WEBVTT

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Prof: Let's turn to the
subject proper.

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I will start by doing the usual
stuff of just refreshing your

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memory on what happened last
time.

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I mean, last time was pretty
heavy, a lot of mathematical

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machinery which I like to work
with.

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Now, you might say,
"I'm going to be a

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psychologist or art historian,
and I don't need all those

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details.

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Just tell me what I need to get
through this course,"

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so that's really a reasonable
question.

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I probably would have the same
attitude if I took a course in

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your field.

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So here it is.

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What I've told you so far is
that the electric field,

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which we understand now as
basically Coulomb's forces,

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allows you to define a
potential because it's a

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conservative force.

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That means the integral of the
electric field on any closed

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loop is 0, and remember that's
what it takes to define a

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potential.

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And the electric force,
which is Q times

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E will also satisfy the
condition.

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That means we can define for
this problem a potential

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V.

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This is called a potential.

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little v this is big
V) V(r

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mv_2^(2 )
V(r _2).

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That is our goal.

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Ah, missed one mistake already.

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Anybody know what's missing
here?

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Student:
>

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Prof: Pardon me, yes?

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What's missing?

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Student:  The charge.

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Prof: Yes, the charge.

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So this should really be
q times V and this

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should be q times
V.

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So let me remind you why it is.

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The force on any charge is
really not just the electric

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field.

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It's q times the
electric field,

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and therefore the potential
difference is associated with

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the work done by the electric
field times q.

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So you've got to remember q
times V is the real

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energy.

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So V is like an
electrical height.

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For example,
if this is the ground there's a

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mountain like that.

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You give me the height.

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The potential energy is not
just the height.

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It's mg times h.

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So let me say you give me
g times h.

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That's like the potential.

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That tells you the whole story
because if you want to lift a 1

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kilogram mass to that height you
do 1 times gh,

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5 kilograms 5 times gh.

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So depending on the mass you
want to lift the work done will

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be different and the potential
energy will be different,

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but this doesn't depend on the
mass you're planning to carry

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around.

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So you take out the mass.

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You can always put it back.

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Similarly the potential
concerns itself with the unit

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charge.

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If q were 1 nothing I
wrote is wrong,

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but in general q is not
1.

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So this is the actual energy.

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So this is the old potential
energy you want.

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So the potential energy is
q times the potential,

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okay?

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So think of V as the
electrical height at a point and

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q is like the mass,
and q times V is

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like mgh.

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Okay, we've got a formula for
V.

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The general formula for
V is that V of

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r_2 minus
V of r_1

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is the line integral of
E⋅dr

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from r_1_
to r_2.

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This minus sign has the
following meaning.

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The electric field exerts the
force E on the unit

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charge,
but if you want to drag the

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charge against the field from
r_1to

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r_2 you apply
force minus E to

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compensate that,
and the work done by you,

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this really work done by you to
move it from

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r_1 to
r_2.

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Let me give you the gravity
example.

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Gravitational force acts this
way, but if you want to lift

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something from here to here you
oppose it.

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Namely it's doing minus mg.

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Your force mg you move
it to height h.

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That's the work you do,
and that you know is the

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potential at the top,
and that's potential at the

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bottom.

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So that's the origin of the
minus sign.

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The electric field wants to go
one way,

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and if you want to take it
against that force and raise it

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electrically speaking the force
you apply is precisely

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-E.

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That ⋅dr is
the work done by you.

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That's what you invest.

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That's what you'll get back if
it rolls downhill again.

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Now if you want V at
some point r minus

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V at infinity that
integral will be minus starting

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from infinity to the point--
these should be all vectors--to

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the point r of
E⋅dr.

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We will now take the convention
that V of infinity is

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equal to 0.

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Remember, potentials are
defined only by the differences

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so you can put the 0 anywhere
you like.

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We'll choose the 0 to be at
infinity.

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Then in this integral it's
going to be 0,

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and if you do this integral
from infinity to r you

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will find for 1 charge it is
q/4Π

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ε_0 times
r for a single charge,

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and r is the distance
from that charge.

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It looks like Coulomb's Law but
a couple of differences.

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There's no 1 over r^(2) and
there is no vector,

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okay?

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The potential due to a charge
here at the point here is a

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number given by this,
and if you've got lots of

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charges that's just fine.

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V due to many charges is
sum over all those charges,

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the value of each charge
(divided) by

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4Πε_0
times the length from where that

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charge is to where you want the
potential.

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That's the principle of
superposition,

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okay?

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The coulomb potential due to
one charge is 1 over r.

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Due to many charges is each
q times its own 1 over

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r times this 1 over
4Πε.

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And finally if you want the
potential difference due to the

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field you integrate it,
and if you want the field from

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the potential you take the
derivative called the gradient,

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which is identical to minus
ix partial

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derivative minus jy
partial derivative.

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You should expect two
derivatives because the force

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you are looking for now is a
vector.

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You're living in 2D so you
cannot say the force is a

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derivative of the potential,
in what direction?

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If you want the force in the
x direction take the

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x derivative.

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If you want it in the y
direction, take the y

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derivative.

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And you put them together into
a single vector and that's the

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electric field.

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So you go from potential to the
field by taking the derivative

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or the gradient,
and you go from the field to

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the potential by doing the
integral.

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Again, the subtlety is the
integral is of a dot product of

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E with a displacement,
because everything is a vector

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now.

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It used to be fx dx in
the old days,

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but f has become a
vector, dx has become a

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vector dr so you need the
dot product.

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Okay, so this is a summary,
but I want to make two course

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corrections from last time.

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First one was when I tried to
prove to you that the electric

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field is conservative I did one
thing which I realized was a

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swindle.

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I didn't realize it myself and
I wasn't waiting for anyone to

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correct me.

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I did not think about it hard
enough.

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So here is the mistake I made.

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Remember what I said.

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I said that's a potential due
to point charge.

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I'm trying to look at it.

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These are different lines of
the electric field.

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Let's say you want to go from
here to here.

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You do some work,
E⋅dr

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along this line.

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The claim was that's the same
work you do if I take another

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path that looks like this,
go this way,

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go that way,
and go that way,

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same work, and that's still a
correct statement.

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It's the same work because here
dr is angular,

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E is radial,
the dot product is 0,

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likewise 0 here.

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At every segment you have here
E is radial,

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dr is radial,
the dot product is the field

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E times that length.

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Here it's the field here times
that length, but the field is

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constant at a given radius so
they're also equal;

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still no wrong statements.

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Then I said,
"What if I do this,

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a little radial,
a little angular and so

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on," and that's also okay.

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Then I finally said,
"What if I have some nice

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smooth path,
which is not made up of radial

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and angular,"
and my claim was,

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well, you take for this
segment, for example,

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you take a radial segment,
then you take an angular

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segment,
then you may take another

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radial segment and another
angular segment.

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So you approximate the smooth
pass by a bunch of radial,

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angular, radial,
angular, and from a long

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distance it all looks the same,
so it should be the same amount

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of work.

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That's wrong.

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I'll give you an example of why
such arguments can be wrong.

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Let's take a triangle.

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This side is 1,
and this side is 1,

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and I want to walk from here to
there.

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So you know I go a distance
√2, but then you say I'm

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going to take a different path.

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I'm going to go horizontal,
vertical, horizontal,

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vertical, equal amounts like
that.

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And these are very tiny
differences.

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You can say it's the same
thing, but this path is of

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length two.

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Understand why?

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All the horizontal parts add up
to 1.

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All the vertical parts add up
to 1, so going zigzag is longer

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by a factor √2.

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So just because things are
approximately the same it

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doesn't mean the answer will be
the same because if you make an

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infinite number of infinitesimal
mistakes you can add up to

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something finite.

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Likewise here,
you've got to make sure that

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when you deform your path from
what was really given to you to

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what you did,
which is radial and angular,

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the work done should be the
same.

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The path length is clearly not
the same in this example,

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and it can all add up,
but the work will be the same.

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That is what I did not show to
you.

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So let me take a segment here
where that's the radial part and

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an angular part,
but the person really wanted it

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on that segment along some
straight line joining the two.

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What I want to show you is that
not only does that approximate

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this,
from a long distance,

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the work done on that plus that
is the work done here,

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okay?

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So for your convenience I'm
going to blow up that picture

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here so you can all see what I'm
trying to do.

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This should be the radial
direction.

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This is the radial direction.

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This is E.

11:47.350 --> 11:50.880
This is some perpendicular
direction, and this is the

11:50.880 --> 11:54.210
direction in which the move
dr was made.

11:54.210 --> 11:55.810
That's the actual path.

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It's going like that,
and you manage to just deform

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it.

12:00.450 --> 12:04.010
So what's the line integral on
the radial path?

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Well, it's the magnitude of the
electric field at that distance

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times the difference in radius
between this and then this end.

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Let me call it delta r,
delta r is the spacing

12:14.798 --> 12:17.408
here between that curve and that
curve.

12:17.408 --> 12:19.208
We agree the dot product is
trivial.

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E is this way.

12:20.379 --> 12:23.189
Displacement is this way,
no cause thetas to worry about.

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That's the work done.

12:25.389 --> 12:29.339
Work done here is 0 because
perpendicular to the field.

12:29.340 --> 12:31.880
So that's the total work done
going like that.

12:31.879 --> 12:33.709
How about going along this line?

12:33.710 --> 12:37.420
It's going to be
E(r)⋅

12:37.424 --> 12:38.544
dr.

12:38.538 --> 12:40.618
When I say E at
r, r is,

12:40.620 --> 12:42.710
of course, not 1 point it's
kind of spread out,

12:42.710 --> 12:45.570
but this is an infinite decimal
region,

12:45.570 --> 12:47.870
so you can talk about the
E at one point and not

12:47.868 --> 12:49.058
worry about the variation.

12:49.058 --> 12:51.798
So don't forget it's an
infinite decimal triangle.

12:51.798 --> 12:54.758
E⋅dr
will be E at the distance

12:54.764 --> 12:57.864
r,
the length of dr and the

12:57.855 --> 13:02.235
cosine of the angle between them
which is this angle.

13:02.240 --> 13:06.960
The length of dr times
cosine is precisely that

13:06.958 --> 13:12.388
distance, Δr and
therefore you do the same amount

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of work.

13:13.460 --> 13:16.130
So basically I am saying I
should wander around.

13:16.129 --> 13:18.769
You may not choose to go radial
angle or radial angular.

13:18.769 --> 13:21.849
If you go at a slant to the
radial the work you do really

13:21.850 --> 13:24.490
depends on how much radial
distance you cover.

13:24.490 --> 13:26.300
That's what
E⋅dr

13:26.304 --> 13:26.974
does for you.

13:26.970 --> 13:31.120
That's why you can deform the
fact anyway and get the same

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answer.

13:31.779 --> 13:33.539
That's the first thing.

13:33.538 --> 13:41.438
Second thing is I forgot to
tell you anything about units.

13:41.440 --> 13:45.920
So the electric field is force
on the unit charge,

13:45.921 --> 13:50.221
so the unit for that is
Newton's per coulomb.

13:50.220 --> 13:54.050
The electric potential looks
like the work done on a unit

13:54.046 --> 13:58.696
charge,
so the units for that is joules

13:58.697 --> 14:03.047
per coulomb,
and the name for joules per

14:03.048 --> 14:04.848
coulomb is a volt.

14:04.850 --> 14:09.940
It means if there's a voltage
difference between two points of

14:09.936 --> 14:14.686
12 volts it's going to take you
12 joules of work to lug a

14:14.690 --> 14:19.110
coulomb from the lower point to
the higher point.

14:19.110 --> 14:23.310
So you guys have to get your
units right, okay?

14:23.308 --> 14:26.948
Now, you might say,
"Why don't you use

14:26.950 --> 14:28.770
units," right?

14:28.769 --> 14:30.319
You know why?

14:30.320 --> 14:35.820
Because I have tenure;
when you have tenure you don't

14:35.816 --> 14:37.356
have to use units.

14:37.360 --> 14:39.630
I park next to the fire hydrant.

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I have tenure.

14:40.360 --> 14:41.200
I tear up the tickets.

