Our Energy Sources, Electricity

Good morning. Today we shall complete a few

more examples on electric fields and then go on to the important concept of electrostatic

potential. So, let us first look at one interesting problem. Suppose you have two line charges.

This is coming out of a problem in your textbook – problem 2.21. You have two line charges;

each line charge has let us say, a charge of rho coulombs per metre. The

line charges are at a distance d apart. Now, we would like to know what is the force per

metre exerted on line charge 2; call this 2, due to line charge 1? Well, in our previous class, we derived what

the electric field was due to a line charge and the answer we got was by taking points

above and points below and integrating from z equals zero to z equals infinity, and what

we got was that the electric field was along the radial direction; it was proportional

to this rho. It was divided by 2 pi epsilon naught and it is scaled as 1 over r. So for

a distance d, this r becomes d. So, it is equal to…along the direction, radial direction,

rho over 2 pi epsilon naught 1 over d. So, this is the electric field. Now, this electric field acts on 1 metre of

charge because I want the force per metre. How much charge is there per metre? Well,

it is nothing but rho itself. So, the force per metre is equal to rho times the electric

field; both are vectors. I already have the electric field. So the answer becomes…the

force is along the radial direction and it is equal to rho square divided by 2 pi epsilon

naught d. So, if you keep two line charges and you…each of them has a charge rho per

metre, this is the kind of force per metre that they exert on each other; which means

that the total force between these two line charges is infinite. So, line charges are

actually very exotic things. You cannot actually create them in practical situations. However, if you talk about wires, say an overhead

wire, power line, well, those lines can be a few metres apart and distance of 100 kilometres;

and therefore such very long closely placed wires are actually a very good approximation

to this picture. So your overhead power lines exert force on each other when they charge

up to different voltages and exert different electric fields. Another example – I would like to know the

force, the electric field due to a disc. A disc has a radius ‘a’ and it is of a height…the

charge…the point where I want the electric fields at height ‘z’. The entire disc

has a charge, let us say sigma coulombs metre square. And I want to know what is the electric

field above this disc, z metres above this disc, and what is its direction. Last time

I did half of this problem. So, let me just repeat that half. What I will do is I will

take a ring. The ring has a radius r, it has a thickness d r. Now, every point on this ring is equally distant

from the point where I want the electric field, because every point has a distance that is

equal to z square plus r square, square root. This is just Pythagoras theorem, because it

is a right angled triangle. So, it is the…hypotenuse square is equal to base square plus height

square. However, the electric field due to each of these charges is pointing in a different

direction. In fact, the electric fields describe a cone. As I move the charge around in a circle,

the electric field moves in a cone around this point. So, how do I add it up and find a net electric

field? What I do is I take diagonally opposite points of charge. If I take two points of

charge that are diagonally opposite, then what happens is that the two directions in

which they point are…will tend to cancel in the x y plane, but will tend to add in

the z plane. Let me try and make that more obvious. I am going to draw the circle looking

from above. This is my ring of charge. I am considering a bit of charge here. The amount

of charge is sigma times the thickness, d r times the theta direction thickness which

is r d theta. At a point up here, the field that it will

cause will be actually pointing somewhere out. I am just projecting it down. So it will

look like this. For example, if I took a piece of charge here, we could point in this direction.

But now, if I took a point diametrically opposite on that ring, same amount of charge, sigma

d r r d theta, now what is going to happen? This charge is going to produce a force. It

is again up here. So, the force due to this is going to look outwards; force due to this

is going to look outwards; but if I project it down, the projected parts are going to

be equal and opposite. They are going to cancel. So, the only electric field that is left will

point straight out. That is, it will point in this picture in the upward direction. So,

what do you get as a result for the electric field? You get d E which is in the z direction;

is equal to this amount of charge sigma d r times r d theta divided by 4 pi epsilon

zero. This gives me the…times 1 over z square plus a square. Radius is, well, I can say

it is r square. This gives me the length of this arrow but I do not want the length of

this arrow; I want this length, the vertical length. So, there is an additional cos theta. I am afraid I am using the same symbol twice.

