In the last module consisting of about five
lectures, we have laid the foundation or
mathematical foundation for getting into the
electromagnetic theory. We will now begin
with a discussion of electrostatics. As the
name suggests that this is phenomena
associated with charges which are not in motion.
So, the we will be talking about electric
field and consequently or a potential which
arises from such a field. We begin with the
basic principal of electrostatics, which is
a charge q it attracts or repels other charges
depending upon the sign of the charge, and
the force between charges is given by what
is
known as Coulomb’s law.
.
So, if you look at this diagram you will realize
that there is a charge q 1 I have kept at
the
position r 1 and there is a charge q 2 at
the position r 2. The origin is absolutely
arbitrary,
because the form of the law does not depend
upon choice of origin. The force between q
1 and q 2 is proportional to the product of
the charges q 1 into q 2 and is inversely
proportional to the square of the distance
namely r between them. The direction the
.direction of the force is along the line
joining the two charges. It is and there is
of course,
a multiplicative constant which is in SI units,
it is usually written as 1 over 4 pi epsilon
0
where epsilon 0 is known as the permittivity
of the free space.
I am assuming that these charges are interacting
in vacuum that is with no medium.
There being there, what happens when there
is a medium in which these charges are
located is something which we will be talking
about much later in these lectures.
.
Now, if F 1 2 is the force on the charge 2
due to the charge 1 then by Newton’s third
law,
the force 1 due to charge 2 is just the negative
of that. That is the famous action reaction
principle, and we have F 1 2 is equal to minus
F 2 1.
..
So, let us summarize what are the properties
of the Coulomb’s law force. Number one to
notice that, the force is inversely proportional
to distance and it is, well second thing is
that if you look at the form then it it is
clear that the force is repulsive if it is
between like
charges and is attractive. If the charges
happened to be dissimilar that is, one is
positive
the other one is negative. The another characteristic
of this charge is, it is a 1 over r
square force, which is essentially a long
range force. Long range forces are those whose
range is essentially infinite. That is the
force really never becomes 0 except in at
infinite
distance.
The other point to notice, that the force
is central force. The characteristic of the
central
forces that it is a force whose magnitude
depends only on the distance between the two
charges and the second thing is its direction
its direction is along the line joining the
two
charges. Along the line joining the two charges
can imply that it is attractive or repulsive
depending upon the mutual signs of these two
charges. The constant or proportionality
which in standard international or S I units
is written as 1 over 4 pi epsilon 0. The 1
over
4 pi epsilon 0 to a great deal of precision
is 8.9874 into 10 to 9, but for all practical
purpose we can take it as 9 into 10 to 9 Newton
meter square per Coulomb square. And
the epsilon 0, itself is the 8.854 into 10
to minus 12 Coulomb’s square by Newton meter
square.
.Now, before we begin the process of electrostatics,
let us look at this force little more in
detail. It turns out that in the nature, we
have basically four types of forces. They
are they
go by the name four fundamental forces of
nature.
.
The weakest of the force is what keeps our
solar system, the planets, all the planets,
the
sun everything at their place and it is because
of the gravitational force this is an
attractive force like electromagnetic force.
However, it is of infinite range. It is a
long
range force. However its strength is very
small. Now, in a relatively speaking the,
there
is a force which is known as the strong force.
I will come to a discussion of that a little
while later. But assuming that strong force
is has a unit 1 then the gravitational force
has
a magnitude which is 10 to the power minus
39 times that of the force which binds the
nuclear the nuclear on together inside a nucleolus.
So, it is an extremely weak force the
next weaker force again in this course will
have not much to do with it, is what is known
as a weak nuclear force.
This is a force which is responsible for beta
decay and it has a range which is fairly small
10 to the power minus 18 meters and it is
called weak nuclear force and its strength
if
gravitational force has strength of 10 to
minus 39. The weak nuclear force has a strength
of 10 to minus 6. Therefore, the stronger
than gravitational force, but you know it
is still
a weak force. The third force is the one with
which we are involved in this course which
.is the electromagnetic force of course, in
this part this module will be talking mostly
about electrostatics. What is electromagnetic?
About it we will come back much later.