14:41.200 --> 14:42.390
I don't do jury duty.

14:42.389 --> 14:45.169
I don't pay taxes,
you don't have to do anything

14:45.169 --> 14:48.479
when you reach--and Alan is
going to tell you the joys of

14:48.484 --> 14:50.204
being a graduate student.

14:50.200 --> 14:53.770
The joys of being tenured prof
are hard to describe,

14:53.769 --> 14:55.029
okay, priceless.

14:55.029 --> 14:58.689
In fact, one of my sons,
when he saw my lifestyle,

14:58.692 --> 15:02.882
said it reminded him of a
jellyfish deep in the ocean.

15:02.879 --> 15:06.819
As soon as the little guy can
crawl he attaches himself to a

15:06.816 --> 15:08.576
rock,
and he never moves,

15:08.578 --> 15:12.458
and eats his own brain for food
because he has no further use

15:12.461 --> 15:13.111
for it.

15:13.110 --> 15:16.700
So apparently that's what I
reminded him of,

15:16.697 --> 15:17.447
anyway.

15:17.450 --> 15:21.480
So you can, if you work hard,
you can be like me,

15:21.480 --> 15:22.070
okay?

15:22.070 --> 15:23.410
You can do that.

15:23.408 --> 15:24.888
Anyway, look,
so you guys have to get the

15:24.894 --> 15:26.704
units,
because seriously speaking if

15:26.703 --> 15:29.603
you don't write any units we
don't know if you're right or

15:29.601 --> 15:31.151
wrong,
right?

15:31.149 --> 15:34.229
For any problem the answer is
19 in some units,

15:34.232 --> 15:34.772
right?

15:34.769 --> 15:36.869
So if you don't give the units
the number is not worth

15:36.865 --> 15:37.295
anything.

15:37.298 --> 15:40.398
Another advantage of units is
you can keep track of your

15:40.399 --> 15:41.189
calculation.

15:41.190 --> 15:43.080
You multiply one quantity by
another quantity.

15:43.080 --> 15:43.990
Everything has units.

15:43.990 --> 15:45.780
You cross out the lengths and
the meters.

15:45.779 --> 15:48.349
You've got to make sure the
stuff you get has the right

15:48.350 --> 15:49.540
units on the two sides.

15:49.538 --> 15:53.608
That helps you also keep track
of calculations.

15:53.610 --> 15:55.210
Okay, so let's go back now.

15:55.210 --> 15:57.600
This is all stuff left over
from before.

15:57.600 --> 15:59.260
So now for the new stuff.

15:59.259 --> 16:05.829
I'm going to tell you the first
part of this lecture is going to

16:05.826 --> 16:10.096
be what are the advantages of
V.

16:10.100 --> 16:10.970
I will just mention them.

16:10.970 --> 16:13.120
Then I will elaborate on them.

16:13.120 --> 16:19.510
The first advantage is the
conservation of energy.

16:19.509 --> 16:24.609
The second advantage is the
computation of E,

16:24.610 --> 16:30.110
and the third advantage is a
lot of good visual pictures

16:30.110 --> 16:32.810
which are very helpful.

16:32.808 --> 16:36.388
So you won't know what these
mean, but this is what we're

16:36.385 --> 16:39.255
going to do so you know where we
are going.

16:39.259 --> 16:40.769
First thing is conservation of
energy,

16:40.769 --> 16:42.809
there's no need to belabor this
point,

16:42.808 --> 16:45.048
I mean, that's why we did the
whole thing,

16:45.048 --> 16:48.528
the whole conservative forces,
gradient,

16:48.529 --> 16:51.429
this, that, all that was to
make sure in the end you can get

16:51.427 --> 16:52.947
a potential energy out of it.

16:52.950 --> 16:56.400
And as you know from the roller
coaster problem if you can do

16:56.399 --> 16:59.909
kinetic plus potential = kinetic
plus potential it saves you a

16:59.907 --> 17:00.997
lot of trouble.

17:01.000 --> 17:03.470
In other words,
in this room right now there's

17:03.471 --> 17:06.591
a lot of electric fields,
a lot of charges everywhere

17:06.586 --> 17:09.326
applying a field here,
and I want to carry a charge

17:09.330 --> 17:11.440
from here and I want to drag it
over here,

17:11.440 --> 17:14.620
or maybe the charge was shot by
a cyclotron or something.

17:14.618 --> 17:18.808
It came flying out at some
speed here, and it comes here a

17:18.806 --> 17:21.736
little later,
and I want to know how fast

17:21.743 --> 17:22.923
it's moving.

17:22.920 --> 17:25.570
One way to do that is to follow
the charge as it moves.

17:25.568 --> 17:27.678
At every instant find the force
on it,

17:27.680 --> 17:30.250
the acceleration from that,
change in velocity,

17:30.250 --> 17:33.300
add up all the changes you'll
get the final velocity,

17:33.298 --> 17:35.798
but as you know with the law of
the conservation of energy you

17:35.800 --> 17:37.360
can skip all the intermediate
stuff.

17:37.358 --> 17:39.718
You just have to know the
initial kinetic energy,

17:39.723 --> 17:41.993
initial potential energy,
final kinetic energy,

17:41.990 --> 17:43.320
final potential energy.

17:43.318 --> 17:45.778
They are equal,
if you sum them- so you can

17:45.784 --> 17:47.314
find one number from it.

17:47.309 --> 17:48.919
One of the numbers you can find.

17:48.920 --> 17:52.080
In particular if you didn't
know the final kinetic energy

17:52.083 --> 17:53.103
you can find it.

17:53.098 --> 17:56.538
So that's the first advantage,
conservation of energy.

17:56.538 --> 18:01.918
Second thing is that it makes
life easy when you want to

18:01.922 --> 18:04.762
compute the electric field.

18:04.759 --> 18:07.359
So this is an example I did,
again,

18:07.358 --> 18:11.598
near the end of the class,
and I think it was needlessly

18:11.595 --> 18:14.445
complicated,
so I will try to repeat the

18:14.454 --> 18:18.504
derivation a little more quickly
because it's a lot easier than I

18:18.497 --> 18:19.947
gave the impression.

18:19.950 --> 18:24.960
So we are trying to find,
let us say, the electric field

18:24.964 --> 18:30.074
you'll do a dipole -q and
q separated by a

18:30.071 --> 18:32.171
distance 2a.

18:32.170 --> 18:35.390
If you did it with forces,
with forces I didn't even try

18:35.391 --> 18:36.681
to do it everywhere.

18:36.680 --> 18:41.400
I remember I did it here and I
did it there.

18:41.400 --> 18:43.100
The force here was like this.

18:43.099 --> 18:44.449
The force there was like that.

18:44.450 --> 18:51.640
Then I said if you do a lot of
work you'll get all these lines.

18:51.640 --> 18:54.510
What makes life difficult is
that if you go to this point

18:54.509 --> 18:56.589
here,
this I explain to you last

18:56.593 --> 18:59.623
time, this guy will repel it
with some force,

18:59.618 --> 19:02.058
this guy will attract it,
and you've got to add the two

19:02.058 --> 19:03.368
arrows and get a new arrow.

19:03.368 --> 19:06.768
That's the net electric field
at that point.

19:06.769 --> 19:11.029
So you've got to add arrows and
that's more difficult than

19:11.027 --> 19:15.507
adding the potential which is
just a number form this guy and

19:15.509 --> 19:16.479
that guy.

19:16.480 --> 19:19.580
Then once you've got the
potential you find the field by

19:19.584 --> 19:20.774
taking derivatives.

19:20.769 --> 19:22.999
You want E_x
take the x derivative,

19:23.003 --> 19:23.663
the minus sign.

19:23.660 --> 19:26.010
If you want E_y
you take minus

19:26.007 --> 19:27.127
dV/dy.

19:27.130 --> 19:28.580
So what's the E here?

19:28.578 --> 19:32.698
You remember it's
q/4Πε

19:32.699 --> 19:37.869
_0 times
[1/r^( ) (r^( ) is

19:37.872 --> 19:41.802
that length) −
1/r^(-)].

19:41.799 --> 19:42.409
That's it.

19:42.410 --> 19:45.860
No arrows, nothing,
that's the total V.

19:45.859 --> 19:48.219
This is r^(-).

19:48.220 --> 19:52.910
Let's define r to be if
you want the distance from the

19:52.914 --> 19:53.624
center.

19:53.618 --> 19:56.888
And I want you to imagine this
is long way off.

19:56.890 --> 19:59.780
This picture is not drawn to
scale.

19:59.779 --> 20:04.349
You should imagine these two
points are like this and I'm

20:04.349 --> 20:05.819
somewhere there.

20:05.818 --> 20:08.618
That's when the dipole
approximation's working.

20:08.619 --> 20:09.599
This is not an approximation.

20:09.599 --> 20:11.419
This is exact.

20:11.420 --> 20:13.150
So let's do the following then.

20:13.150 --> 20:16.500
We come by the denominators.

20:16.500 --> 20:22.090
You get r^( ) times
r^(-). Numerators

20:22.093 --> 20:26.123
r^(-) −
r^( ).

20:26.118 --> 20:29.928
Now, you can see that r^(
) and r^(-) differ

20:29.925 --> 20:33.395
from r because of this
little guy a.

20:33.400 --> 20:36.530
If you use the law of cosines
you know it's that squared plus

20:36.527 --> 20:39.547
that squared minus 2ar
cosine theta or something.

20:39.548 --> 20:43.018
In the bottom we ignore that
difference and we write

20:43.023 --> 20:46.843
everything as 4Πε
_0r^(2).

20:46.838 --> 20:49.708
In other words r^( ) I
approximate with r,

20:49.707 --> 20:51.947
r^(-) I approximate with
r.

20:51.950 --> 20:54.560
So this is not the exact answer.

20:54.558 --> 20:58.468
There are corrections here that
look like 1 plus maybe a number

20:58.465 --> 21:01.805
a divided by r and also
a^(2) divided by

21:01.805 --> 21:05.975
r^(2),
but these are all negligible.

21:05.980 --> 21:08.380
In the numerator you can again
say, "Hey,

21:08.376 --> 21:11.466
why don't you replace this by
r and this by r

21:11.467 --> 21:12.797
then you get 0?"

21:12.799 --> 21:13.549
It is true.

21:13.548 --> 21:17.118
If you replace everything by
r then you do get 0 and

21:17.121 --> 21:20.941
that's because in the difference
the big guy cancels and what's

21:20.939 --> 21:24.629
left over in the difference is
what's going to keep this from

21:24.634 --> 21:25.624
vanishing.

21:25.618 --> 21:28.788
So we have to find the
difference in that distance and

21:28.791 --> 21:29.751
that distance.

21:29.750 --> 21:34.150
So for that what you do you
draw a line here.

21:34.150 --> 21:36.790
Now, you cannot draw a real
perpendicular here,

21:36.788 --> 21:40.058
but if these things are very
far and these lines are almost

21:40.057 --> 21:42.647
parallel you can draw a
perpendicular there.

21:42.650 --> 21:48.110
Then if that angle is theta,
theta is the angle then it's

21:48.112 --> 21:53.482
very easy to show that this
distance is simply 2a

21:53.476 --> 21:59.226
cosine theta because if theta is
this angle then you can say

21:59.230 --> 22:04.890
cosine of this angle here will
be by whatever complimentary

22:04.887 --> 22:08.787
angle is 2a cosine theta.

22:08.788 --> 22:13.798
Therefore in the top I have
2a times cosine theta,

22:13.798 --> 22:20.428
and that I will write as dipole
moment p cosine theta

22:20.433 --> 22:25.723
over 4Πε
_0r^(2).

22:25.720 --> 22:26.610
That's what I was trying to get.

22:26.609 --> 22:27.589
Yep?

22:27.588 --> 22:29.088
Student:  Sorry,
which angle is theta?

22:29.088 --> 22:36.638
Prof: Theta is the angle
measured from here.

22:36.640 --> 22:39.850
Let me draw this.

