5.4 Parallel Plate Capacitor from Office of Academic Technologies on Vimeo.

**5.04 Parallel Plate Capacitor**

Capacitance of the parallel plate capacitor.

As the name implies, a parallel plate capacitor consists of two parallel plates separated by an insulating medium. I’m going to draw these plates again with an exaggerated thickness, and we will try to calculate capacitance of such a capacitor. As we have seen earlier, when we consider two conducting plates, something like this, and if we connect these plates to the terminals of a battery such that the upper one, with the positive terminal of the battery, and the lower one to the negative terminal of the same battery, then the charges will move through–since they continuously repel one other, like charges–will move through these conducting paths, and then the furthest distance that they can go away from each other is the boundary of this surface, which is the lower surface of the upper plate, where the positive charge is, and the upper surface of the lower plate, where the negative charge is. And, by this way, the capacitor is going to get charged to a certain q value. Let’s say the magnitude of this charge is Q, therefore, we will end up with plus q on the upper plate and minus q along the surface of the lower plate.

Let’s give some dimensions to this capacitor. Let us say that the separation distance, between the plates is d. d is the separation distance between the plates. And, also, let’s say, that A represents the plate area. Now, if we look back at the procedure to calculate the capacitance of a capacitor, the first step was, assuming that the capacitor is charged to a certain q value. Therefore, by connecting these plates to the terminals of a battery, we charge it to a certain q value. And that’s our first step.

And the second step, we will calculate the electric field between the plates of this capacitor by applying Gauss’ law, which is integral of E dot dA over a closed surface, S is equal to net charge in coulombs inside of the volume surrounded by this closed surface divided by the perimeter of free space. And here, the insulating medium between the plates is air, therefore, the permittivity of air is Epsilon 0.

All right, well, in order to apply Gauss’ law to calculate the electric field between the plates we’re going to introduce a Gaussian surface. Let’s say our point of interest, between the plate and an arbitrary location is this point, p. The Gaussian surface that I will choose for this case will be in the shape of a rectangular prism. Such that it’s lower side will pass through this point, p, over here. And I will choose that box in such a way that it will completely enclose the surface of this upper plate. Therefore, it is going to be something like this. And, the upper surface of this Gaussian surface will be inside of this conducting plate. So it’s going to be a box, a hypothetical box, of this shape. And let me try to draw each surface of this box.

Now, once we charge the upper plate positively and the lower plate negatively, then we’re going to be generating electric fields originating from the positive plate and entering into the negative plate, therefore filling the region between these two charged conducting plates. That’s the direction of the electrical field. Now our box, which is a rectangular prism, is going to have six sides. Therefore, we can separate the surface integral into the sum of integral over each surfaces of this box. In other words, we can express this whole integral as integral over the, let’s say, front surface, of this box of E dot dA. And, let’s write these in explicit form as we go along. For the front surface, which is this surface that we’re talking about, the dA incremental surface area vector, is going to be perpendicular to that surface. So, it will be something like this.

And since electric field is always pointing in a downward direction, the angle between these two vectors will be 90 degrees for the front surface. And this is, of course, for the region of front surface between the plates, if we go to the region which is, let’s say, above the lower surface, and in that region the electrical field is 0, because inside of the conducting medium, as we have seen earlier, the electric field is always 0.

So here inside of this upper plate, there’s no electric field. Therefore, we can express this front surface integration as E magnitude dA magnitude of cosine of 90. Okay, and, since cosine of 90 is 0, obviously there will not be any contribution from the front surface integration for this lower region. And for the upper part, the electric field is already 0.

And, moving on, we can look at the symmetrical part of this surface, which is the back surface. And in the back surface, dA is perpendicular to the surface, so it’s going to be, then, pointing like this. And an electric field, again, for the below region, it is in downward direction. We’ll just show that with the dash lines. And the angle between them will be 90 degrees. And if we go up and inside of the upper plate, again, the electric field is 0 there. Therefore, integral over the back surface of the Gaussian surface, again we have EdA cosine of 90 degrees. And that, too, will give us 0, since cosine of 90 is zero.

And, moving on, let’s consider the side surface, in this case, the right hand side surface. And for that side surface, again, the electric field is perpendicular to the surface for an incremental region, therefore it’s going to be pointing like this,. Again, in that region the electric field is always downward direction, the angle between them will be 90 degrees, and this is for the region between the plates. And if you go to this region, which is inside of the upper conducting plate, again, we will end up with zero because the electric field there is 0.

So for the right-hand side surface, we will have EdA cosine of 90 degrees, and let’s aggregate this as RHS, right-hand side surface. Again, no contribution from this integration because cosine of 90 is 0. Similarly, moving to the left-hand side surface, plus integral over the left-hand side surface, we will have the similar type of situation because, in this case also, angle between the electric field vector and the incremental surface area vector, for the left hand side, which is this point over here, will give us 0 degrees. And it will give us 90 degrees, and the cosine of 90 is just 0.

