Prof: Okay,
normally I would ask in a small
class if there is something you
didn't follow from last time.
I'm afraid to do that now,
because it's a big class and I
don't know how many things you
follow or didn't follow.
What I will do first is write
down a very quick summary of the
main points from last time.
 
So you should ask yourself,
"Did I follow all those
things?"
 
And if your answer is yes,
then you are fine.
Because I talked about many,
many things,
but you don't need all of that.
 
So what I'm going to write down
right now is the absolute
essentials of last lecture,
okay?
That's going to be needed for
what I do next.
First point was:
everything is made of atoms.
You know that.
 
And the atom has a nucleus.
 
In the nucleus are some things
called protons,
some things called neutrons,
and outside are some things
called electrons.
 
That's all the atomic structure
we need.
Then we say certain entities
have a property called electric
charge.
 
The symbol for electric charge
is q, and you can put a
subscript to say who you are
talking about.
So you can say q for the
neutron is 0.
A q for the electron is
-1.6 times 10 to the -19,
and it's measured in coulombs.
 
The q for the proton is
really--let me put it this way,
q for the proton is a
positive number,
so these minus signs cancel.
 
Now, the importance of the
coulomb is that if anything has
some coulombs on it,
it will interact with anything
else that has some coulombs on
it.
So that if you have two
entities and this one has a
charge of q_1
coulombs,
that one has a charge of
q_2 coulombs,
and the distance between them
is r,
then the force is
q_1q
_2_ over
4Πε
_0r^(2).
 
I'm purposely not putting all
the vector signs on F
because it takes too long,
but you all know what the
answer is.
 
Namely, if you want the force
on 2,
due to 1, will be repulsive if
q_1 and
q_2 are of the
same sign and point to the
direction joining them.
 
Yep?
 
Student:  Shouldn't it
be 1 over r^(2)?
Prof: Yes, thank you.
 
There is another force law
which is r^(2),
not so famous,
called Hooke's Law,
but you're absolutely right.
 
That's the difference between
being Newton and being Hooke.
Hooke is known for the
r^(2)^( )law.
Newton is known for the 1 over
r^(2)^( )law.
So this is a very important
thing.
If you see anything wrong you
should stop me,
because when I go home,
they're going to play the video
for me to watch,
and I'm going to see that
r^(2).
 
There is nothing I can do until
some voice from the back says,
"Hey, send it
downstairs",
and I appreciate that a lot,
okay?
It's good, so never hesitate to
do that, plus sign,
minus sign, symbols;
anything that goes wrong.
It also tells me that you're
following me.
So for all those reasons you
should not hesitate to correct
anything, and you should not
think that you don't follow it
because it's your fault.
 
Probably it is.
 
Sometimes it's my fault as was
demonstrated now.
Okay?
 
So this is the force law.
 
You need only one other
ingredient.
That's the superposition
principle.
You need that ingredient
because we're not going to be
talking only about two charges.
 
We're going to be talking about
many charges.
And the question is:
what will they do when they're
all present,
and it's a great blessing that
we have the principle of
superposition that says that if
you've got,
say, three charges,
say 1,2 and 3,
you want to know the force on 3
due to 2 and 1.
 
You can find the force that 1
would exert.
You can find the force that 2
would exert.
We have two vectors,
and you can add the two vectors
to get the net force.
 
In other words,
the interaction between pairs
of charges is insensitive to the
presence of any other charges.
They go about their business
exactly the same way.
That is a principle that is not
deduced by logic.
You cannot say,
"Of course,
it had to be that way."
 
That's not true.
 
It doesn't have to be that way.
 
In fact it is not that way if
you make supreme,
I mean, extremely exquisite
measurements,
but they occur really at the
really quantum level.
For classical electromagnetic
theory it is actually an
experimental fact that you can
superpose.
All right, if you combine those
things you can calculate
anything and everything that we
will deal with for sometime,
but that's summary of--this is
all I said last time,
okay?
 
Now, I didn't do one thing,
which is to emphasize to you
that the force of gravitation
divided by the electric force is
some number like 10 to the -40.
 
I put this twiddle here meaning
not exactly.
There are factors of 1 and 2 or
10 missing,
but 10 to the -40 is roughly
the magnitude of this ratio,
and it was done by comparing
the force between electron and a
proton.
 
It didn't matter what the
distance was because everything
goes like 1 over r^(2),
so when they take the ratio
that cancels out,
but this was really the ratio
of things like mass of the
electron,
mass of the proton,
divided by 1 over
4Πε
_0;
q of the electron and
q of the proton.
If you put all these numbers in
you've got that.
I want to mention one thing.
 
It may be interesting for you
to think about.
Look at this nucleus.
 
Nucleus has a lot of protons in
it, and according to Coulomb's
Law they all repel each other.
 
And the neutrons,
of course, don't do anything
because they have no electric
charge.
So you should ask yourself,
"Why are all these protons
staying together inside the
nucleus if I don't see any
attraction between them?"
 
I see why the electrons are
hanging around,
because they're attracted to
the nucleus,
because nucleus is positively
charged,
electrons are negatively
charged, and Mr. Coulomb tells
you they'll be attracted.
 
What are the protons doing
together in that tiny space?
That space is like 10 to the
-13 centimeters.
So have you ever thought of
that, or does anybody know why
the protons are together?
 
Yes?
 
Student:  The strong
nuclear force?
Prof: There is a strong
nuclear force.
Okay, that's the answer.
 
In other words,
but that answer will lead to
more questions,
but I'll first state the
answer.
 
There's a force even stronger
than the electric force.
You see, this gives the
impression electric force is
very strong.
 
It is much stronger than
gravity, but there is a force
even stronger than the electric
force that is experienced by
protons and by neutrons.
 
In other words there is another
charge which is not electric
charge, which the protons are
endowed with and the neutrons
are endowed with.
 
And the force due to that
charge, if you like,
it's not called charge but it's
a similar thing,
is much, much stronger.
 
It had to be,
because you have to beat the
electrical repulsion.
 
So then you can ask yourself,
"Well,
in that case,
how did we ever find the
electrical force,"
because here's another force
even stronger than the
electrical force,
maybe a thousand times stronger.
 
Then why wasn't it completely
overshadowing the electrical
force?
 
Yep?
 
Student:  It's
irrelevant over larger
distances.
 
Prof: Yes,
so let me repeat what he said.
It has to do with what's called
the range of the force.
In other words,
look at two different
functions, okay?
 
