9.11 Displacement Current
Earlier we have studied an interesting circuit which was consisting of a resister and a capacitor. We called that circuit as “RC circuit”. Recall an RC circuit. In this circuit, we had a power supply, which generates ε volts of EMF, a switch, a resister, and a capacitor connected to that in series form. We have seen that, when we turn the switch on, current, i, emerges from the positive terminal of the power supply, and flows through the circuit, and enters into the negative terminal, during what we call “the charging process”.
Of course, when we look at what happens as soon as we turn the switch in “on” position, the positive charges are continuously repelling one another at the positive terminal of this power supply. They will see a path to move along, therefore they move along this path, and get collected on the upper pate of this capacitor. In the meantime, the negative charge, also in a similar way. moves along this available path, and gets stored in the lower plate of the capacitor.
As the capacitor is getting charged, the amount of charge at the plates becomes larger and larges, and they generate a repulsive force to the incoming charges. Once the capacitor is fully charged, then they generate large enough, or high enough repulsive force to the incoming charges. At that instant, they cannot go any further, and the charging process stops. And since the charges do not move anymore, then the current in this circuit drops to 0. So, during the charging phase, therefore, as soon as we turn the switch on, as you recall, we start with the maximum current and, as the charges flow and get stored in the plates of the capacitor, then the current decreases, and eventually goes down to 0. The behavior of the current was, as you recall, was an exponential behavior, and it was equal to i-max times e to the –t, over τC , which was the capacitive time constant.
Anyway, the interesting point that we mentioned at that time also with the circuit was if we take an ammeter and place it just right before the capacitor, and right after the capacitor, during charging phase, these two ammeters will read exactly the same value.
And of course, when we turn the switch off, once the capacitor is fully charged, then the capacitor, and if we turn it off and then connect these two ends to one another through a middle branch, as you recall, then the capacitor discharges through that middle branch, by taking this ε out of the circuit. In that case, we end up, again, with a current, and it will flow in the opposite direction of flow to this original one. In other words, in that case, it will be in counterclockwise direction. Again, at that time, the behavior of the current was exactly similar to this case. But we ended up with a negative sign, indicating that the flow is in the opposite direction.
But at that time, also, these ammeters will read exactly the same value. And we said that, again when we were studying these circuits, this is in a way, odd because a capacitor is a device which consists of two conducting plates, separated by an insulating medium. In other words, this gap over here is an insulator, and as you recall, insulators do not let the charges move easily around. They do not conduct electricity. But here, in this case, since these two ammeters are reading exactly the same current value during the charging or discharging phases telling us that somehow this medium between the plates of the capacitor is behaving as if it is a conducting medium, as if it is a piece of wire which is carrying the same current. Because the current comes over here and something happens, and it simultaneously appears at this end also.
Let’s try to understand this phenomenon. Let’s assume that we have a parallel plate capacitor over here, and the plates are in the form of circular plates. So if I drew this in a larger scale, here, it’s something like this. That’s the upper plate, and we have the lower plate. And here is the wire connected to the upper plate, and here is the wire connected to the lower plate. This plate is getting charged positively during the charging phase, and the other one is getting charged negatively.
So as the capacitor is getting charged, of course, the region is getting filled with an electric field, originating from the positive plate, and entering into the negative plate. Therefore, we’re generating a small, electric field package in the region between these two plates.
Okay. We say that the ammeter that we connect over here, and the ammeter we connect over here, after the negative plate, is reading simultaneously the same current value as the capacitor is getting charged. It means that this region is behaving as if it is conducting the current. So in other words, it is in a way behaving as if this current, i, is flowing through this insulating medium. We know that, if we have a current in the medium, then we’re going to end up with a magnetic field that it generates. Those field lines are in the form of concentric circles, about the wire.
Using right-hand rule, for a current like this, holding the right-hand thumb in the direction of flow of current, then the corresponding magnetic field lines are going to be circling about this current in the form of concentric circles, like this. The direction of those field lines are going to be in clockwise direction, along these planes. Of course, these planes are perpendicular to the direction of the flow of current.
Well, let’s look at this picture from the top cross-sectional view for this region. So let’s say top view of the region. For that region, when we’re looking from the top view, we will see that the electric field is going in to the plane. Therefore, we have an electric field which is going into the plane. Again, from the top view, that electric field is filling this region, a cylindrical region actually, the region between these two. So the top view will be like a circle, like this — that’s the cross-sectional view — and the E field is in to the plane, and it is increasing during charging phase.
