Good morning. Last few lectures we have been
developing the theory for electrostatics and
we have reached up to Poisson’s equation
and energy in the field; and I did a problem
last time. Today I plan to solve several problems
which will work out the various parts of the
equations we have derived. Last lecture I
had started a problem and stopped in the middle.
Let me repeat it. You were given a sphere
of charge; the charge density was rho of r.
That is to say, the charge did not depend
on theta. It did not depend on size it just
depended on the radial distance. The sphere
has a radius a, and you are told that electric
field 
E of r is a constant. It is a constant inside
sphere.
So in other words, if I plot it, the electric
field, the electric field is up to this distance
constant and then it falls away. This is what
I have been given. I do not know the value
E naught; but in terms of this information,
I want to pin down what is rho of r. I am
also told, total charge is equal to some Q.
So I want to solve this problem. Basically,
I want to find E of r and I want to find rho
of r and I want to find phi of r.
So in order to do this, you need to use the
equations of electrostatics and since the
problem is spherically symmetric, the first
idea that comes to us is, let us use Gauss’s
law. Well, if you use Gauss’s law, what
you get is that integral 0 to some r of 4
pi r square rho of r d r. This 4 pi r square
is the surface area of the sphere of radius
r. So, 4 pi r square; d r is the volume of
a shell multiplied by the charge density rho
of r. This is equal to Q enclosed which must
be equal to epsilon naught times E naught.
That is what Gauss’s law tells us. At least
it tells us for 0 less than r less than a.
If you took a Gaussian surface that was bigger
than the sphere, then you would get a different
equation. The epsilon naught...sorry, E of
r times...sorry, here I have made a mistake.
Let me just repeat charge enclosed is equal
to epsilon naught times E of naught times
the area and the area is nothing but 4 pi
r square.
Now, when you go outside the sphere, you have
that E of r times 4 pi r square, is equal
to charge enclosed divided by epsilon naught.
But this charge enclosed is integral 0 to
a 4 pi r square r prime square rho of r. What
I mean by that is, I have a radius r and I
want to know what is the electric field coming
out at this radius. The surface area of that
radius is 4 pi r square times the electric
field. So this is the flux. The flux is equal
to charge enclosed divided by epsilon naught.
The charge enclosed there is no charge here.
The charge is only inside the sphere of charge.
So the...only this portion corresponds to
Q enclosed. So that is, this equation integral
0 to a 4 pi r square rho of r d r, luckily
we are given this information. We are told
that the total charge of the sphere is Q.
Then there is 1 over epsilon naught. So outside
the sphere, we have E of r is equal to Q over
4 pi epsilon naught r square. It is a very
strange result which again I am sure, you
are familiar with. The result is following.
If I have a spherically symmetric charge distribution,
the charge distribution could be anything.
For example, I could have lots of charge out
here, relatively little charge in the middle;
but as long as it is spherically symmetric,
if I measure the electric field further away
than this sphere, the electric field, I feel
there is the same electric field. I will feel
if this entire charge was concentrated at
the center. Because you look at this, this
is Coulomb’s law. So regardless of the detailed
shape of rho of r, the answer is Q over r
square is the normalization constant.
So, we already know quite a bit. Let me put
down what we know. We have figured out what
electric field is because up to r equals a,
the electric field is constant, E naught.
Beyond r equals a, electric field is falling
off as Q over 4 pi epsilon naught r square.
Now of course we do not know if the electric
field is continuous there or not, but in the
lack of knowledge, I have just drawn it as
continuous.
So, I have almost already obtained the electric
field. Now, let us go further on and see if
we can obtain other quantities. Now going
back to this equation here, when I use Gauss’s
law, inside this sphere, I got an equation
that said the enclosed charge up to a radius
r. Let me put r primes here to indicate this
is dummy variable. The enclosed charge between
radius 0 and radius r, this charge alone gave
me epsilon naught E naught times the surface
area. This is not very useful because I really
want to know what rho of r is.
Whereas, this is giving me what is called
an integral equation in rho of r and that
is not an easy thing to solve. The way out
of this is to convert from Gauss’s law to
the divergence theorem. The divergence theorem
told us divergence of E is equal to rho over
epsilon naught. Let us apply that and see
if we get more information. In spherical coordinates,
divergence is 1 over r square del del r of
r square E of r and this divergence is equal
to rho of r divided by epsilon naught.
