Good morning. Today, I want to discuss two
or three very important topics. Let me put
down what I want to teach so that we will
see what we have to cover and what we end
up covering. I want to introduce the concept
of electric flux. Following that I want to
introduce the concept of 
divergence of a vector field and along with
divergence, I will introduce the divergence
theorem. From this we will obtain the most
important result namely Maxwell’s first
equation. The equation we are trying to work
towards is
divergence of flux is equal to charge density.
It is a very important equation and the root
to that equation is through a fair amount
of mathematics.
So, what I am going to try and do is minimize
the amount of mathematics by drawing lots
and lots of pictures. So let us make a start.
We already know a few things. We know that
for point charge, the electric field is equal
to the charge divided by 4 pi epsilon naught
times r minus r prime divided by mod r minus
r prime cube. This is Coulomb’s law and
it is the starting point of everything because
that is the only experimental observation
that we start with.
We want to take this observation and go from
this observation to something more useful.
So, let us make a start. Let us assume that
we have a co-ordinate system and we have a
charge Q at the origin. We know that the electric
field therefore point outwards on straight
lines in all directions. So, the electric
field is outwards in all directions.
Now, supposing I draw a sphere around this
origin. Let the sphere have a radius r. So,
it is a sphere of uniform radius r and I want
to calculate how much electric field is leaving
the sphere. So, what I mean by that is, I
want to, I want to integrate over the area
of the sphere. I want to calculate the electric
field d A. Now there is a problem in defining
this integral. The surface of a sphere is
not a flat plane. You cannot write d A as
d x d y. That is not what it is. Secondly,
the electric field is a vector; it has a direction.
So what exactly do we mean when we say area
integral of the vector field E over the area
of the surface? We have to do something because
the answer of this integral will depend on
how we define these different pieces. So,
what I mean when I say this is, I mean, take
a piece of this sphere. The electric field
is going right through the sphere outwards
because we know the electric field is radially
outwards. There is an area; so multiply the
area by the magnitude of the electric field.
That is what I mean here. So, what I am really
saying right now is integral of the area,
over the area, magnitude of the electric field
times d A.
Well, I know something about the electric
field. I know that the electric field is given
by Q over 4 pi epsilon naught r square; but
how do I write down area? Well, in a spherical
coordinate system, the coordinates on the
surface are defined by this angle theta and
if I drop a perpendicular down onto, so this
little area has an image on the x y plane.
This is x, this is y, this is z. Then the
angle with respect to the x axis is called
phi.
So the coordinate system in spherical coordinates
is r theta phi. Sometimes it is called psi.
So, if I use such a coordinate system, then
the area here is this distance multiplied
by this distance. Now, this distance corresponds
to drawing a circle at constant phi but varying
theta. In that case, this circle actually
goes all the way down to the South Pole and
comes back. This is what is called a great
circle. When we do geography, these circles
are very common. They are the circles of longitude.
Now such a circle, if you want to know what
this distance is, you draw lines from the
origin connecting the two ends of this region.
There is an angle between them. That angle,
I will call delta theta. The main angle is
theta itself. So, this length is delta theta
times distance. So it is r delta theta. You
would have learned this from your geometry.
If you take any circular arc, you know that
if this is theta, the distance on the arc
is r theta... So the same idea is, if this
is delta theta, this is r delta theta. If
you go to the other side, that side has a
projection and in this projection, theta is
constant. That is, this line is sitting at
constant theta but varying phi.
And if you want to know the length, here you
have to take the centre of this circle and
draw lines. The angle between these two lines
- that angle is delta phi. And so this length
is equal to this radial length times delta
phi. But what is that radial length? Well,
this angle is theta, this diagonal is r. So,
this length is r sine theta. So, it is r sine
theta delta phi. So, now I can write down
what d A is. It is multiplied by r d theta
times r sine theta d phi. So, this is d A
and this is E. So I am multiplying E by d
A.
Now, there is a very strange thing that happens.