So, let me call this cos phi. Phi is this angle, whereas theta is this angle. So, now

I want to integrate this d E z around the circle. So, I get integral zero to 2 pi d

E z integrated in theta alone. The theta integration gives me 2 pi times sigma d r times r divided

by 4 pi epsilon naught 1 over z square plus r square times cos phi. Now, if you look at

cos phi, cos phi is nothing but z over the hypotenuse. It is equal to z divided by square

root of z square plus r square. So when you put it all together, what you

get is d E r…d E z is equal to sigma r z d r divided by twice epsilon naught z square

plus r square to the power of 3 halves. This answer still depends on d r and we want the

answer for the electric field due to a disc. That is we want to add up rings of different

values of r all the way from r equal zero to r equals a. So the electric field finally

is in the z direction and it is the integral zero to a of sigma r z divided by twice epsilon

zero z square plus r square to 3 halves d r. This is essentially the same equation.

It is the same equation that I came up with for solving the electric field due to a plane.

I ended the last lecture that way. That is because if I take the radius of this disc

to infinity, the disc becomes the plane. So if I replace this a by infinity, then I have

the answer for the plane. Well you can see, this is a very simple integral

to solve because you have got r d r up there and z square plus r square is your dependence.

So you define u is equal to z square plus r square; d u is equal to twice r d r and

you get the answer z hat. I can pull the z out. I can pull the sigma out, over twice

epsilon naught; 1 more factor of 2, integral z square to z square plus a square of du divided

by u to the power of 3 halves. How I got that was I removed the common…constant

pieces out, twice r d r is d u. So, r d r is d u divided by 2. The denominator has z

square plus r square to the power of 3 halves. z square plus r square is nothing but u. So

it became u to the power of 3 halves. This is easily solved. So you get, this is equal

to z hat sigma z over twice epsilon naught, u to the power of minus half divided by minus

half. The half cancels out; so you get minus u to the power of minus half going from z

square to z square plus a square. So the answer finally becomes

that the electric field is in the z direction. It depends on sigma z over twice epsilon zero

times 1 over z minus 1 over square root of z square plus a square. If you compare with

the expression for the plane, the plane is nothing but the disc with a becoming large.

When a becomes very large, z square plus a square is huge; and this huge number is in

the denominator, which means 1 over z square plus a square square root goes to zero. So,

this term will be missing and you will just get sigma z over 2 epsilon naught z, and z

itself will cancel out. So, this was the result I had given you earlier

which was that if you have a plane, an infinite plane of charge, whose charge was sigma and

you went a distance z from this plane and tried to find out what the electric field

was, well, it did not matter how far you went out. However far you went out, the answer

was always z hat sigma over 2 epsilon naught. Now, how can this be? There seems to be something

very wrong with this. How can it not matter how far from the plane you are? The electric

field is always the same. Well, there is a reason for it. The reason is actually not

that difficult. If you look at a particular theta from z, it strikes the plane on a circle;

and if you ask how much charge is there between theta and theta plus d theta, the bigger angle

is theta plus d theta, it will be the charge that is there in this ring. How much charge

is that? Well, it is 2 pi times this radius times the thickness. So, it will be…if this is theta and this

is z, this height is z sine theta. So, it will be…the d Q would be equal to 2 pi times

z sine theta times d of z sine theta. Now, what is the interesting about this is that

it is proportional to z square. So the amount of charge that is present increases the further

out you go; amount of charge between theta and theta plus d theta. The amount of charge

is not constant. If I were 1 metre away, I get so much charge. If I am 10 metres away,

between 30 and 31 degrees, there is 10 times more charge; no sorry, 100 times more charge. However, how far away is this charge? The

distance is z square plus z square sine square theta, is proportional to r square; square

root is proportional to z. You can pull out z square out of the square root. So, you get

an answer that is proportional to z. So the amount of force exerted by this charge force

goes like d Q divided by r square. Well, d Q is proportional to z square, r is proportional

to z. So, r square is proportional to z square equals constant. So, what it means is the further out you go,

the more of the charge of the plane you see. How much more? It quadruples for every doubling

of your distance. At the same time, the distances increased and therefore the effect of that

quadruple charge has become one-fourth. So, two effects cancel which is why no matter

how far away from a plane you go, the force is the same. So the answer is, force remains

constant. Now I hope these kinds of examples give you

a taste of how one does simple integrals to calculate the electric field. Now, in a certain

sense, Poisson’s equation is all you require if you knew where all the charges are because…sorry,

not Poisson’s equation, Coulomb’s law is all you require if you knew where all the

charges are, because Coulomb’s law tells you that the force is equal to 1 over 4 pi

epsilon zero integral rho of r prime d v prime divided by… So, if I knew exactly what charge density

was everywhere, I just do this integral and I get the answer; end of electromagnetic theory.