One function looks like 1 over
137 times 1 over r^(2).
Other function looks like 10
times e to the -r
over r_0
divided by r^(2);
r_0 is some
length, and the length is
roughly 10 to the -13
centimeters, or 10 to -15
meters.
 
Which force are you more
impressed with is the question,
okay?
 
If r is much bigger than
r_0,
say 10 times bigger,
you've got e to the -10
on the top.
 
e to the -10 is a small
number.
e to the -3 is like 1
over 20.
e to the -10 is 1 over
20 cubed.
So then what will happen is,
this force, even though there's
a number 10 in front of it,
will be negligible compared to
this one.
 
Whereas if r is much
smaller than
r_0,
say, r is .1 of
r_0,
so you can forget this number,
e to the minus point is
roughly 1,
then this 10 will dominate the
1 over 100.
So what happens is,
if you're sitting inside the
nucleus,
the nuclear force is
numerically strong,
not only because of the number
in front of it,
but also because this
exponential factor has not
kicked in.
Therefore, the protons feel an
attraction for each other due to
the nuclear force that is
stronger than the repulsion due
to the coulomb force.
 
And the neutrons attract the
protons just as much as the
protons attract the protons with
respect to the nuclear force.
So neutrons are actually good,
because when you throw an extra
neutron into an atom you don't
add to the coulomb repulsion,
but you add to the overall
attraction that they all feel
for each other due to the
nuclear force.
So neutrons are like the glue.
 
As you add more and more
protons, you will find that
you've got to add more and more
neutrons to compensate the
coulomb attraction.
 
But a time will come when the
nucleus becomes so big,
that even if you add enough
neutrons,
the repulsion between protons
from one end of the nucleus to
the other is now becoming
comparable to the attraction due
to the nuclear force,
because nucleus has become so
big this factor is no longer
negligible.
See, over long distances a
coulomb force will always
triumph,
because no matter what the
prefactors are,
the exponential factor in front
of the nuclear force will always
weaken it.
That's why you cannot have
nuclei beyond some size.
If you make them any bigger,
the nuclear electrical
repulsion between the distant
parts of the protons,
due to the distant protons in
the nucleus,
cannot be compensated by the
short-range attraction.
So it's the range of the
interaction that is a
significant factor here.
 
So all the strong forces are
strong, but at short distances
the coulomb force is not that
strong, but it falls like 1 over
r^(2).
 
Now gravity,
on the other hand,
is exactly 1 over
r^(2)^( )with a number
that's much,
much, much smaller than this,
but then we saw the other day
why gravity managed to survive,
because this is
q_1q
_2 and the
q's can be added
algebraically and cancel each
other.
Whereas if you've got
m_1m
_2_ over
r^(2)^( )there is no way
to cancel the m.
 
This is how the different
forces manage to survive for the
different reasons,
okay?
Nuclear wins in the nuclear
zone, but dies very quickly
outside the size of a nucleus.
 
Electrical forces falls more
slowly at like 1 over
r^(2).
 
They dominate atomic physics,
but once you formed an atom
you've got pretty much an
electrical neutral thing,
and once you've got many,
many atoms making up our
planet,
then all that remains is the
gravitational attraction between
a planet and another planet.
Okay, so these are examples of
different forces and why they
were found at various times,
because they all dominate under
different circumstances.
 
All right, so today I'm going
to start with my new stuff.
So all you need to know,
really, if you want to do your
problem sets and your homework,
is Coulomb's Law,
you know, how to stick the
numbers in the Coulomb's Law.
And the only thing I didn't
mention is this 1 over
4Πε
_0 is 9 times 10
to the 9^(th).
 
So today we're going to do
something which is a part of a
great abstraction and it goes as
follows: So I'm gong to take two
charges,
a q_1 here and
q_2 here,
and I'm going to give them some
locations.
 
So let's say this guy is at
vector r_1,
this one is at vector
r_2.
Now, I will really take are of
all the arrows.
This one is
r_2 -
r_1,
that arrow there.
You can check that I didn't
mess up anything,
because r_1
r_2 -
r_1 should be
r_2.
So let's write the coulomb
force now as a vector.
And you've got to say force on
what, so I'm going to say force
on 2, due to 1.
 
Now, you've got to realize it's
a convention.
So I use a convention:
This is the force on this guy
and the force due to this guy,
with the second later.
Now, let's write it out in
detail.
So that is
q_1 over
4Πε
_0.
I'm going to put the
q_2 here.
Then here I want to write
(r_1 -
r_2)^(2).
 
That's the 1 over r^(2),
but then I have to make it a
vector.
 
So for the vector part,
once you have the magnitude of
the vector,
you should multiply by vector
of length unit 1 going in the
direction of this difference
vector.
 
So you can use any symbol you
like.
One is to say
e_12.
e's always going to be
unit vector going from 1 to 2,
or if you're inclined you can
also write
e_12 as
r_2 -
r_1 divided by
the length of
r_2 -
r_1.
They're all unit vectors.
 
Now, do you follow that?
 
I mean, are you having trouble
with unit vectors?
Anytime I have a vector
pointing from here to there,
I want to give a magnitude and
direction.
The magnitude in this case is 1
over the distance squared,
but you have to append to it a
vector of unit length in that
direction.
 
That's what makes it into a
vector.
For example,
suppose I want to describe that
vector r,
and it is 7 meters long?
I cannot write r = 7,
because that doesn't tell you
which way it's pointing.
 
So invent the vector called
i, which is a unit vector
in the x direction and I
multiply it by that.
So 7i is a vector
parallel to i and 7 times
long.
 
7.3i is a vector
parallel to i 7.3 times
long, okay?
 
So you need to add a vector of
unit length to the magnitude,
multiply it to get the actual
vector.
So that's that formula.
 
Then if you have many charges,
I'm not going to do that now,
say one more here,
it'll exert a force on
q_2,
but you've got to add to the
force due to 1.
 
I'm not going to do that.
 
I'm just taking two guys.
 
Now, I'm going to formally
write this as equal to the
electric field at
r_2 times
q_2,
and this is called a field.
You see, if you look at this
thing,
all I've done is rewrite the
expression as something that
involves a charge of
q�_2 and
everything else that involves
the q_1 and the
distance from
q_1 and
q_2.
 
So it looks like there's no
real content to giving this
object a name,
but it's a very profound
notion, so I've got to tell you
the story that goes with it.
In Coulomb's Law you say that
q_1 and
q_2 exert a
force on each other,
okay?
 
And the force depends on the
charge and the distance between
them.
 