You have learned it in school, you have learned
it in college; but it is still strange, namely:
there is r here, there is r here which means
r square, and there is r square in the denominator
which means r cancels out. So, when you do
this integral, integral over the area E d
A, it becomes equal to Q over 4 pi epsilon
naught integral over d theta sine theta d
phi. Now, what are the limits of that integral?
If you look at the diagram, this is x, this
is y, this is z. This is some general point.
The angle made with the z axis is theta. The
angle made with the x axis of the projection
is phi. So clearly, theta goes from 0 when
it is aligned to the z axis all the way to
the South Pole. So, 0 to pi. I work in radians.
So pi radians is the same thing as 180 degrees.
What about phi? When phi is along the x-axis,
phi is 0 and the maximum it goes, it goes
past the y-axis, past the minus x-axis, past
the minus y-axis and finally it comes right
back to the x-axis. So, that is 360 degrees
or 2 pi radians.
In scientific work, we always work with radians
because that is the natural unit for angle.
So the limits of this integral are known.
Theta goes from 0 to pi, pi goes from 0 to
2 pi. Now, this is a very easy integral to
do. The integral on phi is trivial because
sine theta does not depend on phi. So, that
becomes this 2 pi. This can be written as
2 pi Q over 4 pi epsilon naught integral 0
to pi sine theta d theta. Well, integral 0
to pi sine theta d theta is 2. I will leave
it to you to work it out.
So, this answer finally becomes integral over
the area E d A is equal to 4 pi Q over 4 pi
epsilon naught which is Q over epsilon naught
for all r. Regardless of how big a sphere
you work with, regardless of the radius, the
answer is the same. When you integrate over
the area E d A, the answer does not depend
on r, does not depend on how big a sphere
you use; it only depends on how much charge
was inside.
Now, this idea is related to another idea,
an idea that we have already talked about,
which is, if I put a charge Q, I can draw
direction of E in all directions. For example,
a direction E here, a direction E here, direction
E here, direction E here. Now I can draw arrows
and draw a line through the arrows. So, if
I take lots of arrows and just keep drawing
them and then draw a line right through the
arrows, this is called a field line. The place
where you would have seen field lines most
often would have been in magnetic fields where
if you put a bar magnet with a north pole
and a south pole, you draw, you draw magnetic
field lines and you draw them going from the
north pole to the south pole. The same idea
holds for electric fields. In fact, the dipole
is nothing but a magnet.
So, you can draw field lines and the field
lines leave in all directions. For a single
charge, the strength of the electric field
does not depend on the direction. It only
depends on distance. So, you get straight
lines in every possible direction that are
just leaving the charge. And if you say, that
the amount, the number of field lines that
will leave a charge is proportional to the
angle of the region... so if I, if I define
a certain angle, a cone, and I say I will
put one field line per degree square, then
I will have the whole lot of field lines,
approximately 360 degrees, in a certain sense,
squared. It is not quite 360 degrees squared;
but you will have a large number of field
lines going in all directions and if you ask
how many field lines will cut through a sphere
of radius r, the number of field lines is
always the same because no matter how big
a sphere you drew, all these field lines are
going to cut it and it is going to be the
same number; same large, but finite number.
In a certain sense, this integral is measuring
the same thing because the electric field
is closely related to these field lines we
are drawing. The electric field is actually
a measure of number of field lines per metre
square. If you drew it in this way, if you
made field lines very dense and then you drew
them and you took any area and you counted,
we found out how many field lines are coming
out and you counted them all up, the electric
field would be proportional to the number
of field lines per metre square and then you
multiply it by the surface area which was
metre square. So, the total number of field
lines came out and in a certain sense this
statement, this is the same statement as drawing
field lines. Keep that in mind because it
is a very important idea.
So, that was for a charge sitting at the origin
and a sphere centred on the origin; but supposing
I put my charge at the origin but I did not
centre my sphere at the origin. I put my charge
off-centre. This is the centre of my sphere.
It has a radius r but the charge is sitting
somewhere else, sitting inside the sphere,
but is not at the centre. Now the charge is
still giving field lines that are straight
but something different is happening here.