The only problem is, we usually do not know where charge is. Let me give an example. Supposing

you have a metallic ball and I put 1 coulomb of charge on this ball. Well, the charges

are free to move around because it is a conductor and we know that inside a conductor, charges

can move, which means the charge could be anywhere on the surface and it could be anywhere

inside. So, we cannot use this formula. This formula

requires us to know exactly where the charge is; whereas in practice, what we know is the

total charge and we know the shape of the conductors but we do not know exactly where

the charge is. See, even though we have done quite a bit in writing down Coulomb’s law,

we have not actually got to the really useful parts of electrical engineering. In electrical

engineering, the most useful components are inductors and capacitors and a capacitor is

nothing more than a metallic, pair of metallic plates which are kept close

to each other. So, in order to tackle this problem, we need

to introduce some new concepts. The first concept I want to introduce is work done in

order to move a charge. When I was studying in college, I think work was probably the

most difficult concept I ever encountered. Angular momentum was about equally hard; but

those two concepts drove me quite crazy. So, let me spend a little time trying to understand

this word. You should have learned it thoroughly in mechanics, but I did mechanics too. I did

not understand. See, the real confusion that comes is, as human beings, when we carry a

heavy ball or we carry a weight and we just stand there…I am just holding this chalk

and just standing there, we feel we are doing work. We feel our body is burning fuel with

oxygen, producing carbon dioxide so that our muscles can hold up this piece of chalk. Therefore,

we are doing work and we are quite correct. Our bodies are doing work. However in physics, the word work means something

else. The word work does not mean holding up a stationary object. If you have a stationary

object that is not moving, no matter how much force there is on that object, no matter how

heavy the object is, we are not doing work. A weightlifter, when he lifts the weight and

reaches the top, is not doing any work. I mean, he will probably hit you if you told

him that he is not doing any work but he, according to physics, he is not doing any

work. He is just standing there. All the work he did was not lifting that weight, but keeping

it stationary above his head; may have caused lot of problems to his muscles, but he was

not doing any work. The weight was not going any higher. So, work according to physics and according

to engineering means motion of objects against applied fields; and the words are

all important. You must be moving. If you are not moving, there is no work. It must

be an applied field. For example, supposing I have an object that is under some strain.

May be, I have got a compressed spring and I move the whole object. I am not doing any

work. If I am…if I move the object but the object’s internal stresses are the only

forces present, I am not doing any work. The forces have to be applied from outside and

the motion has to be against this field. For example, supposing force due to gravity

is downwards and my weight lifter has got his weight and he starts walking. The poor

fellow is doing a lot of…exerting a lot of effort, but he is not doing any work because

the weight is neither going up nor down; and up and down are the only directions that are

against the applied force. Going sideways does not get you anything. It may get you

an Olympic medal, but it will not get you any work. So, now that we know something about work,

how do we calculate work? We have to use that definition. So, there is motion and a motion

is against a field. So, if I move, if I have a gravitational field m g and I move in some

direction, the part of that movement, that is sideways, does not count. That part of

the motion which is like the weight lifter walking around does not do any work. It is

only the part of the motion that is parallel to the direction of the field that does the

work. So, the amount of work that is done, d W is

given as the force dot the vector distance travelled. So, if this is d s and this is

force m g, you take the dot product of the two. If you take the dot product of the two,

then, only the part that is parallel to the force counts. The part that is 90 degrees

to the force does not count and that contributes to d W. This is the work done by the force

on the object and when work is done on an object, usually

it implies kinetic energy. So, the object starts moving faster and faster

and as you know from this concept we can say, since total energy is conserved, the object

has lost potential energy

and the potential energy was nothing but m g h and gained kinetic energy. So, this is

exactly what we have to do in electricity and magnetism as well, because we have forces.

Now the force is called the electric field, instead of gravitation and we have to find

out how much work charges do when they move in an electric field – it is the same idea.

But let us see how the idea works out. So now I have a charge capital Q. I know that

if there is any other charge small q, the force on that charge is along the line joining

capital Q to small q and pointing away from capital Q. Now I decide that I want to move

this charge from the point A to a point B. It can be on any path, whatever other I like.

I want to know, is there any concept of work done? Is there any concept of kinetic energy

of this charge and does this charge gain or lose any potential energy? We know that in gravitation the concept is

there. If this was the Earth, this was the satellite, the force would point the other

way and if the satellite decided to move like this, it would have picked up kinetic energy

and it would have lost potential energy. Is there a similar concept in electricity and

magnetism? Well, at every point on this orbit, we can work out what the electric field is.