Now, at that moment, we have as if a current, let’s say this is our current, i, as if it is flowing in the direction of this E field. The corresponding magnetic field that it generates is going to be in the form of concentric circles. We can draw several of them, like this. Using right-hand rule, they’re going to be rotating in clockwise direction. So we’re going to end up with magnetic field lines rotating in clockwise directions. Therefore, the corresponding magnetic field is going to be tangent to these field lines at the point of interest, like these, let’s say. This is B1, and B2, and so on and so forth.
Now, let’s consider the perfect symmetrical case of this one. In other words, in this case, let’s assume that we have a magnetic field which is increasing into the plane. This is a case equivalent, as if we have a magnet, a bar magnet with north pole facing towards the plane, and we’re approaching that magnet towards the plane.
So since the field lines are emerging from the north pole, therefore, as we move that magnet towards the plane, then the magnetic field strength will increase in to the plane. At that instant, if we just take a closed, conducting loop, let’s say in the form of a circular wire, and place it inside of this region, as we have seen earlier, according to the Faraday’s law, now since the magnetic flux through the area surrounded by this conducting loop is going to be increasing, due to the increase of this B field, B is increasing, then the flux is going to be increasing so we’re going to end up with an induced EMF, from Faraday’s law, and therefore, a corresponding current flowing through this conducting loop, this circular wire. It will show up, from Lenz’s law, it will show up such that it will oppose its cause.
In order to do that, it should generate a magnetic field, which is going to be pointing in the opposite direction to the direction of this increasing, external magnetic field. To be able to have a magnetic field coming out of plane, through this region, then the associated current has to be flowing in a counterclockwise direction, like this.
Again, for current to flow in counterclockwise direction, it will correspond to a case that we should end up, or we should have an induced electric field line, which is going to be pointing along this wire, in a counterclockwise direction. So the associated electric field will be tangent to this field line, at every point along the field line.
Now, when we look at these two symmetric cases, we can say that, yes, indeed, a changing magnetic field is generating an electric field, and also a changing electric field is generating magnetic field. But the direction of these field lines, hence the corresponding fields, do not match exactly because, in the case of the induced electric field lines, this case obeys Lenz’s law. As you can see, the induced current, therefore the associated corresponding electric field is trying to oppose its cause. Whereas in this case, the current is flowing in to the plane, and the associated magnetic field lines are directly from the right-hand rule. So this one obeys right-hand rule, whereas the other one does not obey right-hand rule, it obeys Ohms law.
So it is because of that reason, when we express the Ampere’s law, which is B dot dl, integrated over a closed contour, C, is equal to first μ0 times i-enclosed. That’s just the Ampere’s law part, but the correction to that, which was made by Maxwell, and since we have a right-hand rule associated with the induced magnetic field, over here, then we have a positive sign here and μ0 times ε0, change in electric flux, with respect to time term here.
Now we’re going to show that why, indeed, this ε0, dΦE over dt is equal to current. So before we do that, let’s make the note over here, that this case obeys right-hand rule, whereas the other one obeys Lenz’s law. Therefore, we have, indeed, some symmetry, but it is not 100%.
Okay. Now let’s go back to our capacitor problem. Now, we said that, since these two ammeters are showing the same amount of current before and after the capacitor, indicating that during charging as well as during discharging phases, this insulating medium is behaving as if it is conducting the electricity. Let’s go ahead and try to figure out the electric field generated due to the charges stored to the plates of this capacitor by applying Gauss’s law.
Well, the Gauss’s law, which states that E dot dA, over a closed surface is equal to net charge enclosed in the volume, surrounded by this closed surface, divided by ε0. So we’re going to try to calculate the electric field between the plates of this parallel-plate capacitor. We did this earlier. We said that we can choose a closed surface that encloses all of the field lines. Actually, not all the field lines, but a closed surface that encloses the surface of one of these plates. And one surface, that closed surface, is passing through the point of interest.
For example, if I’m interested with a point somewhere over here that I’d like to calculate the electric field, in this case, I can choose a cylindrical surface, such that the lowest surface of the cylinder is passing through the point of interest. Then the upper surface completely encloses all of the upper plate. When we look at this formation, we see that dA, the surface-area vector, is perpendicular for the upper plate. For the lower plate, it is also perpendicular, like this. For the side surface, also, it is perpendicular, like this.
