distance relative to the axis of the cylinder.

Now, once we determine the magnitude of the electric field, of course its direction is radially outward from positive plate to negative plate, now we can move to the third step, which is calculating the potential difference between the plates by taking the integral, line integral, from positive to negative plate of e dot dl. Now again, this is a path integral, and we will choose the simplest path relative to our electric field vector. And in this case, the simplest path is basically a path, which coincides with the electric field vector.

Well, that will be a radial path, radially outward, and so there I’m going to choose that path, like this, and it will coincide with the electric field vector. And doing that, the angle between these two vectors becomes zero. And one more thing over here by looking at the geometry, this is a radial distance; electric field is radial outward, and we’re choosing this path also in radial direction. Then the incremental displacement vector along this path will be dr, so we will replace dl with dr.

Okay. Choose the path in radially outward direction, then dl becomes equal to dr. Now let’s calculate the potential difference between the plates of the cylindrical capacitor. V becomes equal to integral of E magnitude, which is q over 2 Pi Epsilon 0 hr, that we determined from step, or part, two. q over 2 Pi Epsilon 0 h times r. And instead of dl, we will substitute, or use, dr, incremental displacement vector in radially direction. So the r magnitude, since both E and dr are in the same direction, therefore the angle between them is just 0, cosine of 0. Cosine of 0 is 1, and our variable is r. And r is varying, when we go back to our diagram. The integral is taken from positive plate to negative plate, therefore r is going to vary from inner radius a to outer radius b.

So the boundaries of the integral will go from a to b. And here q 2 Pi Epsilon 0 and h are constant. We can take it outside of the integral, and that leaves us, the potential difference between the plates, is equal to q over 2 Pi Epsilon 0 h times integral of dr over r, integrated from a to b. Integral of dr over r is ln of r. Moving on, then, we will have q over 2 Pi Epsilon 0 h times ln of r, evaluated at a and d, which is going to be equal to q over 2 Pi Epsilon 0 h times ln of b minus ln of a by substituting the boundaries for r. And since ln of b minus ln of a is equal to ln of b over a, we can finally express this potential difference as q over 2 Pi Epsilon 0 h times ln of b over a.

Once we determine the potential difference between the plates, as the last step, we can calculate capacitance of the cylindrical capacitor from its definition, which is the ratio of the magnitude of the charge stored in the plates divided by, or to the potential difference between the plates, which is V. So we will have q divided by potential difference which is, q over 2 Pi Epsilon 0 h times ln of b over a. The charges will cancel in the numerator and denominator, leaving us the capacitance of a cylindrical capacitor as equal to 2 Pi Epsilon 0 h times 1 over ln of b over a.

We can easily see that as in the case of parallel plate capacitor. In cylindrical capacitor also, the capacitance is dependent to the physical properties of the capacitor. In this case, the length of the height of the cylindrical capacitor, and as well as its inner and outer radius.