We just have to draw straight lines. The lines are getting larger as you get closer to the

charge. Now you can see that you are not, you are

hardly moving in the direction of the charge of the field. So if you move on this line,

the amount of work that you will do is only the dot product of the field and the distance.

So, we can work out. I will call this 1, 2, 3, 4, all the way up to point n. So, total

work done on the charge…that is, this is the work that would have made the charge move

faster and faster. It would have given kinetic energy to the charge; is equal to r 1 2 dot

q E…r 1 let us say, plus r 2 3 dot q E of r 2 plus, etcetera. I keep adding them all

up till I get to r n minus 1 n dot q E of r n minus 1. I have said r 1, r 2, etcetera.

I could have easily said r 2, r 3; some point, the electric field on some point along this

arrow. If I add this all up, this should be the total

work done and there is a short hand way of writing this sum. Just as we changed Coulomb’s

law and made it look like an integral, we can make this also look like an integral.

We can say that this whole route that we went on, we are going to call it a path and typically

symbol c is used – contour, c for contour. So I will say that I am going to do an integral.

It is a vector integral along this root c and on this root I am going to do q times

the electric field. That is the force, dot d r. Why do I say this? If I take r 1 2, r

2 3, r 3 4 up to r n minus 1 n, I have actually built up this entire curve. So in a certain

sense, if I want to know how much work I have done by moving on this curve, it is like summing

up all these little pieces; and you know that a sum consisting of very small steps is nothing

but an integral. But it is a different kind of integral. It is what is called a line integral.

It is not your standard kind of integral. Let us write one just to see what we mean

by the difference. For example, if I have x and y and I wanted

to know going from zero to 1, what is the value of f of x, which is equal to integral

zero to 1 sine x d x. That is the kind of integral we have learned in mathematics, very

straight forward. You learned what it is. It is integral of sine x of minus cos x and

you put it between limits and you get that this is equal to 1 minus cos of 1. So, this is an integral along the real line

but I could equally well ask, I want to know the integral of going around a circle. I want

to know the integral as I go along the circle of integral, let us say, sine of x square

plus y square d theta. Or, if you like…let us leave it like that – d theta. Now, what

am I doing? I am saying that I have a circle and I am dividing the circle into lots of

little pieces. Each of these lengths is nothing but r d theta which is equal to d theta because

r is 1. It is a circle with radius 1 and I want to integrate a quantity called sine of

x, square root of x square plus y square. Well, square root of x square plus y square

itself is nothing but r square which is equal to 1. It is a constant. So, I have taken a

trivial example. The point is this also is an integral but

it is not an integral on a real line. It is an integral on some curve in x and y and more

generally, it could be a curving x, y, z, t. It could be any general curve. If you talk

about intervals of general curves, those are what we call line integrals and the particular

example that I gave which is q E dot the variable which I choose d l I say is an example of

line integral; and what it means is if I have a curve, I break the curve into many small

pieces. On each piece I find the vector corresponding to d l. At each piece there is also an electric

field. So, I can form E dot d l. That is now a number and I will just add it up. Sum I

equals 1 to N q E i dot d l i; and if I make these d ls very very small, this sum becomes

an integral. That is what it means and this is so important. I think it is important that

you think about it and you properly get comfortable with it. We will keep coming back to this.

It will come back to haunt us unless we are quite comfortable with

the idea. But what is the result now? The total work

done on the charge which is in a certain sense equal to minus of change…sorry, in potential

energy, we assume total energy is conserved. The energy gained, kinetic energy gained by

the charge must be the loss of potential energy of the charge. It is equal to the integral

from the starting point, the ending point of q E dot d l. Now that is interesting. It is just a definition.

There is one more interesting thing about it. If I have a charge Q and I have a curve

A to B, at any point the electric field points in the radial direction, it points away from

the charge. The distance I move d l does not necessarily point in the radial direction

but you can always break d l into two parts. There is a part which I will call d r and

a part that is the rest of it; and its obvious that no matter what we do, it is only the

part d r that can do any contribution to this integral because dot product only counts the

portion of d l that is parallel to E. So, you can rewrite this equation as going

from integral A to B but q E r d r. That is at each point, instead of keeping the full

direction of l, I am taking advantage of the knowledge that E is always pointing in the

r direction. So the only…the r part of l matters. So I only keep d r. What does this

do for me? Well, if I have got rid of all the other directions, I do not have to keep

A and B as points; I can say going from r A to r B. And furthermore, I know an expression for

E r. So, let me write that out. It is r A to r B q times capital Q over 4 pi epsilon

naught. E r is nothing but 1 over r square d r. Integral of 1 over r square is minus

