Mixed Problems: Set 3 (Capacitors and Dielectrics)

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take a parallel plate capacitor with air between its plates, and slide in a slab of dielectric constant six. If the slab filled the whole gap, the capacitance would become six times, right? But our slab fills only half the gap, across the whole plate area. So does the capacitance become three times, half of six? Keep your guess; the real answer is a surprise, and we will settle it in our second example. This is Set three of the mixed JEE Main problems, on capacitors and dielectrics, and questions like these come up often. So, let's begin.

0:40Set 3 toolkit: capacitance, combinations, energy

First, the toolkit for this set. Capacitance, capital C, is capital Q over capital V, the charge over the potential difference, measured in farad. For two parallel plates with air between them, we call it C naught, the C with a small zero. C naught equals epsilon naught times capital A, over d, where epsilon naught, the Greek epsilon with a small zero, is the permittivity of free space, capital A is the plate area, and d is the gap between the plates. Okay? Now the combinations. In parallel, every capacitor gets the same voltage, and the capacitances simply add: C one plus C two, where the small subscripts one and two just number the capacitors. Right? Now series, one capacitor after the other. Let me ask you this: in series, what is the same for both capacitors, the charge or the voltage? It is the charge, yes, exactly! In series, every capacitor carries the same charge, and the reciprocals add: one over C equals one over C one, plus one over C two. Clear? And here is a trick toppers use. For just two in series, flip it once: C equals C one times C two, over C one plus C two. We call it product over sum, and it saves a lot of time in the exam. Got it? Last tool: the energy stored is called capital U, and it equals one half C V squared, which is the same as Q squared over two C, okay?

2:14Battery on, or isolated?

Now, the first question in every dielectric problem. Is the capacitor still on the battery, or has it been disconnected and left isolated? When a slab goes in, which stays fixed in each case, the charge or the voltage? If it is isolated, the charge has nowhere to go, so the charge stays fixed. If the battery stays connected, the battery holds the voltage fixed. Decide this first, before you touch any formula. With me? Now fill the whole gap with a dielectric of dielectric constant capital K, and the capacitance becomes K times C in both cases. The arrow on screen means becomes. Right? So, when it is isolated, Q is fixed and C grows K times, which means the voltage drops to V over K, okay? On a battery, V is fixed, so the charge grows to K times Q instead. Mixing these two up is a common slip, so be careful. Clear?

3:13A slab that fills only part of the gap

Now, the case in our puzzle. Look at the left picture; it is the only one we need today. The dark lines are the plates, and the shaded block is the slab, with dielectric constant capital K written on it. The two-headed arrows mark the gap d between the plates, and the slab's thickness t, which is smaller than the gap. It covers the whole plate area, but not the whole gap. Okay? So, let me ask you this. Does this slab simply multiply the capacitance by K, the way a full slab does? No, not at all, because the slab does not fill the gap. Here is the course formula, also written under the left picture: C equals epsilon naught times capital A, over d minus t, plus t over K, okay? Read it like this: the air part, d minus t, stays as it is, while the slab's thickness t counts as only t over K, right? And notice, only the total air gap enters the formula. So the capacitance is the same wherever the slab sits, touching a plate or right in the middle. Clear? And here is why the formula looks like this. We cross the air and the slab one after the other, so the layers act in series, like capacitors in series. Keep this in mind, because we will use it as a check in Example two. Got it?

4:34Example 1: three 6 microfarad capacitors in series-parallel

Alright, Example one, an easy one to warm up. There are three capacitors of six microfarad each. One is in series with the other two, which are in parallel. Find the result. On screen, the Greek letter mu with capital F means microfarad. Pause the video and try it first. Done? Okay, let's check. Some students just add all three and write eighteen microfarads. Others put all three in series and get two microfarads. So which one is right? Neither of them, and that is the trap here! Read the circuit carefully. First, the two capacitors in parallel simply add: six plus six, which gives twelve microfarads. With me? Now this twelve microfarad block sits in series with the third capacitor, of six microfarads. Right? So use product over sum. Six times twelve, over six plus twelve; the cross sign on screen means times. Seventy two over eighteen gives four microfarads, and that is our answer. Clear?

5:43Example 2: K = 6 slab filling half the gap gives 12/7 C0

Now Example two, a harder one, and it is our puzzle from the start. Here, a slab of dielectric constant K equal to six, with thickness d over two, covers the whole plate area of a capacitor C naught. Find C, the new capacitance. Pause the video and try it first. Done? Okay, let's check. First, our guesses from the start. Is it six C naught, because K is six? Or three C naught, because only half the gap is filled? It is neither, so be careful here! Use the slab formula. The air part is d minus t, which is d minus d over two, so d over two. And for the slab, t over K is d, over two times six, which gives d over twelve, okay? Put these into the formula: C equals epsilon naught times capital A, over the sum, d over two plus d over twelve. Now d over two is six twelfths of d, so the bottom becomes seven d over twelve. Right? Flip the fraction: C equals twelve epsilon naught capital A, over seven d, which is twelve over seven times C naught. The wavy equals sign means approximately, so that is about one point seven times C naught. So it is not six times, not even three times, but less than double! And that solves our puzzle. With me? Now a quick check, with the layers idea. Air in a gap of d over two gives double C naught, that is two C naught. The slab has the same thickness and K equal to six, so it gives six times that, twelve C naught. Okay? These two layers are in series, so use product over sum, just like in Example one. Two times twelve, over two plus twelve, gives twenty four over fourteen, which is twelve over seven, times C naught. The same answer both ways, so we can be sure. Got it?

7:41Quick revision: series, parallel, battery or isolated, slab

So, quick revision. In series the reciprocals add, and for two capacitors use product over sum; in parallel, just add. An isolated capacitor keeps its charge, and a battery keeps the voltage fixed. And for a slab that does not fill the gap, use epsilon naught times capital A, over d minus t plus t over K, not simply K times C naught. Now open the practice problem on this page, and give it about two minutes. All the best!