The Displacement Current

Transcript

0:00Introduction

Hello everyone! Draw a loop round the wire of a charging capacitor. Ampere's law links the magnetic field round that loop to the current through any surface bounded by that loop. Now stretch that surface like a bag, between the plates, where no charge flows. Same loop, but now the current is zero! So which answer is right? Today we find the missing piece, the displacement current. It comes up often in JEE problems. So, let's begin.

0:29Ampère's law meets a charging capacitor

Here is Ampere's circuital law. The integral sign with a small circle means once round a closed loop. B and d l, with arrows on top, are the magnetic field and a tiny step along the loop, joined by a dot product. Mu naught, the Greek mu with a small zero below, is the permeability of free space, and i is the current through any surface bounded by the loop. Okay? In the figure, on the left, a capacitor is charging: plus q and minus q on its plates, and i c, the i with a small c below, in the wires. The graph on its right is for the next concept. Loop one goes round the wire; the wire cuts a flat surface across it, so the current through this flat surface is i, the full current in the wire. Now bulge the surface between the plates. What current crosses it? It is zero, exactly, because no charge jumps across the gap between the plates. Right? So, one loop, but two answers, i and zero. Ampere's law, as it stands, cannot be complete. Clear?

1:38Maxwell's fix: the displacement current

So what is going on in the gap? No charge flows across it, but something there keeps changing as the capacitor charges. What is it? It is the electric field, yes, exactly! E with an arrow is the electric field, the coloured arrows between the plates in the figure, and it grows as charge piles up on the plates. So the gap has no moving charge, but it does have a changing field. Okay? Maxwell proposed that a changing electric flux acts as a current, and he called it the displacement current. i d, the i with a small d below, equals epsilon naught times d phi E by d t, in amperes. Epsilon naught, Greek epsilon with a small zero, is the permittivity of free space, and phi E, capital phi with a small E below, is the electric flux, the integral of E dot d A over the surface, where d A, with an arrow, is a tiny piece of that surface. Clear? Now, we call it a current. So, is it a flow of charges across the gap? No, it is not a flow of charge, because no charge crosses the gap. But it is a source of magnetic field exactly as a conduction current is. So the field circles the gap at loop two, as it circles the wire at loop one. Got it?

2:59The Ampère–Maxwell law

Maxwell completed Ampere's law: the integral of B dot d l equals mu naught times, i c plus i d, in brackets, where i c is the conduction current, the flow of charge. Or, mu naught i c, plus mu naught epsilon naught d phi E by d t. Okay? Now a parallel plate capacitor: plates of area capital A, charge q, and between them a field E equals q, over epsilon naught times capital A, as you know. Right? Now tell me: is i d in the gap less than i c, the current in the wires, more, or exactly equal? Yes, exactly equal, and here is why. The flux is E times A, so i d is epsilon naught A times d E by d t. Put in E, and epsilon naught A cancels, leaving d q by d t, which is also C times d V by d t, and that is i c. Clear? The wires carry only conduction current, the gap only displacement current, and the total is continuous round the circuit. That solves our puzzle: the flat surface catches i c, the bulged one catches i d, and they agree. And a topper's trick: know the current in the wires, and you know the displacement current. Got it?

4:20Steady state and alternating voltage

Now two special cases. First, a capacitor fully charged by a battery, and still connected. There is a strong field between the plates, so there must be a displacement current, right? No, no, careful, that is the trap! We need a changing field, not just a field. In the steady state, d E by d t is zero; the double arrow means gives, so i d is zero, and i c is zero too. Clear? Second, an alternating voltage, V equals V naught sine omega t. V naught, with a small zero, is the peak voltage, and omega, the Greek letter, is the angular frequency. Then C times d V by d t gives i d equals omega C V naught, cos omega t. With me? So the amplitude of i d is omega C V naught. Keep it ready for example two. Got it?

5:14Example 1: displacement current in a 2 microfarad capacitor

Alright, example one. The voltage across a two micro farad capacitor is rising at five times ten to the power four volt per second; mu F is micro farad, and V s to the power minus one, volt per second. Find the displacement current. Pause the video and try it first. Done? Okay, let's check. We know how fast the voltage rises, so we use i d equals C times d V by d t. Okay? Two micro farad is two times ten to the power minus six farad, times five times ten to the power four. Right? Two times five is ten, and the powers give ten to the power minus two. So i d is ten times ten to the power minus two, which is zero point one ampere. Clear?

6:02Example 2: amplitude of displacement current with AC voltage

Now a harder one, JEE level. Parallel plates of area fifty square centimetres, the raised two meaning squared, are two millimetres apart in air. The voltage across them is V equals two hundred sine, one hundred pi t, volt. Find the amplitude of the displacement current. Pause the video and try it first. Done? Okay, let's check. In S I units, fifty square centimetres is fifty times ten to the power minus four, that is five times ten to the power minus three square metre, and the gap, small d, is two times ten to the power minus three metre. Okay? Now C equals epsilon naught A over d, with epsilon naught eight point eight five times ten to the power minus twelve. The ten to the power minus three cancels, and eight point eight five times two point five gives about twenty two point one, so C is two point two one times ten to the power minus eleven farad. Right? Now compare with V equals V naught sine omega t. What are V naught and omega here? V naught is two hundred volt, and omega is one hundred pi radian per second. Oh no, careful, do not drop the pi! With me? So the amplitude, capital I naught, is omega C V naught: these three numbers multiplied together. Clear? One hundred pi times two hundred is about sixty two thousand eight hundred. Times two point two one times ten to the power minus eleven, that gives about one point three nine times ten to the power minus six ampere; the wavy sign means approximately. So, about one point four micro ampere. Got it?

7:47Quick revision: Ampere-Maxwell law, id = ic, AC amplitude

So, quick revision. The displacement current, epsilon naught d phi E by d t, exists only where the electric flux changes. The integral of B dot d l now equals mu naught times, i c plus i d, and in a charging capacitor, i d equals i c. Fully charged, i d is zero; on an alternating voltage, its amplitude is omega C V naught. Now go and try the practice problems on this page. All the best!