Now, this divergence looks rather strange.
Let me just remind you where it came from.
Divergence in Cartesian says del del x of
E x plus del del y of E y plus del del z of
E z, that is what divergence says; but when
you are talking about non-Cartesian geometries,
something happens which is quite different
from this. As you move in radius, what is
happening is that the area between these angles,
this angle, will be some delta theta. The
length of this arc itself is changing. That
is because this distance is different from
this distance and the arcs are the functions
of the distance from the centre of curvature.
So what happens is this area is small and
this area is large. So, even as you go in
the direction r, the area of your little box
is changing. This does not happen in Cartesian.
In Cartesian if I drew the... I will draw
a cube or a cuboid, so the area here and the
area here are the same. So the result is,
if I was trying to work out divergence, divergence
I got from taking the electric field coming
in and the electric field going out; and taking
the difference in the amount of flux entering
and the amount of flux leaving, since the
areas were the same, the difference in flux
was equal to the same area times E of x plus
delta x minus E of x. And that is why the
area drops out. So you get a derivative that
acts only on the electric field itself according
to the directions.
But in a situation like this, what happens
is this equation is wrong. Delta flux becomes
the area at x plus delta x times electric
field at x plus delta x minus area at x E
at x. So you see, the area itself depends
on x and how does the area depend on x? Well,
you know that for a sphere, if you fix the
solid angle... So, if you fix delta theta
and delta phi and you just ask what is the
area, the area goes as r square and that is
what this r square is. It is saying that it
is not...you cannot pull the area out. The
electric field is scaled by the area over
which it is being measured.
So, you have this r square E which is representing
nothing else but area times E. So, that is
where this equation comes from. You can read
the appendix of the textbook where it derives
it for you, not very well I must say, but
still it derives it; and I do not think the
derivation is very important. So we have this
equation and we would like to solve it.
Now, solving this equation is easy. Let us
take the r square to that side. This E of
r is known to be E naught. Differentiate this.
What you get is E naught times del r square
by del r which is twice E naught r. So what
you get is twice E naught divided by r is
equal to rho of r divided by epsilon naught;
E naught is known, r is known, epsilon naught
is known.
So you can calculate rho of r. 
So 
it says that the electric field may be constant
but charge density is not constant. In fact,
if I plotted charge density on this curve,
I will draw it as a dashed line. The charge
density is going to look like a 1 over r curve
and then it drops abruptly to 0 beyond a.
This is what rho of r is. We have calculated
rho. So now, let us try and use rho to calculate
Q. As you can see, we assumed Q to define
the electric field outside r equals a, but
this Q must satisfy the following equation.
Q must be the volume integral of rho.
In other words, it must be rho of r times,
4 pi r square, d r. This is d V and this is
rho but we know what rho of r is. Now we have
already calculated it. So we can substitute.
So it becomes integral 0 to a times twice
E naught epsilon naught over r times, 4 pi
r square, d r 1. r will cancel and we are
left with 8 pi E naught epsilon naught. I
have taken all the constants out; integral
0 to a of r d r which is a square over 2.
So the final answer becomes epsilon naught
E naught times 4 pi a square 
and that is not surprising at all. Because
we already had from Gauss’s law that surface
integral E naught or surface integral E d
S was equal to Q over epsilon naught which
would have said that E naught times 4 pi a
square was equal to Q over epsilon naught.
So they are saying the same thing; but in
a certain sense, it is saying something useful.
You see, I had drawn two curves here. I had
drawn E naugh t as constant and I had drawn
this curve and I joined them but I had no
real proof that they were to be joined. It
could easily have been that the curve was
here and jumped up and became this.
But that is not possible because by calculating
charge from the rho that we calculated given
E naught, the electric field just outside
the sphere turns out to be exactly an agreement
with what you calculate just inside the sphere.
So it means that...in fact, the curve is continuous.
Now, from this we can calculate potential.
That is easy. All you have to do is integrate
phi of r is equal to minus integral E d l
up to r and the reference point is infinity.
So it becomes integral r to infinity E r d
r. The region outside the sphere we already
know because it is nothing but Coulomb’s
law.
