A 10-cm-diameter parallel-plate capacitor has a 1.0 mm spacing...

Question

A 10-cm-diameter parallel-plate capacitor has a 1.0 mm spacing. The electric field between the plates is increasing at the rate 1.0×10⁶ V/m s. What is the magnetic field strength on the axis?

Accepted Answer

Welcome back, everyone in this problem. We're considering a circular parallel plate capacitor connected to a battery. The plates are separated by a distance of 1.8 millimeters and have a radius of 15 centimeters. The electric field between the plates is experiencing an augmentation at a rate of 1.4 times 10 to the fourth volt meters per second. And we want to determine the magnitude of the magnetic field at a point located on the axis of the capacitor. For our answer choices A is zero. Teslas B is three times 10 to the negative 13 teslas C is four times 10 to the negative fifth teslas and D is nine times 10 to the negative fifth teslas. Now, let's try to make sense of what our problem really wants from us here. So we're talking about a circular parallel plate capacitor could look something like this, eh, and they're connected to a battery could probably look something like that. Ok. And what do we know? We know that they are separated at a distance of 1.8 millimeters not trying to scale, of course, and they have a radius of 15 centimeters. So we could put that here and label that as capital R. So their radius is 15 centimeters. We also know that the, the, the electric field between the plates is experiencing an augmentation at a rate of 1.4 times 10 to the fourth volt meters per second. In other words, it's increasing at that rate. So we could represent our electric field here or change in electric field here on our diagram. OK. And we're trying to find the magnitude of the magnetic field at a point located on the axis of the capacitor. Now, if we look at our capacitor here, if we were to put in red, we could represent its axis using this red dotted line. OK. So at that point located an axis of the capacitor, we could call that point R that distance R. OK. So essentially this is what we're trying to find the magnitude of our magnetic field at our distance are. Now what do we know that links are changing electric field to a magnetic field? Well, we know that Empires Maxwell, Empire Maxwell's law, we know Empire Maxwell's law tells us that a magnetic field can be caused by a changing electric flux. In other words, it's given by this formula that says the closed integral of B with respect to S the displacement vector of our closed loop is going to be equal to the permittivity of free space. Epsilon knot multiplied by the peril of free space mu knot multiplied by our changing electric flux over time. In other words, it cha it changed the electric flux more than it causes a greater magnetic field. Now, the thing about this is that it relates to changing electric flux to an induced magnetic field. But is there some way for us to link a changing electric field with our induced magnetic field? Well, let's think about what electric flux really is. Recall that light magnetic flux, electric flux is equal to our electric field strength. E multiplied by the area of or the the surface that it is going through, which makes sense because a stronger electric field means greater flux and a greater area also means greater flux. Since we're talking about a circular parallel plate capacitor, then our electric flux is actually equal to pi R squared multiplied by our electric field. Now in the same light. OK. So we've expressed electric flux in terms of our electric field. We also know that the closed loop integral of our magnetic field with respect to a displacement vector X can be written as our magnetic field multiplied by the closed loop integral with respect to X where we can treat our magnetic field as a constant. And if we integrate with respect to our displacement vector X, then that is going to give us the closed loop the distance around the closed loop. And since we're talking about a circular parallel plate capacitor, the distance around that loop is going to be its circumference which is two pi R OK. So no, we, we, we have an expression for the closed integral of B with respect to this displacement vector X. And we also have an expression for our electric flux. So we can substitute that to rewrite our magnetic field in terms of our distance from the axis R OK. Because no, that tells us then solving for a bee. That means that B right, let me write that properly. B multiplied by two. Pr is going to be equal to permittivity multiplied by permeability multiplied by our change in electric flux, which would be a change in pi R squared E we know pi R squared is constant. It's our electric field that's changing. So we can write that as pi R squared multiplied by change in electric field with respect to time. OK. No, here we can solve for B because we can make B the subject of our equation. So B because epsilon knot Muno pi R squared te with respect to T divided by two pi R OK. Pi Kan Pi R goes into R squared R times. So our electric or magnetic field B equals epsilon knot mu knot R divided by two multiplied by D EDT. Now we have an expression for our magnetic field in terms of art. No, when we think about it, let's ask ourselves, what do we know about? I remember we said that it is a point located on the axis of the capacitor. So if it's located on the axis of the capacitor, then that means R is going to be equal to 0 m because it's on the axis. And if R is 0 m, then when we put zero into our equation, we'll eventually get that our magnetic field equals zero. OK. So that tells us, then that at a point located at the capacitor at capacitor axis R equals zero, then our magnetic field B is going to be equal to zero teslas. When we look back on our answer choices that would have been answer choice. A thanks a lot for watching everyone. I hope this video helped.

A 10-cm-diameter parallel-plate capacitor has a 1.0 mm spacing. The electric field between the plates is increasing at the rate 1.0×10⁶ V/m s. What is the magnetic field strength on the axis?

Key Concepts

Capacitance

Electric Field

Magnetic Field Induction

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A 10-cm-diameter parallel-plate capacitor has a 1.0 mm spacing. The electric field between the plates is increasing at the rate 1.0×106 V/m s. What is the magnetic field strength on the axis?

Key Concepts

Capacitance

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Magnetic Field Induction

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A 10-cm-diameter parallel-plate capacitor has a 1.0 mm spacing. The electric field between the plates is increasing at the rate 1.0×10⁶ V/m s. What is the magnetic field strength on the axis?