Electric Field of Point Charges

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Six equal positive charges sit at the corners of a regular hexagon. At the centre, their fields balance, and the field is zero. Now I take away just one charge. Less charge, so still no field at the centre, right? Hmm, hold that thought. Soon you will know the exact size and direction of that field. Questions on electric fields come up often in JEE, so let's begin.

0:31What is an electric field?

So, what exactly is an electric field? Bring a small positive test charge near some charges. We call it q naught, q with a small zero below it. It feels a force, F vector. The field, E vector, equals F vector over q naught, the force per unit test charge. The little arrows on top mean both are vectors. Okay? So the unit is newton per coulomb, written on screen as N C to the power minus one. It is the same as volt per metre, V m to the power minus one. Clear? Turn it around: a charge q placed in a known field feels a force, F vector equals q times E vector. Simple, right? So let me ask you this. If we place an electron in a field, does the force on it point along the field? It does not, so be careful here! The electron's charge is negative, so q times E vector points the other way. When q is less than zero, the force F vector is opposite to E vector. This sign slip costs easy marks. Got it?

1:38Field of a point charge

Now, the field of a point charge q, at a distance r from it. E vector equals k times q over r squared, times r hat. Here r hat, the r with a little cap on top, is the unit vector from the charge to the point. Okay? And k is one over four pi epsilon naught, where epsilon naught, the Greek letter epsilon with a small zero below, is the permittivity of free space. In vacuum, k is nine times ten to the power nine newton metre squared per coulomb squared. Right? Now the direction. When q is greater than zero, the field points along r hat, away from the charge. When q is less than zero, the minus sign flips it, towards the charge. With me? Alright, what if there are many charges around? Do we simply add the sizes of their fields? We do not, because fields add as vectors! E vector equals E one plus E two plus and so on, with small subscripts for each charge, and three dots for the rest. Draw every arrow first, then add them, a vector sum. This is called superposition. Clear?

2:46Electric field lines

Next, field lines, our picture of the field. They start on positive charges and end on negative charges, or go off to infinity. Okay? Here is a question that comes up often. Can two field lines ever cross each other? They can never cross. At a crossing point, the field would have two directions, but the field at a point is a single vector. And in electrostatics, field lines never form closed loops either. Right? Where field lines reach a conductor, they meet it at right angles. Got it? And lines are crowded where the field is strong. Also, the number of lines leaving a charge is proportional to the charge; the curly sign on screen means proportional to.

3:31Equal charges at the corners of a regular polygon

Okay? Now, back to our puzzle. Put equal charges q at all the corners of a regular polygon, like our hexagon. What is the field at the centre? Yes, exactly, it is zero! By symmetry, all the fields cancel at the centre. So E centre, with the word centre as a subscript, is the zero vector. Clear? Now remove one charge. Here is a topper's trick: the full set gives zero, so the remaining charges give exactly the negative of the removed charge's field. No need to add them one by one! With me? The removed positive charge, on its own, would push away from its corner. So the answer points away from the empty corner, right? Oh no, careful, look at that minus sign! The remaining charges give the negative of that field, so for positive q, the field points towards the empty corner. And its size is k q over a squared. Okay? Here, small a is the distance from the centre to a corner, and for a hexagon, that equals the side. So taking one charge away made a field appear, pointing to the gap, and that solves our puzzle!

4:42Example 1: field 30 cm from 2 micro C, force on an electron

Let's try our first example. Find the field thirty centimetres from a two microcoulomb charge, and the force on an electron placed there. The Greek letter mu before the C means micro, ten to the power minus six. Pause the video and try it first. Done? Okay, let's check. First, let us change to metres. Thirty centimetres is zero point three metre, so r squared is zero point zero nine metre squared. Okay? So E equals k times q over r squared: nine times ten to the power nine, times two times ten to the power minus six, over zero point zero nine. The top is eighteen thousand, so this gives two times ten to the power five newton per coulomb. Right? Now the electron. Its charge has size small e, one point six times ten to the power minus nineteen coulomb. So F equals small e times capital E, which gives three point two times ten to the power minus fourteen newton. Clear? And the direction? The field points away from this positive charge, but the electron is negative. So the force is opposite to E vector, towards the charge. Got it?

5:57Example 2: +q, +q, -q, -q on a square, field at the centre

Now a harder one. Charges plus q, plus q, minus q and minus q sit at the corners A, B, C and D, in that order, of a square of side a. Find the field at the centre. Pause the video and try it, and here is a hint: pair up the opposite corners. Done? Okay, let's check. Each corner is half a diagonal away, that is, a over root two. So r squared is a squared over two, and each charge gives a field of two k q over a squared. With me? Now the directions. Plus q at A pushes away from A, and minus q at C pulls towards C, okay? Opposite charges on opposite corners, so these two fields cancel, right? They do not cancel, they add up! Away from A means towards C, so both point along A C, and the total is four k q over a squared. That was the trap, okay? Now look at the charges at B and at D, it is the same story. Both fields point along B D, towards D, and the total is again four k q over a squared. Right? Now, the diagonals of a square are perpendicular, the upside down T sign on screen. Two equal fields at right angles give root two times one of them. So E equals four root two k q over a squared. Clear? And the direction? It lies halfway between the directions towards C and towards D, so it is perpendicular to A B, pointing towards the side C D, okay? That's it, well done!

7:35Quick revision: kq/r squared, vector sum, polygon rule

So, quick revision. The field is force per unit positive test charge; for a point charge it is k q over r squared, away from plus, towards minus. Fields add as vectors, and F equals q times E, opposite to the field when q is negative. Equal charges on a regular polygon give zero at the centre; remove one, and you get k q over a squared, towards the gap. Now go and try the practice problems on this page. All the best!