No contribution from this integral too. And, now, we’re left with the upper and lower surface. And if we look at the upper surface, we can easily see that for the upper surface, that surface is completely inside of this conducting medium and the electric field is 0 in that region, therefore EdA cosine of Theta, the angle between them, is going to give us 0 because the electric field is 0 in that region.

And, finally, we’re going to end up with the last surface, and that is the lower surface. And, there, we have EdA. And, for the lower surface, we can easily see that, which is now we’re talking about anywhere along this lower surface. And let me just try to shade that region, to be able to see clearly. For that surface, electric field is always downward direction, and the area vector is also always perpendicular to the surface, so the angle between the electric field vector and the incremental surface area vector will be 0 degree. And the cosine of 0 is going to just give us 1. And this is the integral over the lower surface of that rectangle prism shape of Gaussian box.

Now, as long as we are on this lower surface, we are going to be the same distance away from the charged surface of this upper plate. Therefore, the magnitude of the electric field along that surface will always be the same. Since it is the same everywhere along that surface, we can take it outside of the integral, since it’s going to be constant, and this whole thing–summation through all these six surfaces–will eventually give us the integral over the closed surface of this rectangular prism, and that’s going to be equal to net charge enclosed, inside of the volume of this box, Gaussian box, divided by Epsilon 0.

Now if you continue, we will have electric field times integral over the lower surface of dA is equal q enclosed over Epsilon 0. Integral of dA over the lower surface means, as we have seen earlier, that we look at all these incremental surface elements, all these “dA”s along this lower surface, and we add them all together through the area, through the surface of this surface of interest. And when we do that, we will end up, whatever the total surface area of that region is, or of that plate is. But, since this plate completely encloses the upper plate, we can easily see that the area of that shaded region will be equal to whatever the plate area of this capacitor is. So this is going to give us nothing but A. E times A will be equal to, therefore, q enclosed over Epsilon 0.

All right, now, the left hand side of the Gauss’ law is done, we will look at this q enclosed. q enclosed, by definition, is the net charge inside of the region surrounded by Gaussian surface. Gaussian surface is this purple box. And inside of that region, when we look at the net charge, since that box completely encloses the charged surface of the upper plate, then q enclosed will be whatever the charge that we have along that plate. And, that is plus q. So, q enclosed is equal to q, and from there the electric field is going to be equal to q over Epsilon 0 A. That is the magnitude of the electric field between the plates of this parallel plate capacitor.

Well, when we look at here, the charge stored in the capacitor is a constant quantity. Epsilon 0 permittivity of free space, which is a constant quantity, and the surface area of the plate of the capacitor is a constant quantity. So all these quantities are constant, therefore the electric field between the plates of a charged parallel plate capacitor is constant. In other words, it doesn’t make any difference; where ever, or whatever the point that we go between the plates, we will have this same strength of electric field for a parallel plate capacitor, which is charged to a certain amount of charge.

Now, since we completed the step two, we will move to the step three. And, it says that we will calculate the potential difference between the plates of the capacitor. From integral of E dot dl integrated from positive to negative plate. This is a path integral; it means that we need to choose a path in order to take this integral. We have already calculated the electric field, and we will choose the most appropriate path, the easiest path, to calculate this path integral as the potential difference between the plates.

And to do that, lets go back to our figure. The easiest path is a straight line of path, and we’re going to choose that from positive to negative plate. Okay. And let’s do that, and lets choose that path in the form of a straight line. And I will show that path in dashed lines, from positive to negative plate. The dl is an incremental displacement vector along this path. And so if you show that, dl is going to be a vector along this path, l, pointing in downward direction.

Now, the electric field is pointing in downward direction, and dl is along this path pointing in downward direction, therefore the angle between them will be 0 degrees. So, V, the potential difference between these two plates, will be equal to integral from positive to negative plate of E magnitude dl magnitude time cosine of the angle between these two vectors, from this net product, and that angle is 0 degrees, relative to the path that we chose. Cosine of 0 is 1 and, as we have seen, electric field magnitude, which is this quantity over here, and that is constant. We can take it outside of the integral, then V becomes equal to E times integral from positive to negative plate of dl.

Well, again, if you go back to our figure, integral of dl means we’re adding all these incremental “dl”s along this path. And then, when we do that, we will end up with the length of this path. And that is the separation distance between these two plates, and we called it as d. So this integral is going to give us the separation distance, which is d, and the integral, the explicit form of electric field is equal to q over Epsilon 0 A. So the potential difference between the plates, if everything is in [Inaudible 00:23:05] system, will be equal to this much of volts.

Once we determine the potential difference between the plates, the last stop is calculating the capacitance from its definition, and its definition was the ratio of the amount of charge stored on the capacitor plate to the potential difference between the plates. Therefore, that’s going to be equal to q over, divided by, q over Epsilon 0 A times c. The charges will cancel in the numerator and the denominator leaving us C is equal to Epsilon 0 times A over d.

If we look at these quantities over here, we see that the capacitance is directly dependent to the geometrical properties of the capacitor. In this case, for this parallel plate capacitor, it’s plate area and the separation distance between the plates. So C is equal to permittivity of free space times the plate area of the parallel plate capacitor divided by the separation distance of the capacitor.