(a) When a circular parallel-plate capacitor is being charged ...

Question

(a) When a circular parallel-plate capacitor is being charged as in Example 31–1, show that the Poynting vector $\vec{S}$ points radially inward toward the center of the capacitor, parallel to the plates.

(b) Integrate $\vec{S}$ over the cylindrical boundary of the capacitor gap to show that the rate at which energy enters the capacitor is equal to the rate at which electrostatic energy is being stored in the electric field of the capacitor (Section 24–4). Ignore fringing of $\vec{E}$ .

Accepted Answer

Welcome back everyone. In this problem. A circular parallel plate capacitor with plate separation D and radius R is charged by a current generating an electric field between the plates. A magnetic field is induced around the capacitor's axis compare the rate at which energy flows into the capacitor. The derivative of you with respect to time with the integral of the pointing vector s over the cylindrical surface between the plates. The plates, a few hints are that the magnetic field at radius R is B equals a half of Muno epsilon NT RD EDT. The surface area of the cylindrical boundary is A equals two PI RD. And we ignore fringing effects and assume uniform fields. A says the rate at which energy flows into the capacitor is too multiplied by the integral of the pointing vector S over the cylindrical surface. B says it's the integral of the pointing vector D sorry C says it's half of the integral of the pointing vector and the D says it's the integral minus mu not the permeability of free space. Now, before we start, let's draw a diagram to help us better understand what's going on. So let's try to represent our parallel plate capacitor. OK. So let's say that this is our capacitor. Oops, let me use a different rubber here. Yeah. And there are few things so far we know about our parallel plate capacitor. First, let me bring this down a bit. First, we know that there is a plate separation of D between the plates. Next, we know that the radius of the plates are OK. So let's say that this is our radius R and it's charged by our current. So we have our current here to the axis of our plate. And let's call that I okay generating an electric field between the plates. So if I were to use a dotted line to represent that here, then between the plates, we have this electric field E OK. And now we also know that a magnetic field is induced around the capacitors axis. So in other words, we could represent that magnetic field here as this circular curve. And we can call that magnetic field. Like our problem statement says b with this information, we want to compare the data which energy flows into the capacitor with the integral of the pointing vector over the cylindrical surface. Let's start by trying to figure out the integral of the pointing vector and to do that, we'll need the pointing vector. Now, what do we know? So let's make a note here. We're solving for integral the pointing vector over the cylindrical surface. So what do we know about the pointing vector? Well, recall, OK. Recall that the pointing vector represents the rate of energy flow per unit area. So it's equal to one divided by the permeability of free space multiplied by the cross product of the electric field and the magnetic field where E is the electric field between the capacitor plates and B is the magnetic field induced to do to the time varying current time, varying current. Sorry. OK. Now, remember we already have an expression for B so we can substitute it into our formula for the pointing vector. Because no, if I were to do that down here, that means our pointing Victor is going to be equal to one divided by Muno E and the cross product of or expression for B A half of Muno epsilon knot RD EDT. OK. Fight OK. Where Phi is the unit vector in the aimal direction and D EDT represents the time rate of change of the electric field. Based on the illustration here, we can tell that E cross B points inward along the radial direction of the circular plate. So that tells us then that if we simplify, S it's gonna be equal to epsilon, no, R divided by two multiplied by D EDT negative R OK. And if this is our pointing vector S then the magnitude of S OK. The magnitude of S is going to be equal to epsilon nod R divided by two D EDT now that we have an expression for S we can integrate the pointing vector over the cylindrical surface. Because if I go back to our diagram here, OK, then what essentially what we're doing? Let me put this in blue is that for our cylindrical surface picture this as our cylindrical surface, then we're going to integrate over this surface. OK. I think I can even put that in here. Let me go ahead and do some highlights, right. So essentially we're integrating here over our surface it. OK. Now, in calculating, in calculating the integral of S with respect to A for energy flow into the capacitor volume note that both S and delta A point inward and that S is constant over the curved surface of the cylindrical volume. So that tells us then OK, that tells us that the integral of with respect to A is going to be equal to S multiplied by A. We already have a formula for a or or formula for the surface area of the cylindrical boundary A. So that is going to be equal to S epsilon knot are E divided by two D EDT multiplied by two pi RD. Now when we go ahead and multiply we get pi epsilon knot R squared de multiplied by the time rate of change of the electric field. OK. So now we have an expression for the integral of the pointing vector over the cylindrical surface between the plate. Next, let's try to relate to the rate of energy stored in the capacitor. OK. So in other words, no, we're solving for DU BT, the derivative of you with respect to T. Now what do we know about our energy stored in the capacity? We recall that the stored energy you is going to be equal to a half of epsilon knot is squared. OK. Multiplied by pi R squared D OK. So now in this case, and that means differentiating you with respect to T we're going to get epsilon knot multiplied by pi R squared multiplied by de multiplied by the rate of change, sorry, multiplied by the time rate of change of the electric field DDT. So now we have an expression for DUDT and the integral of S with respect to A. Now we can compare the both of them. So let's do that look at both of these expressions. Yeah. So we have it here and we have it done here. What do you notice? Well, you probably notice that the both of them are the same thing. They are just arranged a bit differently. So what this tells us then is that by comparing both results, the rate of change of energies, the rate of sorry, the rate of energy stored in the capacitor is going to be equal to the integral of the pointing vector over the cylindrical surface. Therefore, when we look back at our answer choices that tells us that B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Poynting Vector

Capacitor Charging Process

Energy Storage in Electric Fields

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