So, electric magnetic force has a relative
strength of 1 over 137 1 over 137 is a number
which is the magnitude of the fine structure
constant e square over c h cross h cross
being the plank’s constant. So, the electromagnetic
force has a strength relative strength
of e square over C h cross. C is the speed
of light h cross is a plank’s constant and
this
number to a great deal of accuracy is one
over one thirty seven e is the electronic
charge.
.
So, it is something like 10 to minus 2 of
the order of 10 to minus 2, but the strongest
of
them all is what is known as the strong nuclear
force. You all know that inside a
nucleolus there are protons and neutrons.
Neutrons are neutral objects where as protons
are positively charged objects. Now, in a
nucleolus these are bound together. In spite
of
the fact that the protons all have similar
charges and the electrostatic or electromagnetic
force between them is repulsive and neutron
has no charge at all. So, and the
gravitational force is the only other force
that could come in for neutral objects and
it is
extremely weak force, but nevertheless the
nucleolus is bound together.
Now, nucleolus is bound together by a very
special force and its name is strong force
strong nuclear force. It exist between neutrons
protons, protons protons, and neutron
neutron. It is an extremely short range force.
Its range is of the order of 10 to the power
minus 15 meters., which is also called a fermi,
which is also called a fermi and that is
.typical nuclear dimension. Now, this force
is I am taking its strength as equal to 1.
So,
compare to this force strong nuclear force,
the electromagnetic force is about 1 by 100
I
mean 200 of magnitude smaller.
.
Now second thing is that there is a problem
that we come across which I will be briefly
touching upon little later and it appears
that if you go from the classical theory to
the
quantum theory which is known as the quantum
of field theory knowledge of it is not
important for our purpose. But it turns out
that the force between two objects any two
objects is mediated by a third party, which
is normally known as the carrier that is the
crude, explanation of that is, that supposing
you are looking at the interaction between
let
us say two charged bodies, then the picture
is this two charged bodies are continuously
exchanging particles and there by remaining
in touch with each other and the fact that
the
force is propagated from or force exist between
two objects is because of the fact that
they continuously exchange particles and this
particles are known as Bosons.
They incidentally, Boson's the name is after
the famous Indian scientist Satyendra Nath
Bose and so this Bosons are exchanged. Now,
for reasons that I cannot go into in this
course, that if a force if the force between
two objects is of finite range. Now, we already
talked about strong force and the weak nuclear
force. If it is a finite range then the
exchanged particles have masses. That is their
their particles which have mass. On other
hand if the force is of long range namely
of infinite range, as it is for the gravitational
.force and the electromagnetic force. Then
the particles which they exchange happen to
be mass less.
For our purpose the electromagnetic force
is a long range force. So, therefore between
two charged objects there is always an exchange
of a mass less particles or mass less
Boson's. the name of this mass less Boson
is a photon photons are actually quantum of
light, but again we will be probably talking
about it in another course. So, and the photon
is a neutral objects without any mass. So,
basically our picture our picture of the force
between to objects is that they are continuously
exchanging photons and normally
photons are represented by the letters, letter
gamma.
.
So, continuously they are exchanging photons.
Now, another problem that I would like
to point out in this context is that the existence
of this inverse square law force. In long
range force that is perfectly as long as as
long as the charges are static. The however,
supposing the charge, one of the charge let
us say moves. Now, imagine these two
charges are located at great distance. How
does the force between them instantaneously
change?
..
Now, this thing is called action at a distance.
In physic action at a distance means that
the
force if one of the object changes its position
let us say, the force on the second one
because the distance has changed it changes
instantaneously. Now, this is very difficult
to understand because according to special
theory of relativity propounded by Einstein
no signal, no information can ever travel
with a speed greater than the speed of light
in
vacuum which is 3 into 10 meter per second.
Therefore, how does it work that as when a
object moves this information is essentially
instantaneously transmitted to the object
on
which it is exerting a force.
Now, it is in this context that one introduces
the concept of a field associated with a
charged object. We have discussed in detail
the fields scalar and the vector fields, but
think of it very crudely in the following
manner. Imagine it is a crude picture. Let
say
that I am talking about two objects with each
of which I associate a medium and this
medium let us say, is tightly bound to this
two objects. Now when I move one of the
particles it deforms that medium, it deforms
that medium, because the charged body
object is tightly bound to this medium which
we are talking about.