1 over r. So it becomes q, capital Q, over 4 pi epsilon naught times 1 over r A minus

1 over r B. It is a very strange result. What it says

is, supposing instead of going this way, I had done this. I had looped around this Q

many times. I had gone way out, come back, I had gone this way and come back. No matter

what I did, if I finally landed up in B and if I initially started from A, the answer

does not change. The answer only depends on where I started and where I ended. It does

not depend on how I got them; and we have meant such kinds of integrals before. Or,

if we have not, we should have. If you have done any course in thermodynamics,

you know of things called state variables entropy and variables like that that you will

encounter when you do the second law of thermodynamics. They do not depend on the path either. When

you are actually changing the state of a thermodynamic system, you go through some complicated path.

But the initial and final states are defined based on where you landed up. What we are

saying here is something similar. There is some quantity here that…if the

amount of work done on the charge which did not care how you got from point A to point

B; only where you started and where you ended. Now, as it turns out, since we are doing electricity

and magnetism, we are not so much interested on the total work done on the charge but we

are interested in potential energy gained by the charge which is equal to minus of this

quantity, because it is the total work done by the charge; this will give you kinetic

energy. So, if you want potential energy, you have to take minus of that. So it gives

me q Q over 4 pi epsilon naught 1 over r B minus 1 over r A. And typically when we are

talking about point charges, we have a point charge Q and we want to use this formula,

we will take the other point, starting point, very far away. If you take it very far away,

then 1 over r A goes to zero. So you can say, this is equal to q Q over 4 pi epsilon naught

1 over r B. So, you have now got an expression for a state

that depends only on the final location of the charge and it represents potential energy

gained by the charge, when you come from very far away to that point. As before, this is

potential energy. We would like to know potential energy per coulomb. So we define a special

variable, a special field called electrostatic potential. The symbol we use is the Greek

letter phi and the Greek letter phi is nothing but potential energy gained divided by q itself.

That is the potential energy gained per coulomb and it is defined for a point charge source

Q over 4 pi epsilon naught 1 over r. Let me recapitulate. The important thing about

this potential function is precisely that it does not depend how you got to the point

r. However you got to it, this potential function is the same. So, we have…we originally had

electric field which was Q over 4 pi epsilon naught 1 over r cube r. But now, you have

come to a different function phi. It is equal to minus integral to r from far away E dot

d l which is equal to Q over 4 pi epsilon naught 1 over r. This is Coulomb’s law and

this is what we have just worked out. Now, there is a huge advantage to this function.

We may have invented it but this is a vector, this is a scalar, which means that this quantity

involves 3 separate numbers for every point; this quantity involves only 1 number. So, phi is much simpler than E. The other

thing is phi represents something. phi represents potential energy of charge, potential energy

per coulomb of charge. Now you worked all this out for a single charge Q but we know

that any interesting problem is going to have many charges and in fact if you have a metal

we do not even know where the charges are. So what use is this? First things first. We know that electric

field is equal to 1 over 4 pi epsilon zero integral rho of r prime d v prime divided

by r minus r prime cubed and multiplied by r minus r prime. So, there is superposition

working for us; for every little piece of rho d v prime, I can calculate potential,

which means that the potential due to all these different pieces of charge phi of r

must be equal to 1 over 4 pi epsilon naught integral rho of r prime d v prime divided

by… Once again, potential is a much simpler function.

Electric field involves a vector and it involves 1 over r cube times r. It is going as 1 over

r square; potential just involves a simple 1 over r. It is a very very simple function

and in a certain sense, this potential contains everything because if you knew this potential,

then if you went to a nearby point and measured potential and took the difference, then it

must be true that this is equal to E dot d r with a minus sign. It is just coming from

the definition because this is integral from far away to r plus d r and this is integral

from far away to r. And you know, if you take two integrals, you can take the difference

by same r to r plus d r of E dot d l with the minus sign; and for very very small lengths

of integral, it is nothing but the integrand multiplied by the difference of the limits.

So, that is this. So it means that if I know this potential

function and I take the potential function at very nearby points, I have effectively

got a feel for electric field because the difference between nearby points of potential

is nothing but the electric field itself. Of course there is a d r in there but it is

nothing but the electric field. So, you have got a case where you have simplified

the problem. We started with an electric field which was a complicated vector field and we

have now devised a scalar field, much simpler field. It has got a physical meaning to it.

It is potential energy of the charge and it does…it gives you the electric field, after

all. I will continue next time.

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