As you recall, when we were analyzing such a capacitor to calculate the electric field between the plates, we looked at the plate from an exaggerated point of view, something like this. Then we placed the upper surface of the Gaussian surface inside of that plate. Recall that the electric field was 0 inside of a charged, conducting medium because once the upper plate is getting charged, all the charge was being collected on the lower surface of this upper plate so that the inside, there wasn’t any charge enclosed. Therefore, at this region, the electric field right there, is 0.
Of course, for the other surface, we have an electric field at that point, pointing in a downward direction, like this. We said that we can write down this closed-surface integration as the integral over the top surface of E dot dA. There we will have E magnitude, dA magnitude, times cosine of the angle between these two vectors. But for the top surface, since we are inside of the conducting medium, electric field is 0 over there. Then we end up with no contribution from that integration.
Then we take the integral over the side surface, and that gave us E magnitude, dA magnitude. Over the side surface, the angle between electric field vector and the incremental surface area vector is just simply 90 degrees, so we’ll have cosine of 90. But cosine of 90 is 0, so there wouldn’t be any contribution from that surface integration also.
Finally, integration over the bottom surface, of E magnitude, dA magnitude. For that region, we see that the angle between E and dA is just simply 0 degree. That, therefore, gives us cosine of 0 is equal to q-enclosed over ε0.
Okay. Since cosine of 0 is just 1, and as long as we are at the lower surface of this Gaussian cylinder, we will be the same distance away from the source, from the upper plate in this case. Therefore, the magnitude of the magnetic field will be the same, as long as we are on this plane. So we can take it outside of this integral, and we end up with E times integral of dA over the bottom surface, is equal to q-enclosed over ε0.
If we say that, A be the plate area, and since the Gaussian cylinder is completely, fully enclosing the upper plate, therefore its cross-sectional area will be equal to the plate area. Then integral of dA over the bottom surface is going to give us the total surface area of that region, and that is equal to the plate area. So E times A will be equal to q-enclosed over ε0.
Again, q-enclosed is the net charge inside of the region, surrounded by the Gaussian surface, in this case, this Gaussian cylinder. And when we look inside of that cylinder, it encloses all of the charge stored in the upper plate. If you call that charge as “q“, the total charge as “q“, then q-enclosed is simply equal to q. Total charge on the capacitor.
Okay. Therefore, E times A becomes equal to q over ε0. And if we solve for the electric field from there, we’re going to end up with q over A ε0. All right.
So that’s the strength of the electric field between the plates. And, as we have seen earlier, this is not dependent to the position. Therefore, wherever we go between the plates, for a parallel-plate capacitor, we see the same strength for the electric field. But let’s look at this left-hand side, over here. We have the product of electric field time the area.
So E times A, actually, also gives us the flux through the surface area of A, the flux of electric field. Because, if we look at the electric-field flux through the surface in this region, the area of that surface is A, E magnitude is E, while the flux is E dot A and E and A is in the same direction, the angle between them is 0 degree. Therefore E times A times cosine of 0, which is equal to 1, will give us the electric flux. So here, what we have, indeed, is the electric field flux through the surface A.
Well, we know that, as the capacitor is getting charged, the electric field is increasing. As the charge on the capacitor is increasing, the corresponding electric field between the plates of the capacitor will also increase.
If you look at the rate of change of this flux, that will be d(EA) over dt. That’s going to be equal to d over dt of q over ε0. Well, this will also be equal to, then dΦE over dt, which is going to be equal to — since ε0 is constant, we can take it outside of the derivative operator. So we’ll end up with 1 over ε0 times dq over dt. But dq over dt is, by definition, current. From there, if we solve for the current, that will be equal to ε0 times dΦE over dt. And since this is not actually a conduction current, it is happening in a medium through which the electric field is changing, therefore causing a change in electric field flux through some specific area. We call this current as the displacement current, and we denote it with id.
Now, we can understand, in the Ampere-Maxwell’s law, which is B dot dl, integrated over a closed contour, is equal to μ0 times i-enclosed. This is the Ampere’s law part. Net current flowing through the area, surrounded by this closed loop, plus μ0 times the displacement current, and that appears as a result of the change in electric field through the area surrounded by that loop.
So, if the electric field is not changing, but a conduction current, i, is flowing through that region, then we have a corresponding magnetic field. But if we have a region, an insulating medium, such that through which the electric field is changing, therefore that electric field will cause a change in electric field flux, through the area surrounded by this loop. That, too, will generate magnetic fields.
In explicit form, therefore, this is equal to μ0 times i-enclosed, plus μ0, the displacement current, as we have seen now, ε0 times dΦE, over dt.