So phi of r is equal to Q over 4 pi epsilon
naught r, for r greater than a, for r less
than a. The potential phi of r minus phi of
a must be equal to the amount of work done
against a constant electric field. In other
words, it is equal to E naught times r minus,
sorry, a minus r. 
Since E naught itself is nothing but Q over
4 pi epsilon naught a, this is Q over 4 pi
epsilon naught a times a minus r. So phi of
r finally becomes Q over 4 pi epsilon naught
r, if r is greater than a, and it becomes
Q over 4 pi epsilon naught a plus Q over 4
pi epsilon naught a times a minus r, for r
less than a.
So this is the solution we need, if you want
to solve the problem of a charge distribution
that has constant electric field. The important
thing to realize is it is not natural for
a sphere to have constant electric field.
To achieve this constant electric field, we
actually have to have a very singular charge
distribution. Only if the charge distribution
is very spiky will your electric field be
uniform.
Let us do another problem. This problem is
very straight forward but worth understanding.
Supposing I put a charge Q in the middle of
a hollow sphere and the hollow sphere has
an inner radius a, it has an outer radius
b. So this region is air, this region is metal;
and this region is air. So the metal is in
between two regions of air. Furthermore, the
metal region is neutral, that is no net charge.
Now, what I want to know is what charge distribution
can exist inside this metal that will minimize
the stored energy, minimize the energy in
the field due to this charge Q. Now, let us
back up a bit. If I have only a charge Q,
then I knows that the potential is Q over
4 pi epsilon naught r. The electric field
is the radial electric field which is Q over
4 pi epsilon naught r square. So the energy
in the field would have been integral mod
E r square times epsilon naught over 2 d V.
That is, energy density in the field is epsilon
E square over 2 and I would have taken that
integral from 0 to infinity. There is something
sneaky about what I am doing because, if you
remember how we calculated this epsilon over
2, we said that the energy to bring the first
charge in was 0. Yet I have got a single charge
in there and I have calculated energy in the
electric field. So this energy that I am talking
about is the energy that exists around a charge.
That energy was already there. I did not have
to do work to create it. When I put charge
together, to make a point charge of charge
Q, this is the energy in that field. So now,
if I have this energy in a point charge and
I now put a spherical conducting shell around
it, how is that energy going to change? I
am going to do this problem in two ways and
it is worth understanding the two different
ways.
First way, I will use Gauss’s law. Well,
I will...say I have a charge Q and I have
an inner radius a, and an outer radius b,
and I will put an imaginary Gaussian surface,
a sphere of radius r. Now, this is inside
the metallic region and we have already discussed
that inside any metallic region, I cannot
have electric field. So surface integral E
dot d S is 0 because the surface is entirely
inside the metal.
Metal cannot support any electric field. So
that means Q enclosed is equal to 0. Furthermore,
there is another calculation you can do which
is, I can draw two of these Gaussian surfaces
and I can talk about the region between them.
In that case, it says surface integral S 1
minus surface integral S 2 is equal to 0 because
each of them is separately 0, the difference
of them is also 0. So the charge enclosed
inside any such sphere is also 0.
This is the problem. There is no charge in
the metal but there is no charge enclosed.
The charge must be somewhere and the only
place the charge can be is on the inner wall
of the metal. So there is a charge minus Q
on the inner wall and if you draw a Gaussian
surface outside the cylindrical sphere, then
total charge enclosed...well, the charge on
the metal is 0 because metal has no net charge.
Charge at the center is Q. So out here, Q
enclosed is equal to Q. So from Gauss’s
law, what we find is the electric field E
r as a function of r is equal to 
Q over 4 pi epsilon naught r for r less than
a. It is equal to 0 for a less than r less
than b and it is equal to Q over 4 pi epsilon
naught r, for b less than r. It is just coming
out of the application of Gauss’s law.
If I apply Gauss’s law inside the air, I
find the first result. If I apply Gauss’s
law inside the metal, I get the second result.
If I apply Gauss’s law in the air outside
the sphere, I get the third result. So Gauss’s
law solves the problem for us, but I want
to look at this problem slightly differently.
Now we had derived last lecture that the energy
in any cloud of charge cloud described by
rho of r is given by volume integral rho of
r phi of r d V the factor of one half in front.
We derived this by bringing charge slowly
from infinity and building up this cloud and
as a function of time, as we brought the charge
the potential at every point built up linearly.