22:39.848 --> 22:42.818
One can worry about is it this
angle or that angle,

22:42.820 --> 22:45.490
but I'm saying draw the angle
from the middle,

22:45.494 --> 22:46.034
okay?

22:46.029 --> 22:50.169
That angle is theta,
and roughly speaking that angle

22:50.170 --> 22:51.470
is also theta.

22:51.470 --> 22:58.430
Now drop a perpendicular here
and this side is 2a,

22:58.429 --> 23:05.139
therefore this difference is
2a cosine theta.

23:05.140 --> 23:08.360
So now I have V in terms
of r and theta,

23:08.358 --> 23:10.998
but I can write it in terms of
x and y very

23:10.998 --> 23:11.838
easily,
it is

23:11.842 --> 23:15.642
p/4Πε
_0r^(2),

23:15.640 --> 23:17.790
and cosine theta is
x/r.

23:17.788 --> 23:24.498
So the whole thing is
x/r^(3) which is

23:24.502 --> 23:28.852
the same as
x/(x^(2)

23:28.846 --> 23:33.866
y^(2))^(3/2),
right?

23:33.868 --> 23:37.628
r is the square root and
r^(3) is three powers of

23:37.634 --> 23:38.064
that.

23:38.058 --> 23:40.978
Now, you can easily take the
derivative of this guy with

23:40.980 --> 23:43.690
respect to x,
or with respect to y.

23:43.690 --> 23:45.520
Derivatives are very easy to
take.

23:45.519 --> 23:47.139
Integrals are hard to do.

23:47.140 --> 23:49.200
And that'll give you the
electric field

23:49.198 --> 23:52.228
E_x and
E_y everywhere,

23:52.231 --> 23:52.721
okay?

23:52.720 --> 23:55.290
So E_x is
equal to -dV/dx,

23:55.288 --> 23:57.628
and E_y is
-dV/dy.

23:57.630 --> 24:00.870
It's one of the homework
problems is to really calculate

24:00.866 --> 24:01.276
this.

24:01.278 --> 24:15.488
And if you calculate it and you
plot it you will see this thing

24:15.489 --> 24:18.009
emerging.

24:18.009 --> 24:24.009
Okay, in other problems most of
the time it's easy to find V

24:24.011 --> 24:27.111
and then to find E.

24:27.108 --> 24:31.158
For example,
if I have a ring of charge and

24:31.163 --> 24:36.863
I want to find the V here
then let's say the ring has

24:36.856 --> 24:41.486
charge of q and some
radius a.

24:41.490 --> 24:45.510
Then a little guy here produces
a little potential dV

24:45.508 --> 24:49.528
which is the charge dq
(dq is not any distance

24:49.527 --> 24:53.267
it's like the differential)
divided by 4Πε

24:53.272 --> 24:54.842
_0.

24:54.839 --> 24:56.339
And what's the distance?

24:56.338 --> 25:04.558
If that is z then simply
a root of z squared a^(2).

25:04.558 --> 25:07.388
That's due to this little
charge, but every unit of charge

25:07.387 --> 25:09.817
on this ring is equal distance
from this point.

25:09.819 --> 25:10.459
You understand?

25:10.460 --> 25:13.240
It's a ring and I'm
perpendicular to the ring.

25:13.240 --> 25:17.280
So they all contribute the same
number and sum of all the

25:17.282 --> 25:19.812
dq's is just the
q.

25:19.808 --> 25:23.788
So that's the potential on the
axis.

25:23.788 --> 25:31.308
Remember, you can take a
z derivative to get the

25:31.309 --> 25:33.259
field, okay?

25:33.259 --> 25:36.859
Generally it is easier to do
the V.

25:36.858 --> 25:41.048
Either it's a sum or integral
over the charges producing the

25:41.054 --> 25:43.404
V then take derivatives.

25:43.400 --> 25:48.990
But I want to give you one
example in which the opposite is

25:48.989 --> 25:51.589
true,
where in fact it's easier to

25:51.587 --> 25:53.937
get E than to get
V,

25:53.940 --> 25:56.130
and that is--there are many
such examples.

25:56.130 --> 25:57.950
I'm just giving you one.

25:57.950 --> 26:02.050
Let's say there's a hollow
conducting sphere.

26:02.049 --> 26:03.919
It's got charge Q.

26:03.920 --> 26:11.050
So this guy's hollow and
conducting.

26:11.048 --> 26:14.968
I want to find the potential
everywhere due to this,

26:14.970 --> 26:19.660
and the charge Q is
uniformly spread on the surface.

26:19.660 --> 26:22.560
So what's going to be the
potential?

26:22.558 --> 26:24.828
First of all,
by symmetry you can see that if

26:24.829 --> 26:28.079
you pick any point anywhere the
answer's going to depend only on

26:28.080 --> 26:31.130
the radial distance from the
center because it has spherical

26:31.125 --> 26:31.895
symmetry.

26:31.900 --> 26:35.140
But you still have a lot of
hard work to do because you can

26:35.135 --> 26:37.865
divide the sphere into little
rings like this,

26:37.868 --> 26:40.428
slice it at different
latitudes, and find the

26:40.432 --> 26:43.642
potential due to each ring using
this formula I just did

26:43.635 --> 26:46.285
somewhere,
then integrate it over all the

26:46.286 --> 26:48.356
rings and you will get an
answer,

26:48.358 --> 26:53.218
but that's a decent amount of
calculus.

26:53.220 --> 26:58.810
But you don't have to do that
because in this problem because

26:58.805 --> 27:04.385
it's spherically symmetric and
there's a charge Q here

27:04.390 --> 27:09.690
we know from Gauss's Law the
electric field is going to be

27:09.694 --> 27:15.094
purely radial and will look like
Q/4Πε

27:15.094 --> 27:20.314
_0r^(2) as long
as r is outside this

27:20.307 --> 27:21.887
sphere.

27:21.890 --> 27:23.830
So here's one problem where
it's easier to get

27:23.828 --> 27:25.748
E because E is
pretty trivial,

27:25.746 --> 27:26.366
it's radial.

27:26.368 --> 27:28.028
You just need to know the
magnitude.

27:28.028 --> 27:29.538
Gauss's Law tells you the
magnitude.

27:29.538 --> 27:32.718
You put the unit vector back
and you get this.

27:32.720 --> 27:37.130
So as long as you're outside
this sphere it's as if Q

27:37.130 --> 27:40.350
were sitting at the origin and
the potential

27:40.345 --> 27:43.895
V(r),
which is -Q

27:43.896 --> 27:48.096
Q/4
Πε_0r

27:48.096 --> 27:51.736
outside that sphere,
because remember

27:51.740 --> 27:56.290
V(r) minus
V of infinity is equal to

27:56.291 --> 27:59.631
integral of
E⋅dr

27:59.625 --> 28:03.685
from infinity to r with a
minus sign,

28:03.690 --> 28:07.160
and if you just take the field
of a point charge and do the

28:07.164 --> 28:10.284
integral you will get the
potential of a point charge

28:10.280 --> 28:11.360
which is this.

28:11.358 --> 28:17.018
So if I draw the potential as a
function of radius it will look

28:17.015 --> 28:22.665
exactly like that of a point
charge falling like 1/r.

28:22.670 --> 28:24.940
Now, what about for r
less than r?

28:24.940 --> 28:28.060
How do I continue this graph?

28:28.058 --> 28:31.278
Student:
>

28:31.279 --> 28:33.129
Prof: Pardon me?

28:33.130 --> 28:34.040
Student:  Use Gauss's
Law.

28:34.038 --> 28:34.938
Prof: You can use
Gauss's Law.

28:34.940 --> 28:37.610
If you use Gauss's Law inside
what will you get for E?

28:37.609 --> 28:38.369
Student:  0.

28:38.369 --> 28:39.559
Prof: E is 0.

28:39.558 --> 28:41.988
Integral of
E⋅dr

28:41.993 --> 28:45.143
is 0, so potential does not
change inside this sphere,

28:45.141 --> 28:47.221
but don't think potential is 0.

28:47.220 --> 28:48.800
It does not change.

28:48.799 --> 28:52.129
It's stuck at this value.

28:52.130 --> 28:57.200
So throughout the hollow sphere
there's only one potential,

28:57.198 --> 28:57.808
okay?

28:57.808 --> 28:59.248
Electric field is a different
story.

28:59.250 --> 29:02.620
Electric field is,
in fact, 0 inside then falls

29:02.617 --> 29:04.227
like 1/r^(2).

29:04.230 --> 29:07.690
The electric potential is a
constant.

29:07.690 --> 29:10.670
So there are certain problems
like a cylinder of charge

29:10.674 --> 29:13.444
surrounded by another cylinder
of charge where,

29:13.440 --> 29:18.890
again, it's easier to get
E from symmetry arguments

29:18.887 --> 29:23.857
then integrate that E to
get the potential.

29:23.858 --> 29:26.288
Okay, so this is my second
point.

29:26.288 --> 29:29.668
Namely, barring a few
exceptions like this one,

29:29.670 --> 29:32.480
with high symmetry the
potential has the advantage that

29:32.476 --> 29:35.596
it lets us compute the electric
field by first doing a sum of

29:35.595 --> 29:37.825
scalar quantities rather than
vectors,

29:37.829 --> 29:40.379
then taking derivatives.

29:40.380 --> 29:45.370
The third advantage has to do
with being able to visualize

29:45.371 --> 29:46.161
things.

29:46.160 --> 29:51.270
It gives you a landscape and
that's intuitively very helpful.

29:51.269 --> 29:52.949
So here's one example.

29:52.950 --> 29:59.910
Let's take two parallel plates
and put a lot of plus charge

29:59.910 --> 30:05.310
here and equal amount of minus
charge here.

30:05.308 --> 30:07.668
What will the electric field
look like in this region?

30:07.670 --> 30:09.810
Let's say both have the same
density σ

30:09.814 --> 30:11.664
for unit area and they're
infinite.

30:11.660 --> 30:14.800
I'm just showing you a finite
portion.

30:14.798 --> 30:17.448
This guy will produce sigma
over

30:17.452 --> 30:21.392
2ε_0
here and here.

30:21.390 --> 30:25.690
This one, because of negative
charge, will do this;

30:25.690 --> 30:29.940
σ/2ε
_0 on this side.

30:29.940 --> 30:34.190
In the region between them you
can see the two arrows are

30:34.192 --> 30:37.612
additive, so the field will look
like this.

30:37.608 --> 30:40.158
In the region outside the two
arrows are opposite.

30:40.160 --> 30:40.960
This guy's coming in.

30:40.960 --> 30:42.320
That guy's going out.

30:42.318 --> 30:44.078
And remember,
the field doesn't diminish with

30:44.083 --> 30:46.573
distance due to either plate so
they can completely cancel each

30:46.568 --> 30:46.968
other.

30:46.970 --> 30:50.760
So the electric field is
trapped between the two plates,

30:50.757 --> 30:54.817
and has the strength which is
double what you have on either

30:54.818 --> 30:57.228
side if you had only one sheet.

30:57.230 --> 31:02.590
That's the electric field
E.

31:02.589 --> 31:04.199
That's my electric field.

31:04.200 --> 31:05.820
You can ask,
"What's the potential

31:05.817 --> 31:07.777
difference,"
or "What's the potential

31:07.777 --> 31:08.967
at various points?"

31:08.970 --> 31:12.840
This whole thing being metallic
is at one potential.

31:12.839 --> 31:17.039
I say V = 0.

31:17.038 --> 31:19.548
Throughout this line let's say
V is 0.

31:19.548 --> 31:21.938
In fact it's very clear
V is 0 because if I move

31:21.943 --> 31:24.433
horizontally on this line I
don't do any work because the

31:24.426 --> 31:26.286
electric field is perpendicular
to me.

31:26.288 --> 31:30.088
So if V was 0 at one
point it is 0 throughout this

31:30.086 --> 31:30.626
plane.

31:30.630 --> 31:32.160
Yes, I should qualify this
statement.

31:32.160 --> 31:35.700
You cannot pick V equal
to 0 over an extended region.