Now I'm going to say
q_1 produces an
electric field at the location
of q_2 given by
this vector,
and when I multiply by
q_2 it gives me
the force on
q_2.
 
Now, what's the importance of
the electric field?
Whereas q_1
and q_2 exist
only at these two places,
the electric field can be
defined everywhere.
 
It doesn't require a second
charge because you see,
this number you can compute for
any value of
r_2.
 
So the picture we have is that
q_1 produces an
electric field all over space,
and q_2
experiences that field and gets
repelled as a result,
and the force it feels is the
field at that point times
q_2.
 
In general, we say there's an
electric field at this point in
space.
 
If you put a charge q
there it should experience a
force q times the
electric field.
So to understand this,
the charges are only in two
places in our example,
but the field due to
q_1 is
everywhere.
At every point in space I can
compute a field due to
q_1.
 
So the field,
you can see,
will turn into a force if you
multiply it by a charge you put
at the location of the field.
 
So if you've got one charge
here, I claim there's a field
here, there's field there;
there's a field everywhere due
to this guy.
 
How do I know,
because if you put a test
charge it begins to feel a
force.
So one way to say it,
is that the field is the force
on a unit charge you put at that
location,
unit charge because if
q_2 is equal to
1 then numerically E is
exactly equal to the force.
If I go to you and say,
"Find out the electric
field at this point."
 
You say, "Okay,
where?"
I say, "Here."
 
What do you think you have to
do to measure the field?
Yes?
 
Student:  Add up all the
forces on that field from the
different charges,
or add up all of the fields at
that point...
 
Prof: No, okay.
 
Her answer was,
"If I want the field here,
I should find out where all the
charges are,"
right?
 
"Find all the forces they
will exert on this guy,
on the unit charge here,"
right?
But I'm telling you,
that's correct.
That's the theoretical way to
calculate the field at that
point, but suppose you're an
experimentalist and you don't
want to know what produced it.
 
You just want an answer to
what's the field here.
What will you do?
 
Yes?
 
Student:  Place a test
charge.
Prof: Okay, and then?
 
Student:  And then see
what happens to it.
Prof: By `see what
happens', you've got to be
more--so his answer was:
put a test charge,
and see what happens.
 
Now, you've got to be more
precise.
I think you know what you
meant, but I want you to finish
that sentence.
 
Student:  Okay.
 
Well, you can measure the force
by measuring the acceleration on
the charge.
 
Prof: Very good.
 
See, that's what I want.
 
When you said,
"See what happens,"
does it get married?
 
Does it have children?
 
That's not what I meant,
okay?
But you did give the right
answer.
The answer was,
by `see what happens',
you put the charge there,
and you see what acceleration
it undergoes.
 
Then that acceleration is
the--that times the mass is the
force it's experiencing.
 
That should be q that
you place there times E.
If q was 1 that force
itself is equal to E.
If q was 10 you've got
to divide the force by 10 to get
the field there.
 
So, the field is like the sound
of one hand clapping.
People say one hand clapping is
Zen concept, but the field is
like that, because you don't
need two charges to have a
field.
 
You just need one charge.
 
So here is how we understand
that.
You put a charge q.
 
If you go here something has
happened there,
see?
 
You don't need to put a second
charge there to conclude
something has happened.
 
Something really is different
at this location because the
charge q is present.
 
What is different?
 
What is different is that when
this guy was not here and I put
a charge, it just sat there.
 
Whereas when this guy is here
and I put a charge,
it experiences a force.
 
So if you put a 1 coulomb here
it experiences a force which is
1 times the field at that point,
therefore, this charge has
distorted the space around it,
in fact, everywhere.
Now, if you've got many,
many, many charges,
then they will all try to
produce a force on a unit charge
and you should,
like you said,
add up the vectors,
which are the forces due to all
the other charges on that one
location,
on a test charge on that
location.
So the field is the sum of the
field due to all the charges at
that point.
 
Now, Coulomb's Law doesn't work
when charges are moving.
Why is that?
 
Have you any idea why you
cannot use it?
Yes?
 
Student:  The radius is
changing?
Prof: Pardon me?
 
Student:  r is
changing?
Prof: r is
changing so we'll keep changing
it as the charge moves,
but it's only an approximation
when charges move.
 
Do you know why?
 
Yep?
 
Student:  There's a
magnetic field?
Prof: True,
but even the electric
field--his answer was magnetic
field,
but even the electric field is
not properly given.
Yes?
 
Student:  Is the
electric field affected by
something like the Doppler
Effect?
Prof: Not the Doppler,
something else.
If Coulomb's Law were exact,
okay?
Here is what I can do.
 
You take a coulomb and you sit
at the other end of the galaxy.
I have a coulomb here.
 
You know the force my coulomb
exerts on yours because it's
pushing against you.
 
You hold it.
 
Now suddenly I move my coulomb
away from you by a tiny amount.
What will you feel?
 
You will feel the force is
reduced.
Yes?
 
Student:  Relativity
is...
Prof: Yes,
so the special theory does not
allow that,
because I have managed to
communicate with you,
arbitrarily far,
instantaneously.
 
The minute I move this charge,
you know about it,
because your charge moved away.
 
For example,
if your charge was connected to
a spring,
and it had been extended to
some amount because of this
charge,
the minute I move it your
spring will move.
That instantaneous
communication is forbidden by
the special theory,
so it does not happen.
So the correct way to do it,
it'll maybe come later in the
course,
so probably not even at the end
of this course,
but the proper way to do that
is to in the end realize that
the electric field at this point
is not only due to what the
charges are doing now,
but what they were doing in the
past.
Because if some charge at the
other end of the galaxy did
something it takes some time,
namely traveling at the speed
of light,
to carry that information from
there to here.
 
So it's the delayed response to
all the motion in the charges
that you've got to add to find
the field here.
That's what makes the
computation of the electric
field much more complicated.
 
But if you promise me charges
never move then the location now
is the location last year,
the location a million years
ago, then you can use Coulomb's
Law.
Coulomb's Law is good for
electrostatics,
but in real life charges are
moving, so you cannot really use
the formula.
 
Now, in our room,
if you put a charge here and
another charge there,
if you move this,
that guy will move pretty much
instantaneously.
That's because the time it
takes a light signal to go from
here to there is so small,
you may treat it as
instantaneous.
 
So Coulomb's Law is used in
electrical circuits and so on.
You don't worry about the time
of transit because it's too
small,
but over longer distances,
where the time it takes for
light to travel becomes
non-negligible,
you cannot use Coulomb's Law.
It's not wrong.
 