And the amount of work we did was equal to
the potential at any given time t.
So what happened was that, in a certain sense
we were measuring the area under the triangle
and that is where the factor of 2 came. Because,
this rho times phi in a certain sense measures
the area of the rectangle because this axis
is time and not rho but the way I define my
derivation charge was brought in in time.
So, in a certain sense this also represents
rho. So, this was the derivation we did and
we obtained that the total energy was one
half rho times phi d V but in our current
problem. We do not know rho. So, we would
like to get a better expression for this.
Let us see if we can do that.
Now we have one expression from Gauss’s
law which tells us that divergence E is equal
to rho over epsilon naught. So, I can substitute
for this rho by epsilon naught times divergence
of E. Let me do that. So, this becomes equal
to one half volume integral epsilon naught
divergence of E times phi d V. Now this divergence
can be combined with phi. The reason is, phi
times divergence of E, can be written out.
It is equal to phi times del E x del x plus
phi times del E y del y plus phi times del
E z del z.
Now I know from my calculation that when I
have a product of terms and I take the derivative
of the product, this is one of the terms it
comes out. So, I can write each of these as
del del x of phi E x minus E x del phi del
x plus del del y phi E y minus E y del phi
del y plus the third term del del z phi E
z minus E z del phi del z. That is because
if I take the derivative of this it gives
me this term plus this term.
Now this term this term and the corresponding
z term will give me. I will take the epsilon
naught out. So, these 3 terms will give me
divergence of phi times E because, you can
see that is what it is phi times E the x component
is phi E x the y component is phi E y the
z component is phi E z and this term will
become minus E dot grad phi. Well, we are
almost done because I can now use a divergence
theorem on this piece.
And here is what I get. Your book has the
same derivation. So you can check with it
what you get is epsilon naught over 2 surface
integral phi E dot d S applying the divergence
theorem here plus epsilon naught over 2 volume
integral E dot E d V. That is because minus
grad phi is nothing but E. If we go far away
enough phi times E is going to go to 0. So
this term you can drop. And so this integral
which involves knowing both charge and potential
can be reduced to a simpler form. It is equal
to volume integral epsilon naught E square
over 2 d V.
And the interesting thing is you can give
an interpretation to this. Here the energy
in the field was due to the fact that we brought
charge in against a potential. But here, the
charge is no longer visible. The potential
is no longer visible. There is only a field
here and we are saying that at every point
in space due to the presence of field, you
have a certain amount of energy present epsilon
naught E square over 2. This particular form
is very convenient and we are going to use
it just now to solve our other problem.
But it will keep getting used so keep it in
mind. I want to say supposing, I knew that
there is some charge distribution rho in a
metal, then I can apply Gauss’s law again
and what would I get, I would get that E of
r is equal to Q over 4 pi epsilon naught for
a less than r… sorry b less than r and for
r greater than less than a that is for regions
inside and regions outside its nothing but
coulomb’s law.
In the region of the metal I am going to assume
some charge density I do not know what it
is. So, it is going to be Q plus integral
a to r rho of r prime d V, okay? Divided by
4 pi epsilon naught r square that is what
Gauss’s law will give us. Because it is
going to be charge enclosed and what is the
charge enclosed its Q plus. Really I should
write this as 4 pi r square, d r. I was writing
it as dV to make it clear integral a to r
rho of r prime 4 pi r prime square d r prime.
This is for a less than r less than b. This
is completely general. If I knew rho, this
is true. But let me write down the energy
associated with this field. I will call it
u it is going to be equal to integral 0 to
a E r square epsilon naught over 2 d V plus
integral a to b epsilon naught over 2 d V
plus integral b to infinity E r square epsilon
naught over 2 d V. So, there are 3 parts the
part that corresponds to the inner air region.
The part that corresponds to the middle region
and the part that corresponds outside and
regardless of what ever charge, I put down
for the metal this region and this region
are fixed.
That is the energy in the field is not a function
of what I chose for the charge distribution
inside the metal any charge distribution gives
me the same coulomb’s law in region one
and region 3. The only thing that the charge
distribution can change is region 2. So, I
can write u is equal to u 1 plus u 3 these
2 regions plus u 2 and this is fixed and this
is function of rho the second thing. That
is very obvious from this integral is that
I am talking about something that is always
positive or it was 0 E square cannot be negative.