31:35.700 --> 31:39.130
You have freedom to pick it to
be 0 at one point,

31:39.130 --> 31:42.470
but I'm telling you if you pick
the one point in this plane you

31:42.471 --> 31:45.601
can navigate throughout the
plane without changing V

31:45.596 --> 31:48.126
because you move perpendicular
to E.

31:48.130 --> 31:49.940
So V is 0 in the lower
plate,

31:49.940 --> 31:53.930
but it won't be 0 anywhere
else, but you can see that may

31:53.933 --> 31:56.703
be V = 1,
that may be V = 2,

31:56.702 --> 32:01.532
V = 3,
upper plate maybe V = 4

32:01.530 --> 32:02.460
volts.

32:02.460 --> 32:07.080
So lines of E are
vertical and lines of constant

32:07.083 --> 32:09.313
V are horizontal.

32:09.308 --> 32:11.218
This is really like the
gravitational problem.

32:11.220 --> 32:14.340
An electric field raining down
on you is just like the force of

32:14.336 --> 32:14.836
gravity.

32:14.839 --> 32:17.259
This might as well be gh.

32:17.259 --> 32:18.629
This might as well be g.

32:18.630 --> 32:20.780
You multiply a mass you get
mg.

32:20.778 --> 32:23.778
You multiply by the q
you get the force on the charge.

32:23.778 --> 32:27.178
It's straight down and the
potential just increases with

32:27.182 --> 32:27.742
height.

32:27.740 --> 32:29.220
Yep?

32:29.220 --> 32:32.450
Student:  Can you just
go over real quick,

32:32.454 --> 32:36.374
please, how you do the calculus
to get the constant voltage

32:36.365 --> 32:37.305
_________?

32:37.309 --> 32:38.539
Prof: Oh, okay.

32:38.538 --> 32:41.688
Think of it as a plate coming
outside the blackboard.

32:41.690 --> 32:43.360
Student:  For the hollow
sphere.

32:43.358 --> 32:45.248
Prof: Oh,
for the hollow sphere.

32:45.250 --> 32:49.060
The potential is always the
work done to go from infinity to

32:49.059 --> 32:50.739
wherever you are,
right?

32:50.740 --> 32:54.310
So when I come from here what
do I think is at the origin?

32:54.308 --> 32:56.828
I think there is a point charge
at the origin,

32:56.829 --> 32:59.519
so I pushed up against that
until I come here.

32:59.519 --> 33:00.709
So I do that amount of work.

33:00.710 --> 33:04.300
Once I'm in here there is no
electric field inside,

33:04.299 --> 33:06.049
so
E⋅dr

33:06.048 --> 33:08.218
is 0 from then on,
but don't forget the V

33:08.221 --> 33:09.751
that you got up to that point.

33:09.750 --> 33:10.860
You don't drop it.

33:10.859 --> 33:13.439
You keep on adding 0 to it.

33:13.440 --> 33:14.530
So it is 0 everywhere.

33:14.528 --> 33:18.478
You add 0 to all points here
because you can go from here to

33:18.479 --> 33:20.219
here, no electric field.

33:20.220 --> 33:28.350
So the whole sphere has the
same potential as the surface.

33:28.348 --> 33:31.488
So now, here I've got this
potentials and it gives you a

33:31.487 --> 33:34.737
picture that you're running
uphill and these are the steps

33:34.739 --> 33:36.849
you've got to climb to go
uphill.

33:36.848 --> 33:42.218
And if you've got 4 volts here,
if you take a coulomb and you

33:42.222 --> 33:47.422
let it drop it'll pick up 4
joules by the time it hits this

33:47.416 --> 33:48.846
lower plate.

33:48.849 --> 33:50.019
That's the meaning.

33:50.019 --> 33:53.409
That's the work done by the
electric field just like gravity

33:53.412 --> 33:56.692
is the potential difference
which is the work done on unit

33:56.690 --> 33:59.680
charge times the charge which is
1 coulomb here.

33:59.680 --> 34:03.210
If it was 10 coulombs you do 40
joules of work.

34:03.210 --> 34:06.600
If an electron is released
you've got to be a little

34:06.604 --> 34:07.274
careful.

34:07.269 --> 34:11.309
By definition the charge of an
electron is negative.

34:11.309 --> 34:15.779
See, if an electron is released
here in fact it'll go like this.

34:15.780 --> 34:18.140
You don't have any analogs in
gravity.

34:18.139 --> 34:19.449
Gravity everything falls down.

34:19.449 --> 34:22.139
Suppose some things had
negative mass and some things

34:22.135 --> 34:23.165
had positive mass.

34:23.170 --> 34:25.950
All guys with negative mass
you've got to nail them to the

34:25.949 --> 34:27.949
ground otherwise they'll just
float up.

34:27.949 --> 34:31.719
That happens in electricity,
so any electron you leave here

34:31.724 --> 34:32.574
will go up.

34:32.570 --> 34:35.970
So the force on the electron is
-q times E,

34:35.972 --> 34:38.222
so everything has got a minus
sign.

34:38.219 --> 34:42.829
So an electron going from here
to here will in fact lose

34:42.833 --> 34:43.593
energy.

34:43.590 --> 34:47.180
We will lose potential energy
and it will gain and turn in

34:47.179 --> 34:48.249
kinetic energy.

34:48.250 --> 34:51.720
So an electron falls through 1
volt.

34:51.719 --> 34:57.869
The work done is charge of an
electron times 1 volt which is

34:57.869 --> 35:00.789
1.6 times 10^(-19) joules.

35:00.789 --> 35:03.319
That happens to be called an
electron volt,

35:03.315 --> 35:05.295
just a convenient thing to use.

35:05.300 --> 35:08.860
When the charges that are
pushing around are not coulombs

35:08.856 --> 35:12.726
but this tiny electron it's good
to know in electron volts.

35:12.730 --> 35:15.900
For example,
in the hydrogen atom the

35:15.903 --> 35:21.283
coulomb potential is negative
because the charge at the origin

35:21.282 --> 35:26.662
is a proton q (electron
has got some -q) divided

35:26.661 --> 35:30.631
by 4Πε
_0r.

35:30.630 --> 35:33.170
That's the potential.

35:33.170 --> 35:38.340
And in quantum mechanics only
at certain radius you can have

35:38.338 --> 35:39.388
electrons.

35:39.389 --> 35:42.399
So the lowest electron is
somewhere here,

35:42.396 --> 35:46.006
and they say it is minus
whatever, some number of

35:46.005 --> 35:49.685
electrons, 13 point something
electron volts.

35:49.690 --> 35:55.020
That means if you want to drag
an electron from this hole and

35:55.021 --> 36:00.351
liberate it from the atom you've
got to do 13 times 1.6 times

36:00.351 --> 36:02.841
10^(-19) joules of work.

36:02.840 --> 36:06.910
So as if this were 13 volts
down compared to infinity.

36:06.909 --> 36:13.719
It's a hole in the ground 13
point 4 or 6 volts deep.

36:13.719 --> 36:17.859
So if you want to liberate that
electron you have to furnish

36:17.856 --> 36:19.536
that amount of energy.

36:19.539 --> 36:25.639
Okay, now going on with the
visualization let me say--let's

36:25.643 --> 36:32.593
take another case we can all do,
which is a single point charge.

36:32.590 --> 36:36.900
Here's a point charge.

36:36.900 --> 36:39.280
The electric field due to that
looks like this;

36:39.280 --> 36:49.730
we've done it a million times.

36:49.730 --> 36:54.710
Now, what about the electric
potential?

36:54.710 --> 36:58.540
For a point charge the
potential is 1/r.

36:58.539 --> 37:03.399
That means it's a constant on a
circle of radius r.

37:03.400 --> 37:05.910
That means if you pick any
circle, or a sphere,

37:05.909 --> 37:08.859
I'm sorry, V is one
number here,

37:08.860 --> 37:11.030
V is a different number
there,

37:11.030 --> 37:13.830
a different number there.

37:13.829 --> 37:16.599
So that's the whole surface
surrounding the charge on which

37:16.601 --> 37:18.131
V is some fixed number.

37:18.130 --> 37:19.530
It depends on that radius.

37:19.530 --> 37:22.820
You go further out it's a
different number depending on

37:22.824 --> 37:25.454
that radius and it falls like
1/r.

37:25.449 --> 37:30.519
You'll notice once again that
the lines of constant potential

37:30.521 --> 37:34.411
are perpendicular to the lines
of the field.

37:34.409 --> 37:36.659
That was true there.

37:36.659 --> 37:38.099
That is true here.

37:38.099 --> 37:40.309
In fact, let me do one more
example.

37:40.309 --> 37:45.159
If you take a dipole negative
and positive,

37:45.159 --> 37:54.379
and the field lines go like
this, if you compute the

37:54.378 --> 37:57.538
potential,
if you go very,

37:57.539 --> 37:59.779
very close to this guy,
I think you all seem to

37:59.775 --> 38:02.155
understand if you're very near
one charge forget about the rest

38:02.163 --> 38:02.783
of the world.

38:02.780 --> 38:04.380
It dominates everything.

38:04.380 --> 38:07.180
Potential would look like
circle centered here,

38:07.184 --> 38:10.724
but as you go further out the
shape will change and it will

38:10.721 --> 38:11.881
look like this.

38:11.880 --> 38:14.860
A similar thing on the other
side then there will be this

38:14.855 --> 38:17.715
huge infinite sphere that
becomes plainer that goes all

38:17.724 --> 38:19.004
the way to infinity.

38:19.000 --> 38:22.450
These are contours of constant
V.

38:22.449 --> 38:27.169
If you really plot them you'll
again find contours of constant

38:27.168 --> 38:31.728
V or equipotentials are
perpendicular to the electric

38:31.733 --> 38:36.773
field lines,
perpendicular to E.

38:36.768 --> 38:39.798
So I want to explain to you
that this is a very general

38:39.804 --> 38:40.314
result.

38:40.309 --> 38:44.509
This is not something true in
these isolated cases of nice

38:44.509 --> 38:45.319
symmetry.

38:45.320 --> 38:46.020
Why is that?

38:46.018 --> 38:48.638
So let's understand why that is
true.

38:48.639 --> 38:52.249
You remember that the
V_2 -

38:52.253 --> 38:55.283
V_1 is -
integral of

38:55.279 --> 38:59.229
E⋅dr
from 1 to 2.

38:59.230 --> 39:01.680
So let's say
V_2 and

39:01.684 --> 39:04.894
V_1,
2 and 1 are very close.

39:04.889 --> 39:08.919
In that case let's call it
delta V = -E

39:08.920 --> 39:10.860
⋅dr.

39:10.860 --> 39:13.400
There's no integral to do
because I've picked two points

39:13.400 --> 39:16.080
which are near by an infinite
decimal amount dr,

39:16.079 --> 39:18.309
so
E⋅dr

39:18.311 --> 39:21.451
is equal to delta V which
is a minus of

39:21.451 --> 39:24.801
V_2 -
V_1.

39:24.800 --> 39:27.010
What does that mean?

39:27.010 --> 39:30.450
That means if you start at some
point,

39:30.449 --> 39:32.999
and you move a distance
dr, you take a step

39:33.001 --> 39:35.341
dr,
and you find the change in

39:35.342 --> 39:37.532
potential,
the answer depends not only on

39:37.534 --> 39:41.634
the length of your step,
but on which direction you move.

39:41.630 --> 39:45.180
Because of this dot product the
change in potential will be the

39:45.181 --> 39:47.991
magnitude of the electric field
at that point,

39:47.989 --> 39:53.389
the magnitude of the step
length and the cosine of the

39:53.389 --> 39:55.529
angle between them.

39:55.530 --> 39:58.880
So if the electric potential is
replaced by height in two

39:58.882 --> 40:01.792
dimensions,
you know, top of Sleeping Giant

40:01.791 --> 40:04.201
and the hill is going up and
down,

40:04.199 --> 40:06.549
up and down,
you're at some point.

40:06.550 --> 40:08.440
There's a height function
h of x and y

40:08.440 --> 40:10.260
instead of V of
x and y,

40:10.260 --> 40:12.530
and I want to take one-step,
and I can ask,

40:12.530 --> 40:15.140
"What are the consequences
to my height?"

