Leakage through a Dielectric

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take two capacitors filled with the same dielectric. One is tiny, the size of a coin. The other is huge, with plates a metre wide. Charge both to the same voltage, disconnect the battery, and leave them on the table. Slowly, both lose their charge, even though no wire joins the plates. So which one loses its charge faster? Hold that thought; by the end of this video you will answer it in one line. Leakage through a dielectric comes up in JEE problems on capacitors from time to time. So, let's begin.

0:38A dielectric that leaks

So far, we treated every dielectric as a perfect insulator. But no real material is perfect, right? Every real dielectric has a resistivity, rho. Rho, the Greek letter that looks like a small p, is measured in ohm metre: the capital omega stands for ohm, and the m is metre. Okay? So what happens when the capacitor is charged? There is a voltage across the plates, so a small current flows through the dielectric, from one plate to the other. We call it the leakage current. Clear? Now, the dielectric itself acts like a resistance, R, between the plates. From school, you may remember that resistance is resistivity times length, over area. So tell me, for the leakage current, what is the length of its path, and what is its area? Yes, exactly, the current crosses the gap, so the length is d, the gap between the plates. And it flows through the whole plate area, A, okay? So R equals rho times d, over A, when the dielectric fills the whole gap. Got it? Now a quick trap. Suppose we double the plate area. There is more dielectric between the plates, so there is more resistance, right? No, no, wait, look at the formula: the area sits in the denominator. Double the area, and R becomes half. The wider path lets more current through, just as a wider pipe lets more water through. With me?

2:03The product RC

Now bring in the capacitance. With the gap completely filled, C equals K times epsilon naught times the area, over d. K is the dielectric constant, and epsilon naught, the Greek epsilon with a small zero, is the permittivity of free space. Okay? Now compare it with R, the resistance. When we doubled the area, the resistance became half. So what happens to C when we double the area? Correct, C doubles, because the area sits on top in the capacitance. So R falls exactly as much as C rises, and the product, R times C, does not change at all. In fact, R C equals rho times K times epsilon naught. We will prove this ourselves in Example one. Right? Now look closely at this result. Where is the area? Where is the gap? They are not there at all! R C depends only on the material: its resistivity and its dielectric constant. So R C is independent of the shape of the capacitor. Clear?

3:08The leakage current

Next, how big is the leakage current? Put a voltage, V, across the plates. The dielectric is simply a resistance, so I equals V over R, where I is the leakage current. Got it? Now here is a neat move. We just saw that R times C equals rho times K times epsilon naught. So how can we write R using C? That is correct, we just divide both sides by the capacitance. So R equals rho times K times epsilon naught, over C, as you see on screen. Okay? Now put this into the current, which is V over R, and see what we get. I equals C times V, over rho times K times epsilon naught. So at the same voltage, a bigger capacitance means a bigger leakage current. Our puzzle needs this. With me?

3:58Self-discharge of an isolated capacitor

Now let us go back to our puzzle. Charge a capacitor, then disconnect the battery, so the capacitor is isolated. No wire joins the two plates. So how can it lose its charge? Through its own dielectric, exactly! The leakage current flows from plate to plate inside the capacitor, so the capacitor slowly discharges itself. We call this self-discharge. Right? And how fast does it discharge? That is set by the time constant, tau, the Greek letter you see on screen. Tau equals R C, which equals rho times K times epsilon naught. The bigger the time constant, the slower the discharge, okay? So now, the answer to our puzzle. The tiny capacitor and the huge one are filled with the same material, so they have exactly the same time constant. At the same voltage, the huge one leaks a bigger current, but it also has more charge to lose, and the two balance perfectly. Same material, same tau, whatever the size or the shape. Isn't that neat?

5:03Example 1: prove RC = rho K epsilon0 for a leaky capacitor

Time for the first example. Show that R C equals rho times K times epsilon naught, for a parallel plate capacitor filled with a leaky dielectric. Pause the video and try it first. Done? Okay, let's check. First, the resistance. The leakage current crosses the gap, d, through the plate area, so R equals rho times d, over A, right? Next, the capacitance. The gap is completely filled, so C equals K times epsilon naught times the area, over d. Clear? Now multiply them. R C equals rho times d over A, times K times epsilon naught times A over d. So what cancels? Yes, exactly, both the area and the gap cancel, top and bottom. What is left is R C equals rho times K times epsilon naught, as required. Two lines, and the proof is done. See how simple that was?

6:02Example 2: self-discharge time for K = 5, about 44 s

Now the second example. Here it is: a capacitor is filled with a material of K equals five, and rho equals ten to the power twelve ohm metre. Find the time constant for self-discharge. Pause the video and try it first. Done? Okay, let's check. But wait, look at the question again. There is no area, no gap, not even the capacitance. So are we stuck? We are not stuck at all, because the time constant does not depend on the shape. Tau equals rho times K times epsilon naught, and that needs only the material. Okay? So, put in the numbers. Tau equals ten to the power twelve, times five, times eight point eight five times ten to the power minus twelve. That last part is epsilon naught. Got it? Now, ten to the power twelve ohm metre is a huge resistivity, a really good insulator. So the time constant must be huge too, maybe days or years, right? Oh no, careful, epsilon naught brings ten to the power minus twelve, and the two powers of ten cancel, giving just one. So tau is five times eight point eight five, which is forty four point two five. Clear? So the time constant is about forty four seconds. The wavy equals sign means approximately, and the s is for seconds. Less than a minute, even for such a good insulator! Right? And a quick unit check. Epsilon naught is in farad per metre: the F with m to the power minus one. Ohm metre times farad per metre leaves ohm times farad, which is a second. So the units agree. With me?

7:42Quick revision: R = rho d/A, RC, leakage current, tau

So, quick revision. First, a leaky dielectric filling the gap has resistance R equals rho times d, over A, okay? Second, R C equals rho times K times epsilon naught, independent of the shape. And third, the leakage current is C times V, over rho times K times epsilon naught, and an isolated capacitor discharges with the time constant rho times K times epsilon naught. Now go and try the practice problem on this page. You can do it. All the best!