So, when I take the integral of something,
that is either positive or zero, the integral
is always positive or worst case zero. So
I have that the total stored energy is equal
to some fixed part plus one more part that
is either 0 or greater than 0. If you have
studied anything about gravity gravitational
systems and friction and all those kinds of
mechanics problem, you know that systems like
to reach a state of lowest energy.
If you want to ask what is the lowest state
of energy for this system, the lowest state
of energy would be if u 2 were 0 that would
be the lowest energy state, it cannot be negative.
So, best it could be 0, but what does it mean?
If it is 0, it means that E r equals 0 for
all a less than r less than b E r is 0 everywhere
inside the metal which is nothing but what
Gauss’s law told us.
Gauss’s law gave us this result. But, the
result of Gauss’s law is really saying nothing
more or less than that if you allow an electrostatic
field problem to stabilize, if you allow the
charges to reach wherever they want to go
subject to constraints, then the charges will
arrange themselves in a minimum energy state.
It is a very important idea and you will find
that a lot of problems can be understood if
you realize that ultimately electrostatics
is about minimum energy states.
Okay, let me do a third problem now and this
third problem is fairly interesting. I have
a ring of radius a the ring has a charge lambda
coulomb’s per meter, this ring, the charged
ring has a central centre of symmetry is the
z axis and at t equals 0. I have put a charge
q and I give this charge a velocity in the
z direction velocity v naught the question
is what happens, okay? What happens to the
charge in time due to the electric field,
let us work it out since the charge is on
the z axis. It will remain on the z axis because
the forces are all symmetric.
So, there is no force trying to send it away
from the z axis. So, we need to know the force
on the charge at any value z. I can work out
the potential phi of z is going to be equal
to. I am at a distance square root of r square
a square plus z square. That is the length
that is this it is given by Pythagoras theorem
z square plus a square. I have one over 4
pi epsilon 0 and the charge the total charge
which is 2 pi a times lambda.
Now this is one case where you can see how
easy it is to work with electrostatic potentials,
we took a lot longer to calculate the electric
field. Whereas, the electrostatic potential
I could write down by just inspection once
I have got phi, I can calculate E the electric
field is along the z direction E z of z is
equal to minus del phi del z is equal to 1
over 4 pi epsilon naught and I will put this
here 2 pi a lambda. Then I have to take the
derivative. There is a minus sign coming from
here and there is a minus sign coming from
the derivative. So, those 2 signs get cancels
out.
So, I get half over a square plus z square
to the power of 3 halves times derivative
of a square plus z square with respect to
z which is 2 z the 2 cancels the half. And
so I get 2 pi a lambda over divided by pi
epsilon naught z divided by a square plus
z square to the power of 3 halves. So, the
electric field assuming that this lambda is
positive is pushing away from the ring and
its pushing away from the ring in both directions
and the electric field along z is 0 at the
origin. If you plot it E z of z, I will plot
it on this plot itself in this figure itself
it will be something like this.
So, it reaches a maximum the maximum value
it reaches is approximately where z is of
order a, that is where z over a square plus
z square becomes the maximum and then it falls
off. And you can see it falls off as one over
z square because for large z. You can neglect
the a square. So, it is z over z cubed z square
to the power 3 halves is z cubed. So z over
z cubed is one over z square. So it is coulomb’s
law very far away. But the electric field
actually grows as you go close, to the come
close to the ring.
Now let us assume lambda is greater than 0,
okay? It is perfectly general there are 2
cases one is q greater than 0 if q is greater
than 0 the charge starts moving and as it
reaches any point it keeps seeing and accelerating
electric field. The force on the charge is
equal to q E z also greater than 0 is the
force along z. So you have m d v d t along
z is equal to F z is greater than 0 you started
with v z that is greater than 0 and it keeps
increasing. So, this is a kind of switch system
where you are in z or in time I should say
it started with a velocity v naught and it
increased and saturated this particle with
charge q escapes. By that I mean the charge
starts leaving speeds up speeds up and goes
away to infinity, it does not return.
Now what happens if q is less than 0? Now
you can see F z which is equal to q E z is
less than 0. Therefore you start with some
velocity. You start slowing down you reach
some maximum amplitude maximum height z and
then you start coming back then you go to
a maximum negative height and you come back.
