Mixed Problems: Set 1 (Charges, Fields and Gauss's Law)

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Two protons sit one femtometre apart, and each carries a really tiny charge. So the push between them must be tiny too, right? Hmm, hold that answer. Today is our first set of mixed problems on charges, fields and Gauss's law, and problems like these come up often in JEE Main. So, let's begin.

0:26Idea 1: two protons, one femtometre apart

So, idea number one is our puzzle: the repulsion between two protons, one femtometre apart. The f m on screen stands for femtometre, okay? We treat protons as point charges in vacuum, and use Coulomb's law. F equals k times q one q two, over r squared; q one and q two, the q with a small one and a small two, are the charges, and r is the distance. And k is nine times ten to the power nine, in newton metre squared per coulomb squared. Clear? Each proton carries the charge e, one point six times ten to the power minus nineteen coulomb, and r, one femtometre, is ten to the power minus fifteen metre. With me? Ten to the power minus nineteen is really tiny. So the force must be tiny too, right? Let us calculate. F equals nine times ten to the power nine, times one point six times ten to the power minus nineteen, squared, over ten to the power minus fifteen, squared, and that bottom is ten to the power minus thirty. Okay? And F comes out to about two hundred and thirty newtons, a repulsion; the wavy equals sign means approximately. The tiny r, squared in the bottom, makes it big. So tiny charges do not always mean a tiny force. Got it?

1:49Idea 2: flux through one face of a cube

Now, idea number two. Here, a three microcoulomb charge sits at the centre of a cube; mu, the Greek letter before the capital C, means micro, ten to the power minus six. Find the flux through one face, okay? Let me ask you this. Gauss's law needs a closed surface, and one face is not closed. So can we use Gauss's law here at all? Yes, on the whole cube, which is closed. The integral sign with a small circle means a sum over a closed surface of E vector dot d A vector; the little arrows mark vectors, and d A is a tiny piece of the surface. That equals q in, the charge inside, over epsilon naught, the epsilon with a small zero, the permittivity of free space. Right? So phi, the Greek letter for flux, through the whole cube is q over epsilon naught. The charge is at the centre, so the six faces are identical, and each gets one sixth: q over six epsilon naught. Clear? For phi face, the phi with the small word face: three times ten to the power minus six, over six times eight point eight five times ten to the power minus twelve. That is about five point six five times ten to the power four, in newton metre squared per coulomb. Got it?

3:10Idea 3: inside a uniformly charged sphere

Now, idea number three. The field at the surface of a uniformly charged solid sphere of radius ten centimetres is nine times ten to the power five newton per coulomb. Find its charge, and the field five centimetres from the centre. Okay? In the figure, panel d shows E against r: dark blue for a solid sphere, orange for a shell, and we skip the other panels. On or outside the sphere, it acts like a point charge Q at the centre: E equals k times capital Q, over r squared, the falling dark blue curve, marked proportional to one over r squared. That sign, like a stretched alpha, means proportional to. Right? At the surface, r equals capital R, the radius, zero point one metre, so R squared is zero point zero one. So Q equals E times R squared, over k: nine times ten to the power five, times zero point zero one, over nine times ten to the power nine. That gives one microcoulomb. With me? Now five centimetres from the centre, half the radius. Half the distance, so by one over r squared, four times the field, right? No, no, wait, that is the trap! Five centimetres is inside, and one over r squared holds only outside. Inside, E equals k Q r over capital R cubed, so E is proportional to r. That is the straight dark blue line, rising from zero. Clear? So at R over two, the field is half the surface value: one half times nine times ten to the power five, which gives four point five times ten to the power five newton per coulomb. Got it?

4:56Example 1: null point between 1 uC and 4 uC, 30 cm apart

Alright, example one. Charges of one microcoulomb and four microcoulomb are thirty centimetres apart. Where is the field zero? Pause the video and try it first. Done? Okay, let's check. The charges are in the ratio one to four, so split thirty centimetres the same way, which gives six centimetres from the smaller charge, right? Oh no, careful, the field goes as one over r squared, not one over r. Both charges are positive, so between them their fields point opposite ways and can cancel, nearer the smaller charge. Okay? Let x be the distance in centimetres from the one microcoulomb charge. With q one as one microcoulomb and q two as four microcoulomb, k q one over x squared equals k q two over, thirty minus x, squared. Right? Cancel k and take square roots: thirty minus x, over x, equals the square root of q two over q one, which is root four, so two. With me? So x equals root one, over root one plus root four, times thirty, which gives ten centimetres from the one microcoulomb charge. So the gap splits in the ratio of the square roots, one to two, and if you got ten centimetres, very good!

6:17Example 2: field at centre of square, one corner empty

Now a harder one, JEE level. Charges q sit at three corners of a square, each side of length a, and we need the field at its centre. Pause the video and try it first. Here is a hint: fill the fourth corner. Done? Okay, let's check. Each charge is equally far from the centre, so each gives the same size of field. So the answer is three times one field, right? No, no, wait, fields are vectors, and these three point in different directions, so we cannot just add their sizes. Here comes the trick. Okay? Imagine one more charge q at the empty corner. Four equal charges, placed symmetrically, so their fields cancel, and E vector is zero at the centre. Clear? So the three real charges must give minus E four vector, where E four, the E with a small four, is the field of that imagined charge. One field to find, not three. Right? The diagonal is a times root two, so half of it, from corner to centre, is a times root two, over two, which is a over root two. So E equals k q over, a squared over two, which gives two k q over a squared. With me? For q greater than zero, the imagined charge would push away from the fourth corner. The minus sign flips it, so the field points towards the empty corner. So the answer is two k q over a squared, towards the empty corner, and if you got that, you did a perfect job!

7:49Quick revision: Coulomb, Gauss, null point, empty corner

So, quick revision. For point charges, F equals k q one q two over r squared; square r carefully. Use Gauss's law on a whole closed surface; inside a uniformly charged solid sphere, E is proportional to r. For like charges, the zero divides the gap in the ratio root q one to root q two. And with one corner empty, take minus the field of the missing charge. Now try the practice problem on this page. All the best!