For example, this part of the sphere will
be close to the charge. Therefore, it will
see a strong electric field. This part of
the sphere on the other hand, we will see
a weak electric field because it is far away.
The question is, if we now did integral E
d A, what would it be equal to?
In a certain sense, this is the central problem
of this lecture. That is, what is E d A, if
I shift the charge off-centre? And the answer
comes fairly easily. Answer is as follows.
Supposing I take a little piece of area on
the sphere. I can make a cone connecting to
the charge. Now roughly, all parts of this
area are a distance...some distance, call
it rho. So, the electric field in this direction
E is equal to Q over 4 pi epsilon naught 1
over rho square. That is the magnitude, alright?
Now the problem is, this electric field is
pointing this way but if you ask what is the
surface pointing, surface is tilted. If I
look at this electric field, look along the
electric field, let us say this is the direction
of the electric field. This surface is looking
like this; surface is not perpendicular to
the electric field; surface is tilted. Parts
of the surface are further in the direction
of the electric field, parts are further behind;
and the question is what do we mean by E d
A when that happens?
In the symmetric case, it was very simple;
everything was pointing more or less the way
we thought it should. So we did not have to
worry about vectors and scalars. But now we
better worry about it. What we mean by E d
A is if I take this same surface and I extend
this cone and make the cone so that it is
a symmetric cone with respect to the charge,
then there is another area A. This is the
symmetric area. When I say E d A, I mean,
take the magnitude to the electric field,
take this symmetric surface area, multiply
them together.
How do we work out such a concept? It is not
so easy. Let me draw it again so that you
can see what I am trying to get at. I have
a charge Q. I have a surface area that looks
like this and I am saying I really want to
work out my result on a surface area that
looks properly symmetric for the cone. That
is, I have cut the cone not in the proper
symmetric way. I have cut it in some other
odd way. How do I decide what fraction of
this area is given by this area?
So this is d A. This is some d A. I will call
it d A normal. So, how do I figure out how
much d A normal is if I know what d A is?
Well, the answer is quite simple because we
have already done it in the previous problem.
What we will do is instead of working on d
A n, I will take a unit radius sphere and
then I will ask how much area does this cone
cut through this unit sphere. It cuts through
some area. I am going to call it d omega and
then this d A n is equal to d omega times
r square where r is the total distance. Why?
Because as you go further and further and
further, both dimensions of this area keep
increasing linearly with r.
So the area must increase this r square. So,
it is the same angular size. If I was sitting
here and I watched this region and this region
and this region, all look the same size but
the actual size is increased by r square.
The analogy I have for this idea is, supposing
I am sitting. I am here and I have put a camera
there and there is an anthill, very close
by. Far away there is a mountain. If this
anthill, if my position, the anthill position
and the mountain are correct, the base would
agree, the peak would agree; so much so, standing
here, the anthill and the mountain look the
same size. This is what is known as d omega.
This is the apparent size of an object. The
d A is the apparent size multiplied by r square.
Now, what is this other fellow, the tilted
fellow?
Well, supposing my mountain was not like this.
Supposing my mountain was like this. Even
then it would seem the same size because as
far as I am concerned, from this angle up
to that angle, the object is there. The fact
that the object is nearer here and farther
there does not affect the apparent size of
an object. Only if I know the depth do I know
anything about the size of the object. So,
that is the difference. d omega is a kind
of standard size. The normal area is d omega
multiplied by square of the distance; and
the d A is the tilted size.
So, what I do here is I say that well, let
me take, correct this. I used r here and rho
there. I should use a consistent unit. So,
I will use rho here again and rho here again.
So, when I work out my contribution to the
flux, I say E d A is equal to Q over 4 pi
epsilon naught 1 over rho square multiplied
by d omega rho square. Again, the rho squares
cancel. Very strange.
Now when I integrate, you will notice what
I am integrating over. Let me redraw the picture.