It is not appropriate when
charges are moving.
However, it will always be true
that if you go to any one point
the force on any charge q
you put there is q times
electric field at that point,
okay?
So the electric field notion
survives because it doesn't
violate relativity.
 
It says if the field here is so
much q will experience
the force q times
E, but the complication
is what is the field here?
 
Well, it's due to everybody
else, and it's not only due to
everybody else right now,
but everybody else from the
dawn of time because things have
been moving and shaking and
sending signals to us.
 
We collect all that and see
what lands here at this instant.
That decides the field here.
 
So that's the computation of
the field, but the response to
the field is very easy.
 
You put a test charge;
q times E is the
answer.
 
So in modern physics,
in theories that are compatible
with the special theory of
relativity we break the force
into two parts.
 
Charges don't immediately
interact with other charges.
Charges produce a field,
and the field may even
propagate outwards at the speed
of light if you make motions,
but another charge at the
location of that particular
point will respond to the field
at that point.
So it's not responding to the
charge right now.
It's responding to the field
this charge produced at its
location.
 
So all of electromagnetic
theory is going to contain two
parts.
 
The first part is,
find the field due to this
charge configuration,
that charge configuration,
maybe due to various currents,
and the second part is,
given the field,
find the response of charges to
the field.
 
So you understand charges play
a double role.
They are the producers of the
field.
They are also the ones who
respond to the field.
If you don't have a charge you
cannot produce field.
If you don't have a charge you
cannot experience it,
you cannot play that game.
 
To gain membership into
electrostatic interactions
you've got to have charge.
 
So neutrons cannot do that,
but they can do other things.
Like I said,
they take part in nuclear
force, in fact,
just as well as protons do.
All right, so let's go back now
to the simplest problem in the
world: the electric field due to
one charge.
The formula is very simple.
 
Let's put that charge at the
origin.
And the electric field due to 1
charge is q over
4Πε
_0,
1 over r^(2) if you're
here,
and that's your r vector.
 
The field at this point is that
magnitude,
and I'm going to write
e_r_
meaning a vector of unit
length in the radial direction.
Once again, I will tell you if
you want, you can write that as
the position vector divided by
the length of the position
vector.
 
They're equivalent ways.
 
If I write it the second way,
you've got to be a little
careful.
 
It'll look like qr over
4Πε
_0r^(3).
 
Don't get fooled into thinking
the field is falling at 1 over
r^(3).
 
It's really 1 over
r^(2)^( )because there's
an r on the top.
 
See, if you write it as 1 over
r^(2) put unit vector.
If you write it as 1 over
r^(3 )put the position
vector.
 
They're all saying the same
thing.
So here you have this formula.
 
If you're a person who likes to
work with formulas this is all
you need.
 
You manipulate the stuff on
paper, and you add different
fields, but people like to
visualize this.
So how do I visualize this?
 
That's the real question.
 
So here's a very popular method
for visualizing this formula.
You know, you've got,
for example--suppose someone
asks you, what's the height
above ground level of a certain
part of the United States?
 
Okay, you've got some mountains.
 
You've got some valleys.
 
Well, somebody can give you a
function that gives you the
height at any point in the
United States,
but it's more interesting to
have some kind of a contour map
that looks like this,
right?
And all these contours are
different heights.
If you've gone hiking,
you can see those maps.
They tell you pictorially what
a certain function is trying to
tell you.
 
So you want a pictorial
representation of this electric
field.
 
It's very easy to write down
the electric field at one point,
namely you take that point,
you draw an arrow there;
E.
 
That E is the electric
field at that point.
So we try to do our best by
saying here is my test charge.
I'm going to pick a few points,
four points,
maybe eight,
and I'm going to tell you what
the field is at those points.
 
At this point it looks like
that.
At this point it looks like
that.
Here it looks like this.
 
That's already telling you
something.
You've got to be very careful
on the interpretation.
This arrow is not telling you
what's happening throughout the
length of the arrow.
 
It's telling you what's
happening at the tip.
You understand the arrow is in
your mind.
It's a vector.
 
It's not really sticking out in
space.
It's a property at a condition
at that point,
but we've got to draw it,
so we draw it that way.
It doesn't tell you the state
of affairs over its length,
but only at its tip,
at the starting point.
Then you can say,
"Okay, what happens when I
go further out?"
 
When I go further out,
say over here,
it's going to be still--if I
put a test charge here it's
still going to be repelled
radially, but a lot less.
So I do that.
 
So I draw arrows at other
representative points and make
them shorter.
 
In fact, the length of the
arrow will be 1 over
r^(2).
 
So you can do this, okay?
 
But now where all do you want
to draw these arrows?
It's up to you.
 
You pick a few points.
 
You go to another radius,
you draw more arrows.
Someone had this clever idea of
doing the following.
You can probably guess.
 
Their idea was when I join all
these arrows like that,
now if you go to a point like
this, what have I gained and
what have I lost?
 
What more information have I
got when I join the arrows?
Yes.
 
Student:  The more
information is that now you know
the direction that it will
continue on forever if no other
forces act on it,
and what you've lost is the
magnitude...
 
Prof: Okay,
let me repeat that.
That's what you guys should
have been thinking in your head.
When I joined these lines--by
the way,
I do want you to anticipate
what I'm going to say,
because if I'm struck by
lightning,
another electromagnetic
phenomenon,
can you even complete my...
 
Student (Chorus):
 Sentence!
Prof: Sentence, right.
 
Okay, now you should be able to
go a little beyond if I'm doing
a derivation.
 
You've got to be following me,
right?
That's very important.
 
It's got to be active.
 
And I sat through a lecture
yesterday for an hour.
I know it's a very long time.
 
This is what,
an hour and fifteen minutes?
The only way you can survive
this is if you somehow make it
an active event.
 
You've got to do something that
keeps you awake during the
process.
 
One of them is to anticipate
what I will do next in a
calculation.
 
That'll make sure also that
you're on top of it,
that'll make sure that you
catch mistakes.
Okay.
 
All right, so here are these
lines.
As she said quite correctly,
previously I knew the field
direction only at the chosen
points,
but now I know it throughout
this line,
but I've lost information on
the magnitude of the field,
because the arrows--there are
no lengths of anything.
These arrows don't have any
length.
In fact, you can keep drawing
more lines if you like.
They go like that in all
directions.
It basically tells you,
hey, the charge is pushing
everything out radially no
matter where you are.
That's the thrust of this
picture.
But, due to the miraculous
property of the coulomb force,
namely that it falls like 1
over r^(2),
there is information even on
the strength of the electric
field,
and that information is
contained in the density of
electric field lines.
And I'll tell you precisely
what I mean.
So, here is the charge.
 