40:15.139 --> 40:18.169
The answer is if the electric
field is pointing in some

40:18.170 --> 40:21.150
direction and you take a step
dr in a generally

40:21.146 --> 40:24.006
different direction this is the
change you get.

40:24.010 --> 40:26.660
So you can ask,
"How do I get the biggest

40:26.664 --> 40:28.854
drop in height,
or the biggest drop in

40:28.849 --> 40:30.029
potential?"

40:30.030 --> 40:33.060
And cosine theta is 1, right?

40:33.059 --> 40:36.079
Because then there's this,
that times 1 with a minus sign.

40:36.079 --> 40:39.629
So the electric field points in
the direction of the greatest

40:39.626 --> 40:41.396
rate of change of potential.

40:41.400 --> 40:44.310
It drops in that direction.

40:44.309 --> 40:47.669
If you release a marble in the
gravitational field it'll roll

40:47.670 --> 40:50.980
along the gravitational field
which is obtained the same way

40:50.976 --> 40:53.046
from the gravitational
potential.

40:53.050 --> 40:54.630
If you want to climb up the
mountain,

40:54.630 --> 40:57.240
suppose there's a tsunami
coming and you want to get to

40:57.237 --> 40:59.907
the top as quickly as possible,
don't panic,

40:59.914 --> 41:04.314
compute the gradient and move
along the gradient,

41:04.309 --> 41:04.929
okay?

41:04.929 --> 41:08.419
Then stop, think again,
recompute the gradient and in

41:08.416 --> 41:12.576
that manner you will get to the
top in quickest time if you can

41:12.576 --> 41:14.786
compute quickly enough,
okay?

41:14.789 --> 41:16.069
But that's the process.

41:16.070 --> 41:19.310
At every point there's a vector
pointing in the direction of

41:19.311 --> 41:20.411
greatest increase.

41:20.409 --> 41:23.739
The electric field happens to
be the direction of greatest

41:23.739 --> 41:27.009
decrease because of this minus
sign in our convention.

41:27.010 --> 41:30.380
Now, let's us say you're on
this mountain and you don't want

41:30.378 --> 41:33.808
to go down and you don't want to
go up, but you still want to

41:33.806 --> 41:34.716
take a walk.

41:34.719 --> 41:37.019
Can you do that?

41:37.019 --> 41:39.699
Yes?

41:39.699 --> 41:43.729
Or are you stuck at one place?

41:43.730 --> 41:45.130
What do you have to do guys?

41:45.130 --> 41:46.470
Yes?

41:46.469 --> 41:47.359
Student:  You move
perpendicular to the gradient.

41:47.360 --> 41:50.730
Prof: You move
perpendicular to the field.

41:50.730 --> 41:54.790
To move perpendicular to the
field the cause theta is 0.

41:54.789 --> 41:57.789
That's why, now going back to
this, this is called Mt.

41:57.789 --> 42:01.819
Coulomb, it's coming out of the
blackboard really tall and you

42:01.822 --> 42:02.552
are here.

42:02.550 --> 42:05.530
If you want to go downhill you
go along the field.

42:05.530 --> 42:08.940
You run away from the hill all
the way to the valley,

42:08.940 --> 42:09.990
to the plains.

42:09.989 --> 42:13.249
If you want to go to the top
you go opposite of the field,

42:13.250 --> 42:17.090
but if you go along this
equipotential at every point you

42:17.092 --> 42:21.282
do not do any extra work so your
potential does not change.

42:21.280 --> 42:25.560
So it's clear from this that
the equal potential is always--

42:25.559 --> 42:30.769
in fact in general the change
in any potential is gradient of

42:30.773 --> 42:32.863
the potential dot dr.

42:32.860 --> 42:35.890
Except for the minus sign it's
the same story that if

42:35.889 --> 42:40.709
dr is parallel to grad
V or anti-parallel to

42:40.713 --> 42:44.863
E you get the biggest
change in voltage.

42:44.860 --> 42:51.860
All right, so this is what you
should understand in general.

42:51.860 --> 42:54.130
What is good about the
potential is you can take a

42:54.132 --> 42:55.992
whole bunch of charges put them
here,

42:55.989 --> 42:58.439
there, there,
there, and you can draw little

42:58.443 --> 43:00.323
contours,
maybe of constant V.

43:00.320 --> 43:03.500
It's just like the graph you
have of the weather,

43:03.500 --> 43:05.340
of the temperature,
for example,

43:05.335 --> 43:08.705
or if you have a topographic
map of the mountain there are

43:08.713 --> 43:11.103
some valleys,
and there are plains,

43:11.101 --> 43:13.241
and there are some
mountaintops,

43:13.239 --> 43:16.829
and these lines of constant
height are like constant

43:16.833 --> 43:19.183
V,
so it helps you visualize.

43:19.179 --> 43:22.239
And then if you're told you
release something here and it

43:22.237 --> 43:25.407
rolled down and came here you
can find the speed here given

43:25.405 --> 43:27.585
the difference of these two
numbers.

43:27.590 --> 43:30.000
You don't have to go through
the intermediate step.

43:30.000 --> 43:32.310
So these are the advantages.

43:32.309 --> 43:37.499
Okay, now I'm going to return
to the conductors briefly to

43:37.498 --> 43:39.408
tell you something.

43:39.409 --> 43:42.729
One other thing that is new
compared to what we did before.

43:42.730 --> 43:45.130
Remember what I told you about
a conductor?

43:45.130 --> 43:49.990
A conductor is one in which the
electrons do not belong to any

43:49.985 --> 43:51.015
one parent.

43:51.018 --> 43:54.428
They run around to the whole
sample, but they cannot leave

43:54.431 --> 43:55.151
the metal.

43:55.150 --> 43:57.270
At the boundary of the metal
there are forces bringing them

43:57.273 --> 43:57.533
back.

43:57.530 --> 43:59.570
Inside they are free to move.

43:59.570 --> 44:02.950
Therefore, if you take a
conductor and you stick it into

44:02.945 --> 44:06.255
a field initially the electric
field will penetrate the

44:06.260 --> 44:07.120
conductor.

44:07.119 --> 44:09.089
The charges,
because they are free to move,

44:09.094 --> 44:09.664
will move.

44:09.659 --> 44:11.049
The negatives charges will move
one way,

44:11.050 --> 44:14.720
positive charges will move the
other way until they set up a

44:14.722 --> 44:18.022
back reaction to this that
blocks the field inside the

44:18.023 --> 44:20.393
metal until the field inside is
0.

44:20.389 --> 44:23.429
So here's a generic conductor.

44:23.429 --> 44:29.159
If you put it in an external
field it will have some charges

44:29.161 --> 44:33.201
on its surface,
will produce a field this way

44:33.197 --> 44:37.927
that will oppose the external
field in which you put it until

44:37.931 --> 44:41.721
at the end there's no field
inside the metal.

44:41.719 --> 44:44.529
Or, if you did not put it in
the field and just took this

44:44.530 --> 44:46.590
material and dumped some charge
on it,

44:46.590 --> 44:49.870
threw some charge on it and
asked where does it go,

44:49.869 --> 44:53.429
well, the charges are all plus
and they don't like each other.

44:53.429 --> 44:56.699
They want to move away from
each other, and they move as far

44:56.704 --> 45:00.094
as they can until they come to
the boundary and they cannot go

45:00.090 --> 45:00.980
any further.

45:00.980 --> 45:03.220
That's where they stop.

45:03.219 --> 45:06.149
That we saw before,
so we know all the charges in

45:06.152 --> 45:08.112
the metal are in the boundary.

45:08.110 --> 45:11.160
There is no electric field
inside because of Gauss's Law.

45:11.159 --> 45:14.409
If I want to find the charge
here I do a tiny sphere and do

45:14.413 --> 45:16.773
integral
E⋅da.

45:16.768 --> 45:19.318
I get 0, and that's the charge
inside.

45:19.320 --> 45:21.750
And I can move this little
microscope all over the inside.

45:21.750 --> 45:23.250
I keep getting zero.

45:23.250 --> 45:26.220
That's why I know there's no
charge inside.

45:26.219 --> 45:27.499
There's no charge inside.

45:27.500 --> 45:28.750
There is no field inside.

45:28.750 --> 45:33.960
Q = 0 inside metal.

45:33.960 --> 45:41.730
This is a solid metal,
and E = 0 inside solid.

45:41.730 --> 45:48.480
The other variation you can
have is suppose they make a hole

45:48.476 --> 45:49.616
in this.

45:49.619 --> 45:52.609
There is no electric field
anywhere here because it's a

45:52.606 --> 45:53.046
metal.

45:53.050 --> 45:55.820
The whole logic is if you've
got a field you're not in a

45:55.822 --> 45:56.782
static situation.

45:56.780 --> 45:58.010
Let the charges move.

45:58.010 --> 46:00.320
They'll keep moving until they
have no reason to move.

46:00.320 --> 46:03.090
That reason will stop when the
field vanishes.

46:03.090 --> 46:03.650
It's quite amazing.

46:03.650 --> 46:06.840
They can always find an
arrangement in which they kill

46:06.844 --> 46:08.114
the external field.

46:08.110 --> 46:09.160
That's the perfect metal.

46:09.159 --> 46:10.699
So there's no field here.

46:10.699 --> 46:16.589
So if you do a Gaussian surface
like this the charge inside that

46:16.590 --> 46:19.560
is 0,
but that does not preclude the

46:19.556 --> 46:24.286
possibility that there are maybe
some plus charges there and some

46:24.289 --> 46:25.989
minus charges there.

46:25.989 --> 46:31.249
Last time I told you that if
that happens then the pluses and

46:31.246 --> 46:36.326
minuses will run around and meet
and destroy each other and

46:36.327 --> 46:39.037
neutralize,
but that's a little subtle

46:39.041 --> 46:42.071
because maybe if they were the
only two things in the world I

46:42.068 --> 46:44.928
can understand that,
but there's also these charges

46:44.927 --> 46:47.287
outside maybe that'll impede
their motion.

46:47.289 --> 46:50.119
But here's one way to prove
that there can be really no

46:50.123 --> 46:51.963
charge even on the inner
surface.

46:51.960 --> 46:54.690
If there were some plus charges
here and minus charges

46:54.688 --> 46:57.568
there--remember Q is 0
inside so you can only have

46:57.574 --> 46:58.454
equal number.

46:58.449 --> 47:02.079
Then electric field lines will
have to leave the plus charge

47:02.081 --> 47:04.361
and terminate on the minus
charge,

47:04.360 --> 47:07.360
so there'll be an electric
field inside the cavity,

47:07.360 --> 47:10.900
and I'm saying that's not
allowed because of the

47:10.896 --> 47:11.796
following.

47:11.800 --> 47:15.640
If there is a field inside the
cavity let's take that path and

47:15.639 --> 47:19.289
another path which are two ways
to go from here to here.

47:19.289 --> 47:22.239
On one path I get a line
integral of E to be not

47:22.240 --> 47:22.460
0.

47:22.460 --> 47:25.610
Second one I get line integral
equal to 0 because there is no

47:25.608 --> 47:28.858
E, but the line integral
cannot vary with the path in an

47:28.860 --> 47:30.330
electrostatic situation.

47:30.329 --> 47:34.509
Therefore, there cannot be any
electric field inside the cavity

47:34.514 --> 47:36.814
due to anything you put outside.

47:36.809 --> 47:41.189
You can have an electric field
inside the cavity if you go into

47:41.186 --> 47:44.926
the cavity and put by hand some
extra charge there.

47:44.929 --> 47:48.229
Then what will happen is that
guy will produce a field and the

47:48.228 --> 47:51.358
field lines will come to the
boundary and terminate on some

47:51.364 --> 47:52.504
negative charges.

47:52.500 --> 47:54.860
Again, there'll be no field
here.

47:54.860 --> 48:02.740
But a hollow cavity with no
charges inside the hole will not

48:02.744 --> 48:07.294
have charges on the edge either.

48:07.289 --> 48:09.849
So now, what else can we say
about conductors?