I have a unit radius sphere. I have another
bigger sphere and I have a charge somewhere,
Q. So I am looking at some cone from the centre
which is at a distance from this Q. This distance
is rho and I am saying that when I draw this
cone, there is a solid angle here and this
is what I call d omega. Now, this d omega
I can integrate over all possible angles.
So, when I do my integration, here is what
I get. I get integral E d A is equal to integral
and my d omega I do using the same kind of
integration ; 0 to pi d theta 0 to 2 pi sine
theta d phi of Q over 4 pi epsilon naught;
rho squares cancelled and d omega is nothing
but this, the same answer.
Let me review because I think it is a confusing
idea. In this picture, I worked out what happened
if you calculated integral E d A for a charge
at the origin, for a sphere centred around
the charge. It is an easy calculation and
when we do the calculation, we find it is
Q over 4 pi epsilon naught integral sine theta
d theta d phi, which this integral gives you
another 4 pi which cancels this 4 pi. So you
get Q over epsilon naught.
When I go away from the centre of symmetry,
my sphere is centred here, the charge is there.
What is happening is that, near points have
strong electric field but they have small
area; far electric points, far points from
the sphere have weak electric fields but large
areas.
How small and how large? The electric field
goes as 1 over rho square, area goes as rho
square. So this is independent of rho. Similarly,
this is independent of rho and so when you
integrate the whole answer, you get an answer
that depends only on Q. It does not depend
on rho and most peculiarly does not even depend
on where the charge is put. Wherever in the
sphere you put it, you get the same answer
and nothing in here actually cared about the
fact that this is a sphere. I was working
with what are called solid angles. So I could
have generalized this result and made it a
general shape. I would have got the same answer.
Now let us do one more calculation. This is
the last one and it gives us the result. So
we have done sphere charge in the centre.
We got integral d A equals Q over over epsilon
naught. You also had sphere charge of centre,
same answer. So, both of these give you the
same answer. Now the third problem to try
is sphere, but charge outside. What does E
d A look like now? Well, the same thinking
we did can be used here. Supposing I take
a tiny cone and puncture this sphere with
it. The cone has its point at the charge and
it has an apparent angular size of d omega
which would be equal to rho d theta rho sine
theta d phi, alright?
Now, this sphere cone punctures this sphere
at two points. There is a point where it enters
and a point where it leaves; where it enters,
the electric field is going into the sphere
and it is, if this distance is rho 1, the
amount of flux entering the sphere would be
Q over 4 pi epsilon naught 1 over rho 1 square
multiplied by size of this, which is d omega
times rho 1 square entering. Now, let us look
there at the dotted line. Flux is leaving.
How much is leaving? Well, it is Q over 4
pi epsilon naught 1 over rho 2 square into
d omega rho 2 square, right? This distance
is rho 2. So this statement is saying exactly
as much charge is entering. The rho 1s cancel.
The rho 2 cancel; exactly as much flux is
entering as flux is leaving. Total amount
of flux contributed by this charge is zero
because whatever is entering left, whatever
is leaving enter.
So if you do this integral now, you get is
equal to 0. So these are the three results:
if you put the charge in the centre, we worked
out trivially integral E dot d A or E d A
is Q over epsilon naught; if you put the charge
on the side anywhere, it was still Q over
epsilon naught; but if you put the charge
outside, it is zero. So we can we can make
this a kind of a general statement. For one
thing, it does not have to be a sphere. It
can be any kind of surface. A sphere is only
convenient for our thinking.
So what is our conclusion? Our conclusion
is that integral over the area...and I am
going to put a circle. A circle means integral
over the entire area of the surface, of any
surface S E d A is equal to one of two things:
0 if Q naught in volume Q over epsilon naught;
if Q in volume, should be a fairly obvious
statement. What we are saying is if you have
any surface and you put a charge, lines are
leaving the charge in all directions. If the
charge is inside the volume, all the lines
are leaving.