I am not talking about a material medium on
this moment I am just this is just to fix
your
ideas. Therefore, this deformation can propagate
and ultimately be communicated to
other objects on which this one is exerting
a force. Now, instead of talking about a
medium we say that with every object a field
is associated. This field is the
.electromagnetic field and when an object
moves the field associated with that object
changes and this information this information
or deformation of the field if you like
progresses with the speed of light. The electromagnetic
theory is completely consistent
with the special theory of relativity.
.
So, let us processed and define the electric
field. That, this is something which we have
discussed earlier and a, the the field that
is associated with a charged object is a vector
field. In this case we are talking about an
electric field. Now, in order to fix to our
ideas
about what is an electric field consider a
small charge a a minute charge I will call
it test
charge. I have a charge, let us suppose this
charge is q.
..
I do not care, how bigger small that charge
is and around surrounding this I have my
electric field of this charge q. Now, what
I do is I put a small charge, let us call
it q test at
a distance let us say r. Now, this charge
as I told you is infinitesimally small. This
is this
is the required. It is the required because
there is field due to the charge q and if
you
bring in a test charge, this test charge will
have its own field which will result in
changing the field of this charge q itself
and this process will go on. Now, if you bring
in
a infinitesimally small charge, we assume
that its field is not strong enough to
significantly change the electric field due
to this charge q.
Now, the we define the field due to charge
q as the ratio of the force that is felt by
this
test charge q. In the, a divided by the magnitude
of the test charge, in the limit of this test
charge going to 0, the magnitude of test charge
going to 0. So, the electric field of the
source charge which I called as q .There is
the force exerted on a unit test charge the
formal definition is E is equal to limit q
test going to 0 of force f on the q test divided
by
the q test. Now, obviously such a definition
does not depend upon the magnitude of the
test charge q itself, q t itself.
..
Now, so if according to this definition if
the electric field at the point P happens
to be E
then if you put a charge. Now, this electric
field is due to some source or sources. I
am
not specifying now what has given rise to
this electric field. Now, suppose you put
a
charge q at this point P, then a charge q
at that point will experience a force given
by q
times E. Now, let us suppose let us suppose
the electric field is generated by a q 1 which
is at the position r 1 then the electric field
at the point P. Remember that the, if there
is a
charge q at P the force between q 1 and q
2 is given by q 1 q 2 divided by the distance
square, the square of this distance and for
the electric field I divide by the charge
test
charge I am putting at P.
So, therefore the electric field at the point
P due to the charge q 1 located at r 1 is
given
by and the point P is at the position r 1
over 4 pi epsilon 0. This is that term with
permittivity q 1 and r minus r 1 that is the
direction of this vector divided by r minus
r 1
cube. You notice I have written r minus r
1 vector at the top and the q at the bottom.
So,
that the dimension actually is 1 over r square.
..
Now, I come to what is known as a super position
principle. Now super position
principle is that is the source of the electric
field is due to multiple charges many
charges, supposing I have got a charge q 1
at r 1 q 2 at r 2 etcetera etcetera. Then
so this
picture for instance talks about only two
charges, but this I can generalize it to any
number of charges. I will charge q 1 at the
position r 1 q 2 at the position r 2 and I
am
still looking at what is the field at the
point P. Now, notice that electric field is
a vector
therefore the charge the q 1 gives rise to
a force the field, which is directed like
this
along the line joining P and q 1 and it has
some magnitude.
Similarly, the force the charge q 2 at P gives
rising the electric field which is directed
like this. You do a vector addition so the
result field that P due to q 1 at r 1 and
q 2 at r 2
is the vector sum of the forces, exerted on
a test charge on a unit test charge kept at
the
point P. Now, if I generalize it to multiple
charges then this becomes E the electric field
at the point p is equal to 1 over pi epsilon
0 some over i q i which is the i x charge
located at the position r i. Vector r which
is, vector r is the position vector of the
point P
with respect to our origin minus r i by r
minus r i cube. Now, this is known as the
super
position principle.
Now, we assume that the super position principle
is valid for the electric field that is the
effect of multiple charges is simply the linear
sound of the effects due to individual
charges. Fair enough. So, we have talked about
what happens to the electric field due to
a
.single point charge or a multiple charges
located at discrete points in space. Now,
I can
generalize it to include continuous charge
distribution. So, let us look at what is meant
by
continuous charge distribution, supposing
I take a curve and in this arbitrarily shaped
curve.