Take a sphere of radius
r and here are all these
lines going.
 
By density of lines,
I mean the number of lines
crossing a surface perpendicular
to the lines,
divided by the area of that
surface.
Because let's make a convention
that we will draw for every
coulomb a certain number of
lines, 32 lines per coulomb,
32 lines are going out.
 
I draw a sphere of some radius,
32 lines cross that sphere.
I draw a bigger sphere,
32 lines cross this sphere
also,
but they're less dense,
because the number of lines per
area will be some number of
lines per charge divided by the
area of the sphere,
which is 4Πr^(2).
 
Do you follow that?
 
If you take a sphere,
first of all,
every portion of the sphere the
area that you have is
perpendicular to the lines.
 
So the area intercepts the
lines perpendicularly.
That's the agreement here.
 
And you see how many are
crossing per unit area.
That is going like 1 over
r^(2).
So these lines naturally
diverge and spread out in space
so that the density falls
precisely as 1 over
r^(2).
 
That has to do also with the
fact you're living in three
dimensions.
 
Only in three dimensions where
area goes like r^(2)^(
)does this spreading of the
density of lines coincide with
the decline of the force.
 
So these lines tell you more
than simply the direction.
They convey to you visually
where the field is strong.
Wherever the lines are dense,
the field is strong.
Wherever the lines are spread
apart, the field is weak,
and it's a very precise
statement.
The only thing not precise is,
how many lines do you want to
draw per coulomb.
 
That's really up to you,
but you've got to be
consistent.
 
Once you give 32 lines per
coulomb, then if you've got a
charge of 1 coulomb you should
draw 32.
If you've got two coulombs,
you should draw 64 lines.
As long as you do that,
the number of lines crossing
per unit area will be
proportional to the field.
But I'm going to make a certain
choice that will make the number
of lines per unit area exactly
equal to the field,
and here is the choice.
 
It's a choice that makes life
simple.
Let us agree that 1 coulomb
gets 1 over
ε_0
lines.
ε_0 is
a number, right?
1 over 4Πε
_0 is 9 times 10
to the 9^(th).
 
This is some number.
 
Maybe 40 million,
so one coulomb gets 40 million
lines.
 
Don't quote the 40 million.
 
It's whatever this thing is.
 
I don't know what it is.
 
It's a definite number.
 
Then, what's the nice thing?
 
If you've got q coulombs
you will have q over
ε_0
lines,
and if you take a sphere of
radius r you'll get 1
over 4Πr^(2)^( )as the
line density,
namely lines per unit area,
but that is exactly equal to
the strength of the electric
field.
If you picked a different
number like 2 over
ε_0 you
will always be measuring 2 times
the electric field.
 
The density will still convey
the electric field,
namely it'll be proportional to
it, but let's make life easy by
making it equal to it.
 
This is just a convenience.
 
Now, we are really set.
 
If you draw pictures this way,
you can go as far as you like
from this charge.
 
Simply take a unit area with
you.
Take a piece of wood 1 meter by
1 meter, put it there,
see how many lines cross;
that is equal to the electric
field at that point.
 
Okay, so this is the way one
likes to visualize field lines.
So I'm going to give you some
examples.
For a single charge you just
draw it that way.
For two charges,
let's take two charges,
a minus charge and a plus
charge.
Let's say that one is
-q, the other is
q.
 
Then you are very near that
charge.
By the way, I'm not going to
draw 1 over
ε_0
lines per coulomb because it's
going to be too many lines,
okay?
I'm just going to draw a few so
you get the picture.
So I'm going to draw four lines
right near the charge.
You can forget about all other
charges in drawing the lines.
Why is that?
 
Yep?
 
Student:  The field only
depends on the one charge.
Prof: Pardon me?
 
Student:  Why does the
field only depends on the one
charge.
 
Prof: Why does it depend
on the one charge?
In principle it depends on
every charge.
Somebody had an answer back
there?
Yes?
 
Student:  Since it falls
as 1 over r^(2)^( )when
you're so much closer to one
than the other then...
Prof: Right,
because the field is 1 over
r^(2)^( )and the 1 over
r for this guy is going
to infinity.
 
1 over r for this guy is
maybe 1 over 1 meter.
It's finite.
 
So when you come arbitrarily
close to a charge,
it is going to dominate.
 
Well, if it's the only thing in
the universe,
we know the lines will look
like this.
At least they'll start out this
way, but soon,
of course, it won't go out this
way forever, because you will
realize there's another charge.
 
Likewise, it's easy to draw the
lines this way.
Remember, the lines are coming
in because if you put a test
charge, it'll be sucked into
this.
Test charge is always assumed
to be unit positive charge,
so the lines will be coming
into a negative charge,
and leaving,
going away from a positive
charge.
 
Now we've just go to do what
the agencies forgot to do,
which is to connect the dots.
 
You do this.
 
You do this.
 
You do this.
 
You do this.
 
Now, at some point you'll have
to think a little harder,
because suppose I go here?
 
How do I know I should draw the
lines this way?
If I take this guy here,
it will repel it.
This guy will attract it,
and I add the two,
and get a line in that
direction, so you really have to
do a lot of work.
 
If you really want this picture
to be exact,
you have to compute the vector
everywhere,
but if you want a sketch,
you're allowed to guess,
and things look like this.
 
So this is called a dipole,
and this is the field of a
dipole.
 
So here's another example.
 
Both guys are plus.
 
Now what do the lines look like?
 
Again, they will start out this
way near the charges,
but now when you come to the
midpoint here there should be no
electric field right in the
midpoint because it's getting
pushed equally from both sides.
 
So the lines,
if you think about how they
will add, they will do something
like this.
Okay, look, I'm not going to do
a good job for a variety of
reasons,
but you can look at your
textbook or any other book to
see what the lines will look
like when you've two plus
charges.
If you go a mile away from
these two plus charges,
what do you think the lines
would look like?
Yep?
 
Student:  Just like a
point charge that's twice as
strong.
 
Prof: Right,
so that's the intuition you
should keep in your mind.
 
If you go very far from a
charge distribution where you
cannot look into the details,
all you will see is some little
dot that has the entire charge
in it,
and the lines will be coming
out radially.
So only when you zoom in you
realize, hey,
it's not a single charge
2q, it is two guys of
strength q.
 