So when you integrate over the surface, you
get an answer. The answer is Q over epsilon
naught. If on the other hand the charge is
outside, for every line that enters, it also
leaves. So, there is no net charge, no net
contribution. So, E d A becomes zero. Since
this is a vector relation, we have to be a
little careful in how we write E d A. I have
been sloppy, intentionally sloppy, because
I did not want to get into that; but now we
better get into it. What we mean is that if
the charge Q is here, the electric field is
in the direction away from the charge E. Now,
the surface may be some other way, may be
like this. That is my d A. As I said, the
way to make this sense, you draw a cone, draw
a unit surface around the charge, find the
actual area cut by that cone.
There is an easier way of saying the same
thing. If you look at this area, supposing
the electric field, instead of being in this
direction were in the direction perpendicular
surface, then our E d A works out very nicely.
That is what we meant by E d A. Any way, we
meant the electric field is going right out
of the surface, straight up; and then you
just multiply E and d A you get the answer.
The problem is that how much of the electric
field should we use? And the answer again
is obvious. You should use the projected amount
of the electric field; either you project
the area along the electric field or you project
the electric field along the area. And we
know all about projection, we have been doing
it.
So really, this point should be really E,
the electric field dot; the little bit of
area taken as a vector. How can you take area
as a vector? That really is quite a complicated
idea. So, let us look at it. Supposing I have
a square. Let us say this is x, this is y
and vertically z. This distance is d x, this
distance is d y. Now, if I want the normal
direction to this surface, I know that this
is the z direction because it is in the x
y plane. z is naturally perpendicular to the
x y plane but there is another way I can say
it. I can say the normal direction; the direction
of this d S is perpendicular to all lines
in d S. By that I mean I have the square.
So, I can draw any line I like that lies in
the square; that is, lies in the x y plane.
Whatever I choose as normal must be at 90
degrees to all of those lines, must be at
90 degrees to all of those lines. I can choose
two particular lines: I can choose the x line
and the y line.
Now, if I want something that is 90 degrees
to two things, there is a natural operator
I know off, which is the cross product.
So if I take x hat cross y hat, that is the
direction that is perpendicular to the x direction.
It is also perpendicular to the y direction
and that is why the normal direction is along
the z direction.
Now the same idea can be used here, what we
call the normal is nothing but the cross product
of lines in that fall inside the d A; and
if you take any two lines of the d A, take
its cross product. That will give you a vector
in the normal direction; but this is actually
very useful because supposing I have a surface
and in that surface I have two vectors. So
my area is not at 90 degrees. My area is somewhat
tilted. So, there is an angle theta here.
This is my d A. Well, the area of this parallelogram
is the base into height and direction is normal.
That is what I want. But what do I mean by
base into height? If this is l 1, this is
l 2, base into height is l 1 l 2 sine theta
because l 2 sine theta is the height. So base
into l 2 sine theta is the area and the direction
is perpendicular to both l 1 and to l 2. But
if you look at l 1 cross l 2, the magnitude
of that is magnitude of l 1 magnitude of l
2 sine theta and the direction is, that is
90 degrees to both l 1 and to l 2, which is
outwards.
So, that is why when you look at any area
and you want to mean by area, it is the cross
product that comes in. The cross product naturally
defines area, the magnitude of area, and it
naturally points in the normal direction.
So, that is how we get our integral. It says
closed integral E dot d S over any surface
is equal to 0 if Q outside V, Q over epsilon
naught if Q in V and this is what we call
the divergence theorem.
Now we have to take one more step to reach
Maxwell’s equation. It is not a difficult
step to take but still has to be done. Now
if you take any volume, first of all, I have
to take actually two steps. So let me take
the simple step. First, supposing I had multiple
charges. So I had Q 1 Q 2 Q 3 Q N. Now I already
know that for any one charge this is going
to happen as volume integral E dot d S equals
zero. If I will call it E 1 Q 1 in V Q 1 over
epsilon naught, if Q 1 if Q 1 naught in V
and in V.
