Overview

Transcript

0:00Introduction

Hello everyone! One small puzzle before we start. Take a four microfarad capacitor, charged to fifty volts, and join it to an empty six microfarad capacitor. The charge spreads over both, and none of it is lost. But is the energy conserved too? Keep your guess; we will settle it at the end. This chapter is about quick JEE Main questions on electrostatics. Let's begin.

0:28How this chapter works

So, how does this chapter work? Electrostatics usually gives two to four questions in a JEE Main shift. Okay? Typically a Coulomb or field calculation, a Gauss's-law flux, a potential or energy question, and capacitors. Right? The real questions, from twenty twenty one to twenty twenty six, are at the end of Chapters one to six. These are original, about two minutes each. With me? Three sets: charges, fields and Gauss's law; potential and dipoles; and capacitors. Let us pack the tools for each.

1:05Set 1 tools: charges, fields and Gauss's law

Set one, charges and fields. Coulomb's law: the force F equals k q one q two over r squared; q one and q two, with small subscripts, are the charges, and r the distance. Here k, one over four pi epsilon naught, is nine times ten to the power nine in S I units; pi is the Greek letter, and epsilon naught, with the small zero, is the permittivity of free space. Okay? This holds for point charges, or for uniformly charged spheres seen from outside, in vacuum. In a medium of dielectric constant capital K, the force is F naught, the vacuum force with a small zero, divided by capital K, right? The field of a point charge: E vector equals k q over r squared, times r hat. The arrow means a vector; r hat, the r with a little cap, is the unit vector away from the charge. So it points away from plus q, towards minus q, and fields add as vectors. With me? Now Gauss's law. The integral sign with a small circle adds E vector dot d A vector over a closed surface; d A vector is a tiny piece of it, along the outward normal. The total flux is q in, with the small word in, the charge enclosed, over epsilon naught. Clear? Now let me ask you this. If there is no charge inside, the flux is zero. Does that mean the field on the surface is zero too? No, it does not, and this is a classic trap! The flux depends only on the charge inside, but the field on the surface is due to all the charges, inside and outside. Got it?

2:49Set 2 tools: potential, energy and dipoles

Set two, potential. For a point charge, V equals k q over r; notice, r, not r squared. Potential is a scalar, so the potentials of many charges simply add, with their signs. Okay? The energy of two charges is capital U equals k q one q two over r. And a charge q at potential V has energy U equals q times V, right? Now take plus q and minus q. At the midpoint, the two potentials cancel, so V is zero. Is the field zero there too? No, it is not, so be careful here! At the midpoint both fields point the same way, towards minus q, so they add up. So V equal to zero does not mean E equal to zero. Clear? Last, the dipole: plus q and minus q, two a apart, with moment p equal to q times two a. Where r is much bigger than a, the double greater-than sign, the axial field is two k p over r cubed, and the equatorial field is k p over r cubed. With me?

3:55Set 3 tools: capacitors and dielectrics

Set three, capacitors. Capacitance, capital C, is Q over V, the charge over the voltage. For parallel plates, C equals epsilon naught times capital A over d: plate area over gap. Okay? In parallel, capacitances add: C one plus C two. In series, the reciprocals add: one over C equals one over C one, plus one over C two. Right? Now, every dielectric question starts here: is the capacitor on a battery, or isolated? In each case, which stays fixed, the charge or the voltage? If it is isolated, the charge has nowhere to go, so Q stays fixed. If the battery stays connected, it holds V fixed. Decide this first, every time. Got it? Now let a slab of dielectric constant capital K fill the whole gap. The capacitance becomes K times C, the arrow meaning becomes. Isolated, V is divided by K; on a battery, Q is multiplied by K, clear?

5:00Example 1: zero-field point of 1 uC and 4 uC, 30 cm apart

Alright, Example one, from Set one. Charges of one microcoulomb and four microcoulomb are thirty centimetres apart. Where is the field zero? The Greek letter mu with capital C means microcoulomb, and c m means centimetres. Pause the video and try it first. Done? Okay, let's check. Both charges are positive. Many students answer the midpoint, fifteen centimetres from each. Is that right? No, that is the trap here. Between like charges the fields point opposite ways, so the zero is between them, but nearer the smaller charge, since the four microcoulomb one is stronger. Okay? Let the point be x centimetres from the one microcoulomb charge. There the fields are equal in size: k times one over x squared equals k times four over thirty minus x, the whole squared. Right? Take square roots of both sides. Thirty minus x equals two x, so three x equals thirty, and x is ten centimetres. With me? The course's shortcut agrees: root q one, over root q one plus root q two, times d, from q one; the tick-shaped sign is the square root. Root one over root one plus root four is one third, and a third of thirty is ten centimetres. Clear?

6:23Example 2: common voltage and energy lost in charge sharing

Now Example two, from Set three, and it is our puzzle. Here, a four microfarad capacitor at fifty volts is joined to an uncharged six microfarad capacitor. Find the common voltage and the energy lost. Mu with capital F means microfarad, and capital V means volts. Pause the video and try it first. Done? Okay, let's check. When the two are joined, what is conserved, the charge or the energy? It is the charge that is conserved, yes! Nothing escapes from the plates, so the total stays C one times V one, four times fifty, which is two hundred microcoulombs. Okay? This charge now spreads over four plus six, that is ten microfarads. So the common voltage is two hundred over ten, which gives twenty volts. Right? Now the loss, by the formula from the course: C one C two, times V one squared, over two times C one plus C two. Four times six over ten is two point four microfarads; fifty squared is two thousand five hundred. So one half, times two point four times ten to the power minus six, times two thousand five hundred, gives three millijoules, the m J on screen. With me? So, our puzzle. The energy is U equals one half C V squared. Before joining, one half times four microfarads times fifty squared gives five millijoules. After, one half times ten microfarads times twenty squared gives two millijoules. So energy is not conserved: three millijoules are lost, and our puzzle is solved. Got it?

8:08Quick revision: vectors vs scalars, Gauss, Q or V held fixed

So, quick revision. Fields add as vectors, potentials as scalars. In Gauss's law, the flux depends on the charge inside only, and Coulomb's law is for point charges. An isolated capacitor keeps its charge, a battery keeps the voltage, and sharing keeps the charge but not the energy. Now open Set one, and give each problem about two minutes. All the best!