Finally, let's take a case
where this is charge 2q.
This is charge q.
 
Let's say this is charge
-q.
Then some lines will go like
this and some lines will run off
to infinity.
 
Here if you go very,
very far away from the two
charges you'll again see
radially outgoing lines,
except there is a charge
q at the center because
2q and -q give you
a net of q.
Okay?
 
So this is the example of a
dipole.
If you've got more charges it
gets more complicated,
so people don't usually draw
them.
There's one example which is
pretty interesting.
If you've got one plate and
another plate,
this contains all positive
charges,
this contains all negative
charges,
then the field here will look
like this,
will go from the positive to
the negative plate,
because if you put a test
charge between them,
it's getting repelled by the
positive plate and attracted by
the negative plate,
so the lines will go from one
to the other.
 
Near the edges they may do
something more complicated,
but in the bulk they will look
like this.
 
 
Okay, so now I'm going to do
one calculation,
which is: what is the actual
electric field due to a dipole?
In other words,
not just the picture here,
where is that?
 
That picture on the left is a
dipole.
I'm going to do it
quantitatively.
So here's my goal:
I want to take a minus charge
here, a plus charge here.
 
This is at x = a.
 
This is at x = -a.
 
So I want to find the field
everywhere.
So today I'm not going to do
the field everywhere,
because later on I'll show you
a more effective way to
calculate it,
but I'm going to calculate it
at a couple of interesting easy
places.
In other words,
we all know the lines look like
this, okay?
 
But I want to go to some
location and find the magnitude
and direction of the field.
 
But today I will only find it
at two places,
one along the axis at a point
x,
and one on the perpendicular
bisector at a point (x =
0, y),
just going to do those two.
So let's see what's the field
here, the field at this point?
The field at that point,
you agree, is going to be
entirely in the x direction
because this is pushing it,
and that is pulling it.
 
So E is going to be
i (unit vector) times
q over
4Πε
_0 times that
distance squared,
which happens to be (x -
a)^(2).
That's the repulsion due to the
charge that's nearer to you.
Then there's the attraction due
to the minus q,
but it's a little further away,
so it looks like a minus sign
but it is (x a)^(2).
 
If a is equal to 0 you
get 0.
If a is equal to 0 the
two guys are sitting exactly on
top of each other.
 
You will not see them.
 
So you see them only because
they're not on top of each
other.
 
This whole thing fails to be 0
because this a is not 0,
and you can understand why.
 
The minute a is not 0
you're closer to one of the two
charges, so that they cannot
really cancel each other.
So you've got to manipulate
this expression,
so I will do that now.
 
q over
4Πε
_0 and you find
common denominator.
I remind you (x
a) times (x -
a) is (x^(2 )-
a^(2))^( )and everything
is under squares.
 
In the numerator you've got
(x a)^(2) -
(x - a)^(2).
 
So what does that give me?
 
iq over
4Πε
_0,
(x^(2 )-
a^(2))^(2).
 
And how about on the top,
can you do that in your head?
This is going to be
x^(2) a^(2)
2xa.
 
You're going to subtract from
it x^(2) a^(2) -
2xa, and the only thing
that will survive will be
4xa.
 
Is everything okay?
 
Student:  Why is it not
x - a squared times x a squared?
Prof: It is.
 
You are saying why is it not
(x a)^(2) times
(x - a)^(2),
right?
It is, because this would be
x a times x
- a, the whole thing
squared.
Student:  Oh!
 
Prof: And this guy is
x^(2 )- a^(2).
So this is classified as a nice
try, so.
But I want you to keep doing
this.
This time I am right,
but you never know,
okay?
 
I don't want you to give up.
 
There is nothing better than
shooting me down,
but this happened to be
correct.
You satisfied though?
 
Student:  I knew you
were right, I just couldn't
figure out why.
 
Prof: No, no, no, no.
 
I don't want to rush through
this.
Anybody have the same problem
with this?
Look, also I am doing it fast
because this is 958th time I'm
doing this calculation,
so if you're seeing it for the
first time, I've got to slow
down.
So let's see which part.
 
Everybody okay with this, right?
 
So I wrote that.
 
Then this is really an x
a times another x
a,
and an x - a and
another x - a,
but without these guys,
I know it's x^(2 )-
a^(2 )because I've got
two of everything.
 
I squared everything.
 
So this answer's actually an
exact formula of the electric
field along the axis of the
dipole.
But normally what one is
interested in is,
when x is much bigger
than a.
When it's much bigger than
a, downstairs you've got
x squared.
 
This could be 1 kilometer
squared, a squared was
maybe 1 millimeter squared.
 
So in the first approximation,
it's not an exact formula
anymore, from now on it is
approximate, in the limit
x is much bigger than
a.
You can see it's going to be
i times q over
2Πε
_0 times
2a divided by
x^(3).
So what did I do now?
 
I took from the 4a.
 
I borrowed a 2a to write
this here and I canceled the 2
with the 4 here to get that.
 
Then on the top I had an
x, and the bottom had
x^(4).
 
I get x^(3).
 
So let me write--can everybody
see this thing from wherever you
are;
the last formula here?
So I'm going to write it as
i times p divided
by 2Πε
_0 x^(3).
So I will tell you what I'm
doing.
So the final formula I had was
electric field E = i
times p divided by
2Πε
_0x^(3),
where p--I'm sorry,
i times p,
no arrow;
p is equal to 2aq.
 
So let's delete some extra
arrows I had.
That's right.
 
You never should settle for
something that looks like that.
So that's p.
 
So p is called the
dipole moment of this dipole,
and it's given by the product
of the distance between the
charges and the value of one of
the charges,
the plus charge.
 
So whenever I give you two
charges, call it dipole,
you can associate with them a
vector.
And the vector is,
if you've got a charge
-q here and a charge
q there separated by a
little vector r, then the
dipole moment is q times
the little vector r.
 
In our example the little
vector r was 2a in
the x direction.
 
So 2a times i
times q is the dipole
moment.
 
So this means electric field is
parallel to the dipole moment
and falls like 1 over
x^(3).
That's the most important part
of the dipole.
A single charge,
the field falls like 1 over
x^(2) if you move a
distance x.
A dipole will always fall like
a bigger power of x,
because to go like 1 over
x^(2) you've got to have
net charge.
 