Similarly E 2, similarly E 3, etcetera, etcetera,
etcetera. But the electric field due to a
set of charges is the sum of the electric
fields. So the loop integral, sorry, the closed
surface integral total electric field dot
d S is really sum on all the charges surface
integral, sorry, E i 
dot d S is equal to sum on charges in V divided
by epsilon naught. By that I mean if I had
a charge Q i that was outside, that particular
charge would not give me any integral E dot
d S. We have already worked that out because
it is zero. Only those charges that are inside
the sphere would contribute to this integral.
So, it is the sum of all the charges that
are in volume V divided by epsilon naught.
Now, this is really the useful version of
Gauss’ law, the divergence theorem, and
we can immediately use it actually to get
answers. But there is another version of this
equation which is also useful and that version
looks as follows. Supposing I take this charge,
take this sphere that has many charges in
it and let us say there is a particular Q
i that is sitting somewhere. I look at a small
volume that contains that particular charge.
Now, what I know is if I do integral over
this delta V, sorry, delta S of E dot d S,
it is going to be equal to sum of charges
enclosed divided by epsilon, except there
is only one charge enclosed.
So, it is going to be equal to Q i over epsilon
naught. That is a very funny thing. If you
look at it saying that if I have lot of charge
everywhere and I look at a small part of this
box only, the charge inside that box is going
to give me this contribution. It is tough
that outside is not going to give me any effect
but if it is a very small box, I should be
able to write this in a different way.
Let us look at what I mean by that. I am going
to go to Cartesian coordinates. So I go to
a cube. When I say the integral over the cube
E dot d S, I mean integral over this surface,
this surface, this surface and behind the
back surfaces. So there are six surfaces over
which I will integrate. Let us say this is
x, this is y, this is z. Now, if you look
at...for example, the y z surface, we will
call this surface 1 and this is surface 2.
If I integrate over the y z surface, both
of them, what do I get? This is the value
at this point is x, the value at this point
is x plus delta x. Similarly, the value is
y and this is y this y plus delta y; this
is z and this is z plus delta z.
So, what I get is over this surface. This
integral becomes E of x y z times delta y
delta z and its flux is entering. I am going
to put a minus sign here. Now E is a vector.
So, which component is going to enter? Well,
the E x component will enter; the E y component
will go parallel; the E z component will go
parallel. So, E y and E z cannot contribute
at all; only E x can contribute. Now, what
about region surface 2? That would give me
electric field leaving. So, I am going to
put a plus sign again, E x; but the value
of x is x plus delta x y z. Again, delta y,
delta z - if I combine these two terms, I
can write this as first. Let me say, plus
4 other terms, this is equal to E x of x plus
delta x y z minus E x of x y z into delta
x delta y delta z plus 4 other terms.
I have to divide by delta x. So, do you see
what I have done? I have taken delta y, delta
z, common. I have multiplied and divided by
delta x and written out the other terms. What
do you get? You get that integral E dot d
S is equal to del E x del x times the total
volume plus 4 other terms; and I am going
to write out the 4 other terms. That is del
E y del y delta V plus del E z del z delta
V. Each of these is two terms, two surfaces.
So, there are six surfaces to a cube. This
piece 
is the scalar quantity and it is called the
divergence and in a certain sense what is
measuring is amount of outward flux leaving
that little volume. So, you can now relate
these two quantities to the previous question
we have which is loop surface integral E dot
d S is equal to charge enclosed divided by
epsilon naught. Now, charge enclosed in a
volume is typically written as the charge
density in the volume multiplied by the volume
because for interesting problems in electricity
and magnetism, we do not have point charges.
Point charges are only where we start. What
we actually have is smeared out charge, many,
many charges. So, we have so many charges
per cubic metre multiplied by the volume.
So, we now have a final equation. Divergence
E is equal to rho over epsilon naught. It
is the same equation as surface integral over
a closed surface. E dot d S is charge enclosed
divided by epsilon naught saying the same
thing; but that is an integral version of
the equation, this is a differential version
of the equation and this is the very useful
version of the equation. It is called the
divergence theorem or it is called Maxwell’s
first equation. It is this equation is so
important that we will keep coming back to
it in the next few lectures.