.
There is a charge which is uniformly distributed.
The uniformly distributed charge is has
a charge, density linear charge density lambda.
In principle this lambda could depend
upon the position on that curve, but we have
assumed that the charges are uniformly
distributed. So, if you take a charge element,
a length element delta l, along that curve
then the amount of charge in that line element
delta l is delta cube which is equal to
lambda times the length delta l. So, one defines
the charge density lambda, linear charge
density lambda as limit of delta l going to
zero delta q by delta l. Now, remember that
in
principle this lambda could vary from point
to point, but I have taken for simplicity
that
lambda is a constant.
This is not implied by this definition. Now,
let us look at how does one express the
electric field the electric field due to a
linear charge distribution. We have talked
about
super position principle. Super position principle
says that if I have many charges then
the electric field at a point due to collection
of charges is equal to the sum of the electric
field due to the constituent charges. Now,
if I have a continues charge distribution,
the
summation goes over to an integration. So,
what I have is this if you refer to this picture.
.This is some arbitrarily chosen origin depend
upon origin. So, take for example, a small
length element delta l. This is the curve,
delta l along the curve.
The amount of charge in this is given by this,
the length element I am calling it delta l
prime and the prime will be my index on the
curve so delta l prime. Therefore, the charge
is lambda times delta l prime of this little
infinitesimally small length element. Now,
I
am interested in calculating the electric
field at a point P located at the position
r. So, I
connect this. This is the vector r I, which
means this gives me the vector position of
point
P with respect to the i f element charge length
element and then I simply add out or in
this case I do an integration. Add a 1 over
4 pi epsilon 0 lambda lambda times d l lambda
times d l prime is the amount of the charge
there. That replaces the q i in our earlier
expression and the strength of the field which
is vector r minus r prime which is just the
r
i prime vector divided by r minus r prime
cube and the integration is over the entire
length there.
Now, remember that even if you take a small
enough, even if you take a small enough
length. They are literally very large number
of charges there, because they discreet
charges that I have in a material are actually
electrons which give rise to the electric
field, but however whatever I am talking about
should not be done at the atomic level
because there are very few electrons and this
continuous limit that I am talking about is
absolutely not applicable there. Now I can
I can extend this, I can extend this to a
distribution of charge on a surface which
is in two dimension.
..
So, here I have given you shown you an irregular
surface. There is an irregular surface
and and there is charge distributed there.
Once again, how do I define a surface charge
density, which I will indicate by sigma. So,
what I do is I look at a small area element
that small area element, I will call it as
d s i prime again because prime is the index
which I am using for coordinates on the surface.
So, if the amount of charge in that area
element happens to be delta q, then I define
a surface charge density sigma as the limit
of
delta Q by delta s as the surface element
delta s goes to 0. So, the total charge would
be
integration of sigma d s over the entire surface.
Now, to find the electric due to such a surface
at a point P, which is located at the
position vector r is exactly we proceed the
same way. Suppose, this element d s i is
located located at this position here, position
some r prime. Now, what I am interested is
this vector and therefore, the field at this
point is given by there is a small error in
the
expression for the electric field given in
this slide. So, let me write it down.
..
The electric field at the position r is given
by 1 over 4 pi epsilon 0 integral over the
surface. Now, sigma which in principle could
depend upon the position r prime d s
prime, so that is your charge element and
the electric field we know varies as inverse
square. So, vector r minus r prime divided
by r minus r prime cube. So, this is the surface
charge distribution. What about the volume
charges?
.
Now, once again the volume charges are this.
So, this is a rho d V. I take a small volume
element d V and the once again I am interested
in calculating the field at the point P,
.therefore the amount of charge that is contained
in this is rho d V prime. d V prime is the
volume element there and once again I used
the same expression.
.
The electric field at the point r is given
by 1 over 4 pi epsilon 0 integral over the
volume.
rho r prime that is the volume density d V
prime which is the element of charge in that
element of volume, times vector r minus r
prime divided by r minus r prime cube. So,
that gives me the electric field due to what
I call as the volume charged distribution.
The
volume charge density rho is defined as limit
of delta V going to 0 delta Q by delta V
where delta Q is the amount of charge that
is contained in that volume element. So, the
total charge Q is you have to integrate over
that entire volume is integral of rho d V.