And so you expect oscillations can we see
that well let us write out the force equation
force equation says m d v z d t is equal to
q E z which is given by we already have the
form for q.
So, m d v d z this is the force rate of change
of momentum is equal to some constants which
are negative lambda is assumed positive q
is negative. So, this whole thing is negative
times a function of z. This is the function
of z you must be familiar with this kind of
equation when you have done the simple pendulum.
This is the standard way of solving such problems
which is multiplied both sides by v z.
So, you get m v z d v z d t is equal to q
E z which is a function of z times v z but
v z is nothing but d z d t 
m v z d v z d t can be written as d d t of
one half m v z square that is rate of change
of kinetic energy. The other side can be written
as well. I wont write it out yet it is equal
to the same expression q E d z d t. Now let
me integrate both sides with respect to time,
that is, I will just multiply through by d
t so what do I get.
I get that d d t of one half m v z square
is equal to minus q d phi d z d z d t which
can be written as minus d d t of q phi. So,
this is your kinetic energy this is your potential
energy. So, you can combine the 2 together
you get d d t of one half m v z square plus
q phi equals 0 or something which is very
obvious i can define a quantity E called m
v z square over 2 plus q phi of z equals constant
this would be my total energy 
this is kinetic energy this is potential energy.
So clearly what is happening is that the total
energy of the particle is remaining constant
as the particle moves in z it can exchange
kinetic energy for potential energy or potential
energy for kinetic energy. But the sum of
the 2 must be a constant this is extremely
similar to Kaplarian orbits - very similar
to gravitational orbits, is exactly the same
problem. So, also the problem of a pendulum,
now given all this what can we say about the
orbits initial kinetic energy was v naught
it was velocity. So, initial kinetic energy
was m v naught square over 2 initial potential
energy was q phi at z equals 0 which we can
work out here.
We already have the form for q of z the final
velocity is one half m v z square and the
final potential is q phi of z. What this really
means is supposing I want to look at orbits
that manage to reach infinity at infinity
this becomes 0. The potential energy due to
the electric field keeps reducing as you go
further and further away. So it is equal to
0 when you reach infinity this piece becomes
what is called escape velocity. The same idea,
as escape velocity for a satellite leaving
the earth, there you would be talking about
gravitational potential energy and you would
be talking about kinetic energy of the satellite.
So, this is the velocity I would not escape
velocity is probably the wrong term or say
it is a terminal velocity or asymptotic velocity.
This is my starting velocity and this is my
starting potential. So, what it means ultimately
is, I am inside a potential. Well, because
my particle is being attracted back by the
charge, that the charged ring, so I can draw
a potential energy curve in z the potential
energy curve in z says that very far away
the particle has 0 potential energy no charges
have any effect very close.
I have particles of opposite charge they actually
have negative potential energy this is my
potential energy. I now take a particle with
some kinetic energy. Let us say I took so
much kinetic energy. This would be my one
half m v naught square if I take it, then
this is my total energy. My system conserves
total energy E equals one half m v z square
plus q phi of z is constant. So, as the particle
moves in z this is not perfectly parallel,
but it should be the total energy of the particle
does not change.
So, the particle initially has a strong velocity.
But as at most further and further in z this
velocity keeps reducing or rather it is kinetic
energy keeps reducing. And finally the kinetic
energy reaches 0 when it reaches 0 the particle
has reached its highest point at that point
it has no velocity left it just has potential
energy. So, it turns round and falls back
so the particle goes up goes up turns round
comes back down goes up turns round and does
an oscillation.
If, on the other hand, I had more energy the
points were at turns or further apart, if
it has less energy the points were the particles
turns, turns back or closer apart. But supposing
the particle had too much kinetic energy,
if it had kinetic energy greater than this
amount, then even no matter how far away it
goes there is always some amount of velocity
left. Some amount of kinetic energy left and
this is the particle that has greater than
escape velocity with respect to the hook.
So, you will get trajectories where particles
leave come back and oscillate and you will
have other trajectories where the particle
just goes away. All of this is very analogous
to mechanics is very similar to what you see
in a pendulum. It is very similar to what
you see in satellites and it is very similar
because the equations are very similar and
it is always important to know the different
parts of physics have the same equations because
it makes it easier to study.