As long as the net charge is 0
the fact that there are two
opposite charges that don't
quite cancel each other,
it always comes from the fact
that the distance between them
is not 0.
 
And the distance will appear in
the numerator,
and that must be the
corresponding distance in the
denominator,
because the formula should have
the same dimensions.
 
That's what turns the 1 over
x^(2) to 1 over
x^(3).
 
Yep?
 
Student:  That only
works if x is a lot
bigger than a.
 
Prof: Yes.
 
So this formula is good for all
x;
this formula is good only for
x bigger than a.
You'll find whenever you're
working with dipoles,
people will always ask you to
find the field very far from the
dipole.
 
So here's a second place where
I can go and find the field.
That's going to be here,
-a, a,
and I'm going to find the field
there.
So that's at a distance
y.
So let's look at this.
 
That's a q here and
-q here.
A q will repel it that
way, and a -q will
attract it this way,
and their sum will be that.
I'm going to compute that sum.
 
So how do I do this?
 
Let's look at this guy here.
 
We know that these arrows have
equal magnitude because this
distance is the same as that
distance.
Therefore, it's the horizontal
part that will remain.
The vertical part will cancel.
 
You see that?
 
We've got two arrows of the
same length with this angle and
this angle equal.
 
The horizontal part will be
additive and the vertical part
will be equal and opposite.
 
So I'm only going to compute
the horizontal part.
So the electric field now will
be -
i (that's to tell me
it's in the negative x
direction) times q over
4Πε
_0 times 1 over
distance squared which is
(y^(2)^( ) a^(2)).
 
This is (y^(2)^( )
a^(2)).
That's also (y^(2)^( )
a^(2)),
but I want the horizontal part
of this,
so I want the cosine
θ.
That is the same as this one.
 
The cosine θ is
a divided by
(y^(2)^( )
a^(2))^(½).
See, you want to take that
force and find this horizontal
part.
 
Then I'm going to put on
another 2 because this is going
to contribute an equal
horizontal part.
So the E,
in the end, is equal to
-iq2a over
4Πε
_0 divided by
(y^(2)^( )
a^(2))^(3/2).
 
That is then -p divided
by 4Πε
_0 divided by
y^(3),
for y much bigger than
a.
I'm sorry, y^(3),
y is really the
distance.
 
So, right, if y is 1
mile and a is 1
millimeter that's essentially
the distance.
If you like you can call it
-p divided by
4Πε
_0r^(3).
Where r is the distance,
if you like,
from the center of dipole.
 
Look, the point of this
exercise is twofold.
One is to show you how to add
vectorially the fields due to
two guys.
 
Another is, to have you
understand,
at least, how to do the
computation at a few simple
places,
where the direction of the
field is not so hard to
calculate.
Actually, one would like to
compute it here,
but it becomes quite nasty.
 
The magnitude is not so hard,
but the direction is hard to
calculate, so we'll find a
shortcut.
But at these two places,
on the perpendicular bisector
and on the axis,
the mathematics is pretty
simple.
 
That's the electric field.
 
Yeah?
 
Student:  Is p as
a vector different from the
p that you crossed out as
a vector over there?
Prof: No.
 
It's the same p.
 
So p as a vector in our
example will be the charge at
either end times the distance
between them.
The vector difference is
2a times i.
So you can write the formula in
terms of the dipole moment.
Okay, so now I'm going to do
the second part of the problem,
which is finding the response
to E.
This was all computing E.
 
Of course there are more and
more complicated examples,
but we did a few simple ones.
 
Next is going to be,
if I give you E, can you
find what will happen to the
charge.
So I'm going to do two examples.
 
One is there are these two
plates I mentioned to you.
This one is all positively
charged.
This one is negatively charged.
 
And I shoot a particle here,
with some velocity
v�_o in the x
direction,
and the field everywhere is
down, and the electric field is
some constant,
-j times some number
E_0.
 
-j because i is
this way and j is that
way.
 
So what will this do is the
question, and where will it end
up?
 
Well, I think you can all tell
that it'll end up somewhere
there.
 
What we're trying to find out
is how much does it fall,
and when it comes out,
what's the direction of this
final velocity vector.
 
Well, the force on this charge
is equal to -q times
E_0 j.
 
The acceleration will be
-qE_0 over
m times j in the y
direction.
So what'll be the position?
 
Position will be from lecture
number one of your Physics 200.
So let's say the starting point
r_0 is our
origin.
 
v_0 is
whatever it was projected in
with,
plus ½
times qE_0
over m,
t^(2) j.
 
Okay, so as a function of time
this tells you where the
position will be.
 
At t = 0,
you are at the origin.
As t increases,
it's moving horizontally due to
v_0 and it's
also dropping vertically due to
the acceleration.
 
Then it's very easy from now on
to compute anything you like.
For example,
when you want to go to that
point, what will be the time?
 
Anybody tell me what the time
will be when I go to that point?
How long will it be in the
region between the plates?
Yep?
 
Student:  The distance
of the length of the plate
divided by initial velocity.
 
Prof: Which velocity
should we take?
Student:  The initial
horizontal velocity.
Prof: That's correct
because t will be
L over v_0
where
v_0�'s
the magnitude of the initial
velocity because x velocity is
never changing.
Acceleration is in the y
direction.
So the time it takes to cross
will be independent of the fact
it's falling in the y direction.
 
So if you put t equal to
all of this you will find out
where it will end up.
 
And that's how you make
pictures on the television.
You've got a bunch of plates,
and then you drive charges,
and if you apply the right
electric field the electron will
land on a screen and make a
little dot.
The screen will look like this,
and you're looking at it from
the other side.
 
It'll glow.
 
Then you want the dot to move
up and down, you can move the
voltage.
 
Then if you want to move it
back and forth you've got to put
another set of plates,
not like this,
but coming out of the
blackboard.
That way you can move the
electron beam in all directions.
That's how you scan the
television screen.
You can also use magnetic
fields, but this is one simple
way using electric field.
 
All right, final thing to
discuss is: what is the force of
a uniform electric field on a
dipole?
So let's take an electric field
in the x direction like that.
It's got a magnitude
E_0 and it's in
the x direction.
 
And in this electric field I
stick a dipole in like that.
Here is the q.
 
Here is the -q.
 
Let's make that a.
 
Let's make that a.
 
So the plus charge will
experience a force like that.
The minus charge will
experience a force like that.
This will be q times
E_0.
That'll be -q times
E_0.
So dipole as a whole will not
feel any net force,
because the two parts are
getting pulled by opposite
amount.
 