..
Let me illustrate some of this with couple
of examples of calculating the electric field.
I
have taken a line charge of length L and I
have sort of placed it such that the origin
is
taken at the center. It is a simple line and
I am interested in calculating what is the
field,
electric field at the point x y point P which
whose coordinates are x y. The line is along
the x direction and of course, the y axis
is perpendicular to it. So, let us look at
this how
does one do this. Consider a charge element
at a distance x prime and having length d
x
prime. Now, what I, do is this, that this
red rho indicates the direction of the force
at the
point P, that electric field at the point
P due to the charges contained in this element
d x
prime which is of course, lambda times d x
prime.
Now, what is this distance r prime. Now, notice
r prime square r prime square is given
by, you can complete a, you can complete a
right angle triangle here. r prime square
is
given x minus x prime whole square. x is the
position x coordinate of P. plus y square
y
is the y coordinate of the P. So, x and y
are fixed where as the x prime will change
depending upon where am I am taking this length
element d x prime. Now, and the
vector r prime is given by x minus x prime
i plus y j.
..
So, therefore the field at the point x y due
to such an element d x prime, which is located
at x prime 0; 0 is the y coordinate is simply
given by our familiar expression, which is
1
over 4 pi epsilon 0. The amount of charge
that is contained in d x prime namely lambda
d
x prime that distance Q remember. The square
of the distance was x minus x prime
whole square plus y square. So, distance cube
is x minus x prime whole square plus y
square rise to the power 3 by 2 times the
position vector here. That is x minus x prime
i
plus y z. Now, so what I have done here is
to write down the x and the y component of
this electric field that is simply writing
this term and this term separate. Now, for
an
arbitrary position x y, this integration cannot
be done in a close form and one normally
you can however, go to computer and sort of
plot what are the e x and e y.
However, let us look at a particularly important
case, let us look at a infinite charge.
Now, if I have an infinite charge 
the instead of wire being located from minus
L by 2 to
plus L by 2. It is from minus infinity to
plus infinity. In this limit these integrals
which
we talked about here. Look at the e x prime
integral e x the e x so if it is from minus
L by
2 to plus L by 2 this is a difficult integral.
Now, suppose it is a minus infinity to plus
infinity, I can reduce this integration very
simply by replacing x minus x prime with
some other variable. Let us say z then the
integral that I get is this.
..
That suppose, I say x minus x prime is equal
to z, then d x prime is minus d z. So, this
integral becomes integral z d z square plus
y square to the power 3 by 2 from minus
infinity to plus infinity and this integral
is 0 because this is an odd integral and however
therefore, the x component vanish. The y component
however, does not vanish.
Remember, that the y component was the so
I will go back a little bit y component is
given by this. If you have the same substitution
x minus x prime equal to z you get y d z
by z square plus y square. So, this is what
I have written down here to the power 3 by
2
this integral you can do this integral.
.
.You can do very easily so by substituting
z is equal to y tan theta. So, E y is equal
to
lambda by 4 pi epsilon 0. Integral from minus
infinity to plus infinity. y d z. So, y is
of
course, fixed coordinate so I cannot change
it d z. So, therefore if z is equal to y tan
theta
d z becomes y sec square theta d theta. So,
let us write down d z is another y so that
gives
me y square sec square theta d theta divided
by I have got z square plus y square and z
is
y tan theta. So, I get y square into 1 plus
tan square theta inside so what I get is y
cube 1
plus tan square theta is sec square theta
so i get sec cube theta.
So, therefore, what I get here is lambda by
4 pi epsilon 0 and I have got a 1 over y there
and integral of 1 over sec cube theta which
is cos theta d theta. Now, notice that I said
z
is equal to y tan theta. So, if z is going
from minus infinity to plus infinity, then
my tan
theta must go from minus pi by 2 to plus pi
by 2 and integral of cos theta d theta is
sign
theta which if I put the limit I get 2 from
this integration. So, I will be left with
lambda
over 2 pi epsilon 0 y.
.
This is the electric field due to a an infinite
line charge. As a second example let us look
at the field of on the axis of a charged ring.
This is the ring of radius R, the ring of
radius
R .This distance I am looking at the field
along the axis which I have taken as the x
axis.
the point P is at a distance x which is the
fixed distance from the origin and this is
this is
where I am interested in calculating the field.