48:09.849 --> 48:13.429
Today I used the fact that the
line integral of E is 0

48:13.431 --> 48:16.661
to tell you that in a hollow
cavity you cannot have any

48:16.655 --> 48:18.085
charges in the wall.

48:18.090 --> 48:27.320
Second thing is every conductor
is an equipotential.

48:27.320 --> 48:33.310
It's an equipotential.

48:33.309 --> 48:33.939
Why is that?

48:33.940 --> 48:39.640
Can you tell me why that is
true if it is not equipotential,

48:39.637 --> 48:41.277
the whole lump?

48:41.280 --> 48:42.900
Yep?

48:42.900 --> 48:44.640
Student:  If it wasn't
equal potential then there would

48:44.635 --> 48:45.655
be an electric field inside
there.

48:45.659 --> 48:46.989
Prof: And how do we know
that?

48:46.989 --> 48:49.279
Student:  Because
there's a potential difference

48:49.275 --> 48:50.075
________________.

48:50.079 --> 48:51.509
Prof: That's right.

48:51.510 --> 48:53.240
In other words,
if you take the gradient,

48:53.239 --> 48:55.229
namely taking derivatives,
you'll get a non-zero

48:55.226 --> 48:57.846
derivative because I'm telling
you V is not constant.

48:57.849 --> 48:59.589
In some direction it's got to
change.

48:59.590 --> 49:01.530
There's an electric field in
that direction,

49:01.525 --> 49:03.095
at least, but that's not
allowed.

49:03.099 --> 49:06.249
So V is a constant,
so take any metal shaped like

49:06.250 --> 49:06.880
anything.

49:06.880 --> 49:12.220
You can associate a single
potential to all of it,

49:12.217 --> 49:12.977
okay?

49:12.980 --> 49:15.000
If it's got charge on it that's
the same,

49:15.000 --> 49:18.340
and what it means if you start
at infinity the work done to

49:18.335 --> 49:20.845
come there,
the work done to come there,

49:20.853 --> 49:24.713
the work done to come there are
all the same because they're all

49:24.706 --> 49:26.356
measuring the potential.

49:26.360 --> 49:29.360
The whole metal is at one
potential.

49:29.360 --> 49:31.510
That allows you to do certain
tricks.

49:31.510 --> 49:33.680
So here's one trick you do.

49:33.679 --> 49:36.439
There are more tricks,
but this is the only one we can

49:36.440 --> 49:37.690
discuss in this class.

49:37.690 --> 49:41.350
So here's an infinite
conducting plane,

49:41.353 --> 49:45.213
a sort of ground,
attached to the ground,

49:45.210 --> 49:48.970
and over this guy I put a
charge q.

49:48.969 --> 49:52.859
And I ask you,
"Find the electric field

49:52.858 --> 49:54.938
everywhere now."

49:54.940 --> 49:58.350
That's your challenge.

49:58.349 --> 50:01.989
Without this guy fields are
just radial, so radial,

50:01.985 --> 50:05.905
radial, but the fields come
near this metal they cannot

50:05.911 --> 50:06.931
penetrate.

50:06.929 --> 50:10.329
They've got to stop at the
surface of the metal.

50:10.329 --> 50:15.089
So the field lines will do that.

50:15.090 --> 50:17.930
They will all terminate on the
metal, and to terminate you need

50:17.925 --> 50:20.485
some negative charge which the
metal will suck out of the

50:20.485 --> 50:20.985
ground.

50:20.989 --> 50:24.709
That's the reason for grounding
it.

50:24.710 --> 50:30.280
So the question is what kind of
charge distribution do you get

50:30.280 --> 50:31.650
on the metal?

50:31.650 --> 50:34.530
How much charge is there on the
metal?

50:34.530 --> 50:37.930
And because there are positive
charge and negative charge here

50:37.931 --> 50:41.111
what's the force of attraction
between this charge and the

50:41.112 --> 50:41.672
plane?

50:41.670 --> 50:44.560
In general that's a very
difficult problem.

50:44.559 --> 50:47.119
If I gave you a potato,
aluminum potato,

50:47.119 --> 50:50.069
put a charge next to it,
it will have a force,

50:50.074 --> 50:52.574
but it's not easy to calculate
it.

50:52.570 --> 50:57.480
But this infinite sheet turns
out very easy to calculate.

50:57.480 --> 51:02.440
The reason is that we can solve
this problem by borrowing from

51:02.436 --> 51:03.896
another problem.

51:03.900 --> 51:06.390
The problem I borrow from is
the following.

51:06.389 --> 51:08.279
Forget this plane.

51:08.280 --> 51:12.990
Continue these lines,
and think of the problem of a

51:12.992 --> 51:17.142
dipole with an equal negative
charge here.

51:17.139 --> 51:20.859
So the lines from them go like
this, right?

51:20.860 --> 51:23.870
And if you slice it in the
perpendicular bisector of this

51:23.869 --> 51:27.309
distance between them you will
hit every line perpendicularly.

51:27.309 --> 51:29.879
You guys remember that part
about the dipole field at

51:29.880 --> 51:30.770
somewhere there?

51:30.768 --> 51:33.718
You cut it right down the
middle.

51:33.719 --> 51:43.349
So this plane was an
equipotential at V = 0.

51:43.349 --> 51:45.939
If it's an equipotential there
is nothing there.

51:45.940 --> 51:48.830
It was just a mathematical
surface on which V was

51:48.831 --> 51:50.201
constant and equal to 0.

51:50.199 --> 51:53.809
If V is constant you can
slide in a conductor there and

51:53.806 --> 51:56.456
nobody will know,
because the field is 0 along

51:56.460 --> 52:01.550
the surface of the conductor,
nothing happens.

52:01.550 --> 52:04.510
So you can reconcile the fact
that conductors are

52:04.510 --> 52:07.230
equipotentials,
and solve problems in which a

52:07.226 --> 52:10.266
charge is in front of a
conductor if you're lucky enough

52:10.273 --> 52:13.493
to find the situation where
that's an equipotential surface

52:13.487 --> 52:16.637
of the same shape as your
conductor then you can slip your

52:16.644 --> 52:17.924
conductor there.

52:17.920 --> 52:22.450
So in this example what happens
is that the electric field here

52:22.445 --> 52:26.675
throughout the plane is exactly
what you would get due to a

52:26.681 --> 52:28.581
dipole minus and plus.

52:28.579 --> 52:32.999
That you can easily calculate
at all points on the plane.

52:33.000 --> 52:35.820
And once you've got the
electric field you remember

52:35.815 --> 52:38.625
σ/ε
_0 is the

52:38.630 --> 52:41.110
electric field so you can find
σ,

52:41.110 --> 52:44.840
the charge density.

52:44.840 --> 52:48.730
And if you integrate the charge
density you will get exactly

52:48.731 --> 52:51.361
-q,
because the number of lines

52:51.362 --> 52:54.852
coming from here have to
terminate on the plane rather

52:54.846 --> 52:57.536
than on this and that'll be
-q.

52:57.539 --> 53:00.369
And finally,
what's the force of attraction

53:00.365 --> 53:02.785
between the plane and this
charge?

53:02.789 --> 53:08.719
Can you make a guess?

53:08.719 --> 53:13.069
What's the force with which the
plane attracts the charge?

53:13.070 --> 53:16.490
I'm asking you to guess,
so it cannot be too difficult.

53:16.489 --> 53:19.309
Students:
Same as that of the other

53:19.313 --> 53:19.923
charge?

53:19.920 --> 53:23.230
Prof: Same as the force
that this negative charge would

53:23.226 --> 53:26.206
exert on this positive charge,
and I'll tell you why.

53:26.210 --> 53:30.500
This guy is looking around,
okay, seeing some forces.

53:30.500 --> 53:34.830
It's due to these negative
charges on the plane,

53:34.833 --> 53:40.093
but they're identical to the
field created by this guy.

53:40.090 --> 53:43.390
So it cannot tell the
difference and therefore if

53:43.389 --> 53:47.579
you're releasing it it'll move
the same way in this problem as

53:47.581 --> 53:49.301
in the other problem.

53:49.300 --> 53:53.780
So it'll be drawn to the plate
with an attractive force which

53:53.782 --> 53:57.672
is q times -q
divided by square of that

53:57.668 --> 53:58.638
distance.

53:58.639 --> 54:02.769
So the summary of what I did
now is that if you solve any

54:02.768 --> 54:07.488
problem, let's say the dipole
problem, and you get some shape.

54:07.489 --> 54:10.009
It's an equipotential.

54:10.010 --> 54:13.150
You're allowed to,
then, replace that with a chunk

54:13.146 --> 54:16.986
of aluminum and keep the field
lines the way they are because

54:16.985 --> 54:20.505
that aluminum will be an
equipotential and it'll happily

54:20.507 --> 54:24.027
sit on equipotentials because
there are no forces on the

54:24.027 --> 54:27.417
surfaces of the conductor moving
the charges.

54:27.420 --> 54:31.150
So the moral of the story is
you can sneak in a solid object

54:31.152 --> 54:35.142
bounded by the surface of an
equal potential for any problem.

54:35.139 --> 54:37.519
So if you can solve certain
problems and the equal

54:37.519 --> 54:40.579
potentials have nice shapes you
can put conductors of that shape

54:40.581 --> 54:41.991
in that precise location.

54:41.989 --> 54:44.759
That'll answer a different
problem in which the second

54:44.764 --> 54:45.764
charge is absent.

54:45.760 --> 54:47.210
You only have the first charge.

54:47.210 --> 54:50.640
It's like a phantom charge,
an image charge.

54:50.639 --> 54:54.509
So it's like this charge looks
into the mirror and sees the

54:54.514 --> 54:55.054
image.

54:55.050 --> 54:58.370
The image is minus of itself
and it's drawn to it,

54:58.369 --> 55:03.049
because the field produced by
this guy and this guy is the

55:03.047 --> 55:07.477
same as the field produced by
all of these and this.

55:07.480 --> 55:19.110
All right, so now we've got a
final problem with energetics.

55:19.110 --> 55:23.780
The question is I want to bring
a whole bunch of charges,

55:23.780 --> 55:25.540
q_1,
q_2,

55:25.539 --> 55:30.609
q_3 from
infinity and bring them to this

55:30.612 --> 55:31.882
arrangement.

55:31.880 --> 55:34.060
So when they're infinitely far
they don't even know about each

55:34.063 --> 55:34.353
other.

55:34.349 --> 55:36.309
They don't feel any force.

55:36.309 --> 55:37.459
So here is my goal.

55:37.460 --> 55:40.360
I've got to do some work,
bring everybody together.

55:40.360 --> 55:41.800
They may all be positive.

55:41.800 --> 55:44.230
They may not want to be in the
same region, but I'm going to

55:44.233 --> 55:44.773
force them.

55:44.768 --> 55:47.808
The question is,
"What work do I have to

55:47.809 --> 55:48.569
do?"

55:48.570 --> 55:50.300
We're going to calculate that.

55:50.300 --> 55:53.940
We do that by saying first
let's take charge 1.

55:53.940 --> 55:56.970
Let it be wherever it has to be.

55:56.969 --> 56:02.949
Then I bring charge 2 from
infinity and put it here.

56:02.949 --> 56:07.189
The work done for that is by
definition q_2

56:07.188 --> 56:10.238
times the potential here,
potential there is

56:10.239 --> 56:12.439
q_1
/4Πε

56:12.436 --> 56:17.096
_0r_12,
r_12  is the

56:17.101 --> 56:19.471
distance between 1 and 2.

56:19.469 --> 56:20.809
Now I nail the two charges.

56:20.809 --> 56:23.379
Nobody is allowed to move so I
don't do further work.

56:23.380 --> 56:29.870
Then I bring a third charge
from infinity to its location

56:29.867 --> 56:30.677
here.

56:30.679 --> 56:32.969
How much work should I do?

56:32.969 --> 56:37.159
The work that I have to do is
the potential energy from

56:37.157 --> 56:40.567
infinity to here due to these
two charges.

