Overview

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Imagine a closed surface, like a balloon, with two charges inside it: plus two microcoulomb and minus three microcoulomb. And just outside it sits a bigger charge, plus five microcoulomb. Does that outside charge change the electric flux through the surface? Hmm, it changes the field everywhere around it, so surely it changes the flux too, right? Keep your guess. Today, we will first see how to use this unit's problem bank, and then we will settle this puzzle with a problem from it. Gauss's law questions like this come up often in JEE, so let's begin.

0:43What is in this chapter

So, first, let us see what this chapter is. It is the problem bank for the whole unit on electrostatics: mixed solved problems, in three parts. Okay? Part A has eight solved problems, at JEE Main level. Part B has six longer problems, at JEE Advanced level. With me? And Part C has six concept checks, on the ideas that are often tested in statement form. And each part ends with one practice problem on the page, for you to solve. Got it?

1:14Part A: a 30-minute test

Now, when should you attempt Part A of this chapter? On the very first day of the unit, just to have a look? No, not on the first day: the course says to attempt it only after Chapters one to seven, okay? That is because its problems are mixed: fields, flux, potential, conductors, capacitors and energy, all together. In the exam, no question tells you which chapter it comes from. Right? So finish every chapter first, and then attempt it as a real thirty minute test: solve each problem yourself first, and only then check the solution. Clear? Now, quick maths. Eight problems, and thirty minutes on the clock. So how much time can you give each problem? Thirty over eight is three point seven five, so just under four minutes for each problem. Keep that number in your head while you solve. With me?

2:09Part C: two statement checks

Now, a taste of Part C, with two checks just as the course asks them. Here is the first one: is Gauss's law true for a closed surface of no symmetry, any odd shape at all? The answer is yes, it always holds. Symmetry is needed only when we want to use the law to find E, the electric field. Remember this, because our first example uses it. Okay? Now the second check. Does the capacitance, capital C, depend on the charge, capital Q, on the capacitor? Put in more charge, and get more capacitance? The answer is no. The charge and the potential difference, capital V, change together. Capacitance depends only on the geometry, and on the medium between the conductors. Statements like this come up often, so read them carefully.

3:02Example 1: flux through a surface with a charge outside

Got it? Now, Example one, from Part A, an easy one, and it is our puzzle from the start. Here, a closed surface contains two microcoulomb and minus three microcoulomb; the Greek letter mu with a capital C means microcoulomb. And a five microcoulomb charge is outside. Find the flux. Pause the video, and try it first. Done? Okay, let's check. Many students simply add every charge they see: two, minus three, plus five, which gives four microcoulomb. Is that right? No, no, that is the trap, because only the charge inside counts. Gauss's law, for a closed surface: phi, the Greek letter for the flux, equals q enclosed over epsilon naught. The q with the small subscript e n c is the enclosed charge, and epsilon naught, the Greek epsilon with a small zero, is the permittivity of free space. Okay? So, the enclosed charge is two, plus minus three, which is minus one microcoulomb. Micro means ten to the power minus six, so that is minus ten to the power minus six coulomb. Clear? Now divide by epsilon naught, eight point eight five times ten to the power minus twelve; the cross means times. Phi comes out about minus one point one three times ten to the power five, where the wavy equals sign means approximately. The unit is newton metre squared per coulomb; the little minus one on the coulomb means per coulomb. And the minus sign tells us the net flux is inward. Right? And now, our puzzle from the start. The outside charge does change the field, E, at points on the surface. But every field line from it that enters the surface also leaves it, so it adds no flux. So, if your guess was no change, very good! The flux depends only on the enclosed charge, while the field depends on all the charges. Got it?

5:02Example 2: dielectric slab in series capacitors, battery on

Now Example two, and this one is from Part B, a hard one. Here, two identical capacitors, each of capacitance capital C, are joined in series across a battery of voltage capital V, okay? Then a slab of dielectric constant capital K fills one of them, with the battery still connected. Find the new voltages. Pause the video, and try it first. Done? Okay, let's check. Before the slab, the two identical capacitors share the voltage equally, V over two each. Now many students say, the battery is connected, so the voltage stays constant, and each one still has V over two. Is that right? No, no, wait, that is the trap. The battery fixes only the total voltage, capital V, across the pair. How that total splits between the two capacitors can change. With me? The slab multiplies the capacitance of the filled one by K, so it now has K times C, while the other still has just C, as before. In series, the equivalent capacitance, C with the subscript e q, is the product over the sum: K C times C, over K C plus C, and the dot on screen means times. Cancel one C from the top and the bottom, and it is K times C, over K plus one. Clear? In series, both capacitors carry the same charge, capital Q, which is C e q times V, the battery voltage. So Q is K times C times V, over K plus one. Right? Now the filled one. Its voltage, V one, the V with a small one below it, is Q over K times C, because voltage is charge over capacitance. Put in Q, and the K times C cancels, so V one is V over K plus one. That is less than V over two, because K is more than one. Okay? And the other one: V two is Q over C, which is K times V, over K plus one. Add them, and you get back V, exactly the battery voltage. So the slab did change the split: the filled capacitor, with the bigger capacitance, now takes the smaller voltage. Got it? And look, Part B's practice problem on the page continues this very problem. By what factor does the total charge drawn from the battery change? Try it yourself, using the charge we just found.

7:24Quick revision: Part A test, Gauss's law, series voltages

So, quick revision. Part A is a thirty minute test after Chapters one to seven; Part B is at JEE Advanced level; and Part C has the concept checks. Gauss's law holds for any closed surface: phi equals q enclosed over epsilon naught, and charges outside add no flux. And with the battery connected, only the total voltage is fixed; in series, the bigger capacitance takes the smaller voltage. Now set a thirty minute clock, open Part A, and begin. All the best!