Now, let us look at what happen, this this
incidentally is cuts the x y plane perpendicular
to that. So, if I now take again as we did
.earlier a length element here, ignore the
width that is shown that is shown for clarity.
If I
take a length element here then this is at
a distance R and the field at the point P
is along
this. The direction is shown by this red arrow.
This distance is equal to for any element
is
equal to square root of x square plus r square.
Let us suppose this is at an angle theta,
this element is at an angle theta with respect
the
geometry that has been shown. Now, this is
the direction of the electric filed due to
this
charge element and I can resolve it into an
x component and a y component.
.
Now, notice the direction of the field d E
is along the line joining the element to the
point
P and the magnitude is 1 over 4 pi z epsilon
0 Lambda d l which is the charge, x square
plus r square. So, remember all the elements
because this is a circle which is
perpendicular to the x y plane. I mean which
is cutting the x y plane perpendicular to
the
x axis. x axis is the axis of that ring so
all the elements are located at the same distance
from the point p and that distance is square
root of x square plus y square. So, what we
do we get now is the x so this is the direction
and if this angle is theta if this angle is
theta then the x component of the field is
d E cosine theta and the y component of the
field is d E sin theta.
So, this is what I have written down lambda
d l by x square plus R square this is this
is a
constant because x is the fixed number R is
the radius cosine theta and now, if you look
at what is cosine theta. The cosine theta
is nothing but x divided by this distance
again.
.So, therefore what I get is 1 over 4 pi epsilon
0. x by x square plus R square to the power
3 by 2 lambda d l. Now, all the numbers here
are constant. They do not depend upon
what is the position of d l.
So, if you now have to integrate I am talking
about x component at the moment. If you
now integrate all those being constant come
out, and you have to integrate over the
length d l which is obviously equal to 2 pi
R, the circumference. So, the x component
that you get of the electric field is Q by
4 pi epsilon 0 x by x square plus R square
to the
power 3 by 2. What happened to the y component?
Look at this, that while the x
component added up while the x component added
up the y component will cancel
because from symmetrically placed element.
Supposing I look at this section the force
will be directed like this. Now, if is directed
like this the x component will still be along
the x axis but the y component will be in
the reverse direction.
So, the y component of the electric field
cancels by symmetry and the field is directed
along the axis, the field is directed along
the axis. So, therefore the electric field
due to a
charged ring on its axis, is given by Q by
4 pi epsilon 0 x divided by x square plus
R
square to the power 3 by 2. There is an interesting
consequence there of supposing the
distance at which you are interested in measuring
the field, is much larger than the
radius. What would you expect? See if you
go far far away. A small range looks to you
like a point charge. So, we must get back
in this limit, the electric field due to point
charge and you can see what actually happens.
See if x is much larger than R then I can
neglect this R square in the denominator and
x
square to the power 3 by 2 will give me x
cube and I am left with x by x cube which
is
simply equal to 1 over x square which is nothing
but my Coulomb’s law. So, we have
been talking about the electric field due
to point charge to begin with. A distribution
of
point charges at discreet points. We talked
about super position principle. Let us recall
back. The vector field concept which we had
discussed a few lectures back.
..
Supposing, I have a point charge positive
charge look at this picture. Now, your field
electric field due to a positive charge, remember
my test charge is always taken as
positive. So, a test charge will be repelled
in this field and the closer the test charge
comes to the source charge which is positive
here, stronger will be the electric field.
Now, clearly since it is a point charge I
expect a perfect symmetry between the forces.
So, this is the way the electric lines of
force, the vector field looks like for a positive
charge. Reverse is the situation for the negative
charge because they should all point
towards the charge.
.
.What happens if I have a positive charge
and the negative charge? The lines of force
must start the forces directed such that the
lines of force are those where if you put
a test
charge there it experiences a force along
that tangent and it must get away from a
positive charge go towards a negative charge.
So, that was the force lines of force due
to
a positive and a negative charge.
.
And if I have two positive charges, it will
repel. And therefore, you notice that the
now
two lines of force can never intersect, because
at the point of intersection then the field
will be ill defined. So, what we have done
today, what we have done today is to discuss
the concept of an electric field. Next time
we will be discussing the nature of the electric
field and talk about concepts such as potential
and consequences, thereof.
.