If the electric field were not
uniform,
namely if it were stronger here
than here,
then of course it will drift to
the right,
but I'm taking uniform electric
field,
and because the charges are
equal and opposite,
the net force on it is 0.
 
But something is not 0.
 
You know what something is
that's not 0?
Yep?
 
Student:  The torque.
 
Prof: The torque.
 
There is a torque because there
is a force here and the force
there.
 
You can imagine they're trying
to straighten out the dipole so
it ends up looking like this.
 
So let's find the magnitude of
that torque.
Magnitude of the torque is the
force times the perpendicular
distance.
 
So if this angle is
θ here,
and you want the perpendicular
distance, it is
asinθ.
 
Then there's another
asinθ from
that one.
 
That's the torque.
 
But since 2q a is
p, it's pE_0
sinθ.
 
You can see this makes sense.
 
If θ was 0,
if the dipole was aligned with
the field,
the torque vanishes,
because if the charges are like
this there is no tendency to
rotate.
 
The biggest torque you get,
if the charges are like this,
then this gets rotated that
way.
That gets rotated that way.
 
You get the maximum torque.
 
Put θ equal to
Π by 2,
you get a torque of p
times E_0.
That's also the reason it's
called a dipole moment.
Now, I'm going to write this as
a cross product,
and I'm assuming you guys are
familiar with the cross product,
right?
 
You take two vectors,
p and E,
the cross product has a
magnitude, which is p
times E times sine of the
angle between them,
and a direction obtained by
turning a screwdriver from
p to E.
 
So p is like this.
 
E is like this.
 
Turn a screwdriver from
p to E,
it goes into the board,
and that's the torque,
and the dipole will then rotate
until it lines up,
or if you don't want it to
rotate you've got to provide a
counter-torque of this
magnitude.
Now, if you take any dipole and
leave it,
you know it will like to become
horizontal,
so there is a certain restoring
torque that tries to rotate it
so it becomes horizontal,
and it's not very different
from a spring,
where if you pull it from
equilibrium,
there is a restoring force that
brings you back to where you
were.
So this dipole is happiest when
it's horizontal.
If you go away from horizontal
the torque brings it back.
So just like for a spring,
if you've got a force which is
-kx we can assign a
potential energy U,
which is ½
kx^(2) so that the force
is equal to
-dU/dx.
This is something I am assuming
you guys know;
the relation between potential
and force.
Relation between potential and
force,
the potential at
x_1 minus the
potential at
x_2 is the
integral of the force from
x_1 to
x_2.
 
This is how potential is
defined.
This is the reason why,
if you knew the potential minus
the derivative gives the force,
and if you knew the force,
its integral will give you the
change in potential.
All right, so now when you do
rotations,
whatever you had for force you
had for torque,
and whatever played the role of
x is played by
θ.
 
In other words,
in rotational dynamics torque
is to force--just like one of
the SAT questions.
Torque is to force as
θ is to x,
and work is just work.
 
If you want to calculate the
work done by the dipole,
but if you like the potential
energy when it's at an angle
θ and its potential
energy when it's at angle 0 is
the integral of the torque
dθ from 0 to
θ.
 
Now, what is the torque?
 
The torque is
-pEsinθ
dθ from 0 to
θ.
There's a minus sign because if
θ tries to
increase, the torque tries to
decrease it.
That's why it has got a minus
sign.
Now, integral of minus
sinθ is
cosθ,
so you will get
pEcosθ -
pEcos 0 which is
-pE,
and that is supposed to equal
to U(θ) -
U(0).
By comparing the two
expressions you can identify
U(θ) to be--
sorry, this is
U(0) minus
U(θ),
therefore
U(θ) is
equal to
-pEcosθ,
or just -p dot E.
 
So that's the final formula you
have to remember,
that the--can we bring it down
here?
When you have a dipole in an
electric field,
it has a potential energy
associated with the angle which
is -p dot E,
and if you draw a picture of
that as a function of
θ it goes like
this.
 
θ = 0 is when the
dipole is parallel to the field.
That's when it has the minimum
energy, -pE.
At 90 degrees, energy is 0.
 
At 180 degrees, it's maximum.
 
And the torque is just
-dU/dθ and
you can see that the torque here
and there are zero,
but this is the point of stable
equilibrium.
That's the point of unstable
equilibrium.
See, if this was a potential
energy like a shape of the
ground,
if you left a marble there
it'll stay there,
but if it moved a little bit,
it would roll down hill.
 
But if you left the marble here
it'll stay there,
but if you move it,
it'll rattle back and forth.
That's a stable equilibrium.
 
That's unstable.
 
So for the dipole when it's
parallel to the field you are
here, and at anti-parallel to
the field you are there.
The difference is when you're
parallel and you move it a
little bit,
it'll have stable oscillations,
whereas if at anti-parallel if
you move it a little bit it'll
flip over completely and come
down here.
All right, so I'm going to
summarize the main points so you
can carry that with you,
okay?
We saw today that we should
think in terms of electric field
from now on, and we no longer
talk about direct interaction
between charges.
 
We say charges produce fields,
and fields act on charges to
move them.
 
The force of a field on a
charge is just q times
E.
 
The field is found by adding
the field due to all the charges
in the universe,
provided they're all at rest,
and you just add by Coulomb's
Law.
So we found the field due to a
dipole allowing the axis and
perpendicular to the axis.
 
We saw the notion of field
lines as a good way to visualize
what's going on in the vicinity
of the charges.
The lines tell you the
direction where the field is,
and they tell you in your line
density,
lines cutting a unit area
perpendicular to them,
the strength of the field.
 
Then I calculated for you the
field of a dipole along the axis
and perpendicular to the axis.
 
There are a lot of formulas,
but one thing you should carry
in your head.
 
When you've got two equal and
opposite charges and you go very
far, the field will go like 1
over r^(3).
The 1 over r^(2) part of
them is canceled,
okay?
 
That's the main point,
and it goes like 1 over
r^(3) times 2 in one
place, and 1 over r^(3)
times 1 in one place.
 
It doesn't matter.
 
The main thing is it's 1 over
r^(3).
Finally, if you take a dipole
and you put it in an electric
field,
it tends to line up because
there's a torque,
p cross E,
trying to line it up.
 
With that torque you can
associate a potential energy by
the usual formula that the
torque is minus a derivative of
the potential energy.
 
That potential energy is
-p dot E.
Some of these things may come
in handy later on.
So you don't have to memorize
them, but they'll be involved
later on.
 
 
 