56:40.570 --> 56:44.770
So that means the work I'll
have to do is equal to

56:44.768 --> 56:47.768
q_1q
_3/4

56:47.768 --> 56:51.708
Πε
_0r_13

56:51.710 --> 56:54.710
q_2q
_3/4

56:54.710 --> 56:59.510
Πε
_0r_23.

56:59.510 --> 57:01.150
You follow that?

57:01.150 --> 57:02.320
That's the work that I have to
do.

57:02.320 --> 57:05.030
The first guy,
q_2,

57:05.032 --> 57:07.282
had to fight only this one.

57:07.280 --> 57:09.710
The third guy has to fight
these two, and the fourth guy

57:09.710 --> 57:11.610
will have to fight these three
and so on.

57:11.610 --> 57:15.980
I'm just going up to three
charges.

57:15.980 --> 57:18.660
From this it should not be too
hard to guess in general if

57:18.657 --> 57:21.557
you've got many,
many charges you do

57:21.561 --> 57:26.221
q_iq
_j/4Πε

57:26.224 --> 57:33.114
_0r_ij
where i goes from 1 to--

57:33.110 --> 57:36.910
let's see, from 1 to n,
where j goes from 1 to

57:36.914 --> 57:41.874
n,
but i cannot be equal to

57:41.867 --> 57:45.907
j and you divide by 2.

57:45.909 --> 57:50.209
So if you're not able to read
this I'll tell you the summation

57:50.211 --> 57:53.111
is 1 over 2,
i from 1 to n,

57:53.114 --> 57:56.644
j from 1 to n,
but with the condition that

57:56.639 --> 57:59.969
i should not be equal to
j times all these.

57:59.969 --> 58:01.769
You can see that's valid here.

58:01.769 --> 58:02.659
Yep?

58:02.659 --> 58:04.029
Student:  V or
U?

58:04.030 --> 58:06.220
Prof: This is the--oh,
I'm sorry.

58:06.219 --> 58:07.719
You're absolutely right.

58:07.719 --> 58:09.189
This is not the potential.

58:09.190 --> 58:10.710
It's the potential energy.

58:10.710 --> 58:13.170
It should be called U.

58:13.170 --> 58:15.590
Did I say--yeah, that's right.

58:15.590 --> 58:16.990
Because
q/4Πε

58:16.987 --> 58:19.147
_0 is the
potential, but I multiply by

58:19.150 --> 58:20.210
q_2.

58:20.210 --> 58:25.060
That is actually an energy,
so these are all energies

58:25.057 --> 58:27.107
denoted by U.

58:27.110 --> 58:30.770
Now, if you have a continuous
set of charges you must replace

58:30.773 --> 58:32.243
a sum by the integral.

58:32.239 --> 58:35.429
Do you see what's going on here
why there's a half?

58:35.429 --> 58:36.879
Because this one counts
q_1q

58:36.882 --> 58:38.392
_3,
then again it counts

58:38.387 --> 58:39.827
q_3q
_1,

58:39.829 --> 58:42.139
but there's only one such thing
so I divided it by 2 and I have

58:42.144 --> 58:44.054
nothing like q_1q
_1,

58:44.050 --> 58:46.710
so you don't allow that.

58:46.710 --> 58:51.190
That's the energy to assemble a
whole bunch of charges together.

58:51.190 --> 58:52.350
So that's the work you do.

58:52.349 --> 58:55.999
If you do that work and you
take your hands off it'll fly

58:55.998 --> 58:59.708
apart and give you back that
energy in the form of kinetic

58:59.713 --> 59:00.433
energy.

59:00.429 --> 59:02.239
Now, let me do another simple
example.

59:02.239 --> 59:05.529
I want to take a sphere of
radius r,

59:05.532 --> 59:08.752
and I want to put q
coulombs on it,

59:08.748 --> 59:13.688
and I want to know how much
work I have to do for that one.

59:13.690 --> 59:17.970
What work do I have to do to
charge this sphere with q

59:17.965 --> 59:18.745
coulombs?

59:18.750 --> 59:23.410
Well, first couple of coulombs
come in they don't run into

59:23.405 --> 59:26.505
anything, so I put some charge
here.

59:26.510 --> 59:29.470
That makes it harder to bring
the next set of coulombs and the

59:29.465 --> 59:30.575
next set of coulombs.

59:30.579 --> 59:34.049
At some intermediate stage,
when the charge on this sphere

59:34.054 --> 59:37.294
is q and I want to bring
in a charge dq,

59:37.286 --> 59:39.356
how much work do I have to do?

59:39.360 --> 59:43.920
When the charge is q the
potential of the surface of the

59:43.920 --> 59:46.490
sphere is
q/4Πε

59:46.494 --> 59:49.074
_0 over R.

59:49.070 --> 59:52.230
The whole sphere is at that
potential and I'm trying to

59:52.226 --> 59:55.726
bring in a tiny more dq
from all sides and smear it on

59:55.733 --> 59:56.263
this.

59:56.260 --> 1:00:00.650
The work for that will be that
times dq.

1:00:00.650 --> 1:00:06.230
Then the total work I do when I
go all the way from 0 to capital

1:00:06.226 --> 1:00:09.586
Q will be a
Q^(2)/2 times

1:00:09.590 --> 1:00:13.310
1/4Πε
_0R.

1:00:13.309 --> 1:00:20.689
Sorry, capital R.

1:00:20.690 --> 1:00:24.110
Again, you see why you get the
half because the last coulomb is

1:00:24.110 --> 1:00:26.430
much harder to bring than the
first one.

1:00:26.429 --> 1:00:27.229
The first one is free.

1:00:27.230 --> 1:00:28.390
It's like a spring.

1:00:28.389 --> 1:00:31.989
When you pull it the first
inches are easier.

1:00:31.989 --> 1:00:34.759
The other ones are harder
because the force is kx,

1:00:34.764 --> 1:00:37.144
so the work done is the
integral of x that

1:00:37.141 --> 1:00:38.481
becomes x^(2)/2.

1:00:38.480 --> 1:00:41.700
Similarly to bring in an extra
charge you've got to worry about

1:00:41.697 --> 1:00:44.917
how much charge is already there
on this sphere that's opposing

1:00:44.916 --> 1:00:46.366
you with this potential.

1:00:46.369 --> 1:00:49.999
You work against that by
bringing dq from infinity

1:00:49.998 --> 1:00:52.978
and you integrate it you get
q^(2)/2.

1:00:52.980 --> 1:00:57.400
So this is a good problem to
explain the difference between

1:00:57.398 --> 1:00:59.148
potential and energy.

1:00:59.150 --> 1:01:03.430
This is the potential and this
is the charge I brought from

1:01:03.432 --> 1:01:07.222
infinity,
and that together is the

1:01:07.215 --> 1:01:14.695
du and integral of the
dU is my total u.

1:01:14.699 --> 1:01:17.079
So one can ask,
"What's the energy to

1:01:17.076 --> 1:01:19.386
produce any arrangement of
charges,"

1:01:19.394 --> 1:01:22.414
and you should be able to
either do the sum or do the

1:01:22.409 --> 1:01:23.279
integral.

1:01:23.280 --> 1:01:27.980
One homework problem will be
how to assemble a solid ball of

1:01:27.981 --> 1:01:30.771
charge, not a sphere,
solid ball.

1:01:30.768 --> 1:01:34.608
It's got no radius and slowly
it grows up to some size and you

1:01:34.612 --> 1:01:38.082
can do similar integrals where
you slap on more and more

1:01:38.079 --> 1:01:41.229
spherical shells on this onion
to build it up.

1:01:41.230 --> 1:01:44.600
Again, if you knew the
potential at any stage you can

1:01:44.601 --> 1:01:48.431
find the work to add a thin
layer to it and integrate it.

1:01:48.429 --> 1:01:56.509
Okay, so final topic is one of
capacitors.

1:01:56.510 --> 1:02:02.300
So a capacitor is like a way of
storing charge and storing

1:02:02.304 --> 1:02:03.224
energy.

1:02:03.219 --> 1:02:04.959
In other words,
if I want to store

1:02:04.960 --> 1:02:07.390
gravitational energy I can do
the following.

1:02:07.389 --> 1:02:11.069
I build this tank,
and maybe there's some water in

1:02:11.074 --> 1:02:14.164
the lake here at the foot of the
tank,

1:02:14.159 --> 1:02:17.489
and I go to the top,
and go to the top and empty

1:02:17.492 --> 1:02:21.822
little buckets until I've got
some water at the big height.

1:02:21.820 --> 1:02:26.110
Then I've done some work and
it'll pay me back when I open

1:02:26.110 --> 1:02:28.820
the tap and let it all rush
down,

1:02:28.820 --> 1:02:33.710
and it may turn a turbine blade
and give me back thermo electric

1:02:33.708 --> 1:02:34.328
power.

1:02:34.329 --> 1:02:36.799
You want to do similar things
electrically.

1:02:36.800 --> 1:02:40.180
You want to invest with some
work and it'll give you back.

1:02:40.179 --> 1:02:43.579
So one thing I can do is I can
take a plus and minus charge

1:02:43.583 --> 1:02:46.873
which love to be next to each
other and separate them.

1:02:46.869 --> 1:02:50.549
Now, I've done some work,
but I cannot walk around like

1:02:50.552 --> 1:02:51.032
this.

1:02:51.030 --> 1:02:52.590
I've got other things to do.

1:02:52.590 --> 1:02:55.300
I've got to find a better way
to store energy.

1:02:55.300 --> 1:02:56.840
How do you shake hands with
people?

1:02:56.840 --> 1:02:58.520
So you have to find another way.

1:02:58.519 --> 1:02:59.809
So here's what you do.

1:02:59.809 --> 1:03:03.949
You take one conductor and you
take another conductor,

1:03:03.952 --> 1:03:07.082
any shape you like,
it doesn't matter.

1:03:07.079 --> 1:03:09.909
They are both electrically
neutral, but they're conductors.

1:03:09.909 --> 1:03:14.819
So you take a tiny charge from
this guy and put it there.

1:03:14.820 --> 1:03:18.950
Then take more and more charge
and start putting it there.

1:03:18.949 --> 1:03:21.389
At any stage,
when you stop,

1:03:21.387 --> 1:03:26.987
you will have a lot of positive
charges on this conductor and a

1:03:26.985 --> 1:03:31.135
lot of negative charges on this
conductor.

1:03:31.139 --> 1:03:34.159
This conductor will be at one
potential because it's a

1:03:34.164 --> 1:03:34.854
conductor.

1:03:34.849 --> 1:03:37.249
That conductor will be at
another potential because it's a

1:03:37.251 --> 1:03:39.151
conductor, but both have unique
potentials.

1:03:39.150 --> 1:03:42.050
They are not the same,
but this whole conductor has

1:03:42.045 --> 1:03:44.415
one potential,
this has another potential,

1:03:44.418 --> 1:03:45.228
V'.

1:03:45.230 --> 1:03:47.540
And because it's harder and
harder to pump the coulombs you

1:03:47.539 --> 1:03:49.729
can imagine that there is a
potential difference between

1:03:49.731 --> 1:03:52.031
them,
because you've got to go uphill

1:03:52.027 --> 1:03:55.637
to pump even more charge into
this already charged object.

1:03:55.639 --> 1:03:59.249
So we define the capacitance as
the charge that you transferred

1:03:59.253 --> 1:04:01.823
divided by the voltage
difference you got.

1:04:01.820 --> 1:04:04.030
So let me call this 0 by
convention.

1:04:04.030 --> 1:04:08.970
That is called the capacitance.

1:04:08.969 --> 1:04:12.459
That's the ability of that
system to hold charge.

1:04:12.460 --> 1:04:14.330
If they're very,
very tiny metal spheres there's

1:04:14.326 --> 1:04:15.196
some amount of work.

1:04:15.199 --> 1:04:19.419
If they're huge metal spheres
you can transfer a lot of

1:04:19.418 --> 1:04:20.588
charge, okay?

1:04:20.590 --> 1:04:21.550
That's the definition.

1:04:21.550 --> 1:04:24.350
It's measured in farads.

1:04:24.349 --> 1:04:26.839
Again, it's got some units
coulomb for volt,

1:04:26.835 --> 1:04:30.295
but again it's got a different
name, named after Mr. Faraday,

1:04:30.304 --> 1:04:31.754
measured in farads.

1:04:31.750 --> 1:04:34.970
Usually you will have
microfarads in a lab.

1:04:34.969 --> 1:04:38.199
Farad is a huge unit.

1:04:38.199 --> 1:04:42.869
So let's find the capacitance
of certain simple arrangements.

1:04:42.869 --> 1:04:47.569
Simplest arrangement is the
parallel plate capacitor.

1:04:47.570 --> 1:04:48.440
Here it is.

1:04:48.440 --> 1:04:50.710
There's one plate.

1:04:50.710 --> 1:04:52.830
Here's a second plate.

1:04:52.829 --> 1:04:54.969
The area of the plate is
A.

1:04:54.969 --> 1:04:57.829
The distance between them is
d.

1:04:57.829 --> 1:05:02.989
And I put plus charges on this
one and I got them from this

1:05:02.985 --> 1:05:03.515
one.

1:05:03.519 --> 1:05:04.349
Do you understand that?

1:05:04.349 --> 1:05:06.649
You don't have to bring your
own charge.

1:05:06.650 --> 1:05:10.370
You remove charge from one
conductor you therefore leave it

1:05:10.371 --> 1:05:13.971
negatively charged and you dump
it on the other conductor

1:05:13.965 --> 1:05:16.655
thereby leaving it positively
charged.

1:05:16.659 --> 1:05:18.899
So you don't need to bring
charge from the outside world.

1:05:18.900 --> 1:05:21.470
You just redistribute the
neutrality so that one is now

1:05:21.472 --> 1:05:22.952
negative and one is positive.

1:05:22.949 --> 1:05:26.179
So Q stands for the
magnitude of charge that you

1:05:26.181 --> 1:05:27.021
transferred.

1:05:27.018 --> 1:05:29.568
So suppose I transfer some
charge Q.

1:05:29.570 --> 1:05:34.660
My job is to find Q
divided by V and that's

1:05:34.655 --> 1:05:36.195
my capacitance.

1:05:36.199 --> 1:05:39.039
Okay, so how am I going to find
the voltage difference between

1:05:39.041 --> 1:05:39.601
these two?

1:05:39.599 --> 1:05:42.909
I think you guys should know by
now voltage difference is the

1:05:42.913 --> 1:05:45.953
electric field times the
distance between the plates.

1:05:45.949 --> 1:05:47.729
There's no line integral.

1:05:47.730 --> 1:05:51.160
E⋅dr
is just E times d.

1:05:51.159 --> 1:05:55.329
E, if you remember,
is

1:05:55.327 --> 1:06:02.467
σ/ε_0
between the plates,

1:06:02.469 --> 1:06:07.679
and sigma is Q/A.

1:06:07.679 --> 1:06:10.689
Charge density is the total
charge over that.

1:06:10.690 --> 1:06:16.600
And that is by definition
Q over C,

1:06:16.599 --> 1:06:24.459
therefore the capacitance of
this parallel plate capacitor is

1:06:24.463 --> 1:06:29.973
ε_0
A/d.

1:06:29.969 --> 1:06:32.679
So that's one of the easier
calculations to do.

1:06:32.679 --> 1:06:37.629
So anytime someone gives you
two objects and says find for me

1:06:37.628 --> 1:06:41.668
the capacitance you will take
charge from one,

1:06:41.670 --> 1:06:46.130
you will put it on the other
one and then you will--

1:06:46.130 --> 1:06:50.790
hey, where is all these
blackboards?

1:06:50.789 --> 1:06:52.729
Ah, bad choice.

1:06:52.730 --> 1:06:55.230
So you guys took all this down?

1:06:55.230 --> 1:06:56.370
Yes?

1:06:56.369 --> 1:06:59.969
Student:  Should that
A be on the bottom for

1:06:59.974 --> 1:07:02.444
the A Q over
________________?

1:07:02.440 --> 1:07:06.710
Prof: Oh, yes.

1:07:06.710 --> 1:07:09.110
σ is Q/A.

1:07:09.110 --> 1:07:11.540
Yeah, this is what comes from
knowing the answer.

1:07:11.539 --> 1:07:13.369
This is what you guys do in the
exam.

1:07:13.369 --> 1:07:17.169
When you turn that page then
all kinds of stuff happens.

1:07:17.170 --> 1:07:19.300
Signs change,
numbers change and the right

1:07:19.295 --> 1:07:21.055
answer appears on the next page.

1:07:21.059 --> 1:07:23.919
That's really what I did
because I knew this formula.

1:07:23.920 --> 1:07:26.860
We're getting there no matter
what, but you're quite right.

1:07:26.860 --> 1:07:29.980
I didn't put the A in
the right place.

1:07:29.980 --> 1:07:32.530
So any way, it's not good to
know the answer every time

1:07:32.534 --> 1:07:34.574
because you're not that alert
doing this.

1:07:34.570 --> 1:07:37.930
Anyway, this is the correct
answer.

1:07:37.929 --> 1:07:42.329
So here's the capacitance of
this parallel plate.

1:07:42.329 --> 1:07:46.089
Here's another problem you
could do.

1:07:46.090 --> 1:07:46.850
These are also common.

1:07:46.849 --> 1:07:51.039
Spherical capacitors are one
sphere of radius a

1:07:51.038 --> 1:07:55.538
surround by second sphere of
radius b and you shove

1:07:55.541 --> 1:08:00.861
charges,
maybe plus on this one and

1:08:00.856 --> 1:08:06.626
equal number of minus on that
one.

1:08:06.630 --> 1:08:08.550
So what's my goal?

1:08:08.550 --> 1:08:11.600
Goal is to find the electric
field in the region between

1:08:11.596 --> 1:08:13.616
them,
right, and integrate it from

1:08:13.621 --> 1:08:15.371
here to here,
and to give the potential

1:08:15.371 --> 1:08:17.501
difference,
and divide by the charge.

1:08:17.500 --> 1:08:21.580
So if I put a charge Q
here you can see that the

1:08:21.577 --> 1:08:25.347
electric field in this region is
just Q over

1:08:25.351 --> 1:08:28.751
4Πε
_0r^(2),

1:08:28.750 --> 1:08:33.620
because by Gauss's Law if I
draw a sphere here that encloses

1:08:33.621 --> 1:08:37.521
charge Q,
and it's isotropic and uniform,

1:08:37.515 --> 1:08:40.285
so it's like that of a point
charge.

1:08:40.288 --> 1:08:43.018
Therefore, the potential
difference between these two

1:08:43.020 --> 1:08:45.700
points will be like the
potential difference between

1:08:45.698 --> 1:08:48.638
these two points with a point
charge of the origin,

1:08:48.640 --> 1:08:54.790
so V_2 -
V_1 will be

1:08:54.788 --> 1:09:00.358
Q/4Πε
_0 times

1:09:00.359 --> 1:09:03.839
[1/a - 1/b].

1:09:03.840 --> 1:09:07.480
Well, we are done because this
is what I call V.

1:09:07.479 --> 1:09:12.909
The capacitance is
Q/V and that's

1:09:12.905 --> 1:09:19.505
going to be 4Πε
_0 divided by

1:09:19.512 --> 1:09:23.052
[1/a −
1/b]

1:09:23.051 --> 1:09:29.421
which is 4Πε
_0ab/(b

1:09:29.421 --> 1:09:32.491
− a).

1:09:32.488 --> 1:09:35.758
So in every geometry if they
want capacitance move charge

1:09:35.761 --> 1:09:39.551
from one guy to the other guy,
and find the field between

1:09:39.545 --> 1:09:42.895
them, then integrate it to find
the potential.

1:09:42.899 --> 1:09:45.619
The potential will always be
proportional to the charge.

1:09:45.618 --> 1:09:48.618
Take the ratio and that's
capacitance.

1:09:48.618 --> 1:09:50.888
I'm going to put this guy to a
test.

1:09:50.890 --> 1:09:52.870
Let me write down clearly the
answer here.

1:09:52.868 --> 1:09:59.048
4Πε
_0ab/(b

1:09:59.051 --> 1:10:01.801
− a).

1:10:01.800 --> 1:10:07.110
What test can I do if I want to
compare it to a parallel plate

1:10:07.105 --> 1:10:08.145
capacitor?

1:10:08.149 --> 1:10:11.189
Can you see how I can relate it
to parallel plate capacitor?

1:10:11.189 --> 1:10:12.099
Yep?

1:10:12.100 --> 1:10:15.910
Student:  Make the a
close.

1:10:15.909 --> 1:10:16.819
Prof: a what?

1:10:16.819 --> 1:10:19.019
Student:  b -
a is close.

1:10:19.020 --> 1:10:21.470
Prof: Right,
you can take b -

1:10:21.467 --> 1:10:24.377
a very close,
but it still may not look like

1:10:24.381 --> 1:10:26.191
a parallel plate capacitor.

1:10:26.189 --> 1:10:28.109
So here's what I say we do.

1:10:28.109 --> 1:10:31.099
Make b and a
astronomical,

1:10:31.104 --> 1:10:33.184
size of the galaxy,
okay?

1:10:33.180 --> 1:10:36.000
And b - a is just
1 inch.

1:10:36.000 --> 1:10:38.340
Then in that case it will
really look like two infinite

1:10:38.340 --> 1:10:38.730
planes.

1:10:38.729 --> 1:10:42.139
The fact that it gets wrapped
around over the length of the

1:10:42.143 --> 1:10:45.503
galaxy is just like saying the
earth is round which is not

1:10:45.497 --> 1:10:47.497
obvious from short excursions.

1:10:47.500 --> 1:10:50.380
In that case we can write
4Πε

1:10:50.377 --> 1:10:51.677
_0.

1:10:51.680 --> 1:10:54.580
a and b are
really equal to some common

1:10:54.579 --> 1:10:58.039
number R and b -
a is the distance between

1:10:58.037 --> 1:11:00.267
the inner shell and the outer
shell.

1:11:00.270 --> 1:11:03.880
That you can see.

1:11:03.880 --> 1:11:05.010
Wait.

1:11:05.010 --> 1:11:08.980
I'm sorry, 4ΠR^(2) is
the area times

1:11:08.984 --> 1:11:12.704
ε_0
over d,

1:11:12.695 --> 1:11:13.575
right?

1:11:13.579 --> 1:11:15.169
So that's one test.

1:11:15.170 --> 1:11:22.360
It doesn't mean it's right,
but if this test failed it

1:11:22.359 --> 1:11:26.699
means you're definitely wrong.

1:11:26.698 --> 1:11:29.888
Okay, now how much energy does
it take to charge up a

1:11:29.893 --> 1:11:30.633
capacitor?

1:11:30.630 --> 1:11:32.980
It's what I did earlier,
but let me do it again.

1:11:32.979 --> 1:11:36.459
If you take two plates or any
two surfaces,

1:11:36.460 --> 1:11:40.080
and there's a certain amount of
charge q on it and I

1:11:40.082 --> 1:11:42.832
transfer delta q the work
I do is,

1:11:42.828 --> 1:11:48.258
the delta q times the
voltage at that stage.

1:11:48.260 --> 1:11:52.780
Voltage at that stage is equal
to q/C,

1:11:52.783 --> 1:11:57.223
which is the q you have
up to that point.

1:11:57.220 --> 1:12:06.470
Therefore, the total work you
do is

1:12:15.170 --> 1:12:18.490
So when you want to charge a
capacitor with a charge q

1:12:18.492 --> 1:12:21.812
on it either you can write it
that way or you can also do the

1:12:21.814 --> 1:12:22.594
following.

1:12:26.201 --> 1:12:28.761
Q/C is a V.

1:12:28.760 --> 1:12:30.660
Multiply top and bottom by
C.

1:12:30.658 --> 1:12:34.168
You can write the energy in a
capacitor either in terms of the

1:12:34.171 --> 1:12:37.741
charge stored or in terms of the
voltage between the plates.

1:12:37.739 --> 1:12:41.869
So we'll resume this next time.

1:12:41.869 --> 1:12:46.999