All language subtitles for LS1U6B1

So you may have noticed when we did
this problem, there were several

points where we had to think about it.

Just, which way do things point?

And we had to arbitrarily say, oh well,
it's in the negative x direction

or the negative y direction.

So sometimes you want a way where
you don't have to do that.

If that really bothers you,

I'm going to show you another
way to do this problem.

However, it's a little bit risky because
it may completely destroy your

understanding of electrostatics.

You may never recover from this.

So if you're really happy with how we
did it in unit 5, just turn it off.

Don't watch.

You don't want to see this.

If unit 5 was just completely
enjoyable to you, then stop.

But if it bothers you, if you say, well,
it's negative, then watch this

at your own risk.

To do it this way--

We're going to try to do it without all
that inspection, all that thinking

about which way things point.

And to do it, you really have to be
really good with unit vectors.

So let's look at our
unit vector again.

For example, let's do r hat 1-2.

Well, we called it 2-1.

Let's do 2-1, r hat 2.

When we looked at it, we had to
just think about which way

does r hat 2-1 go?

But we can also go with a mathematical
definition of r hat 2-1.

Because what was r hat 2-1?

It's a vector in the direction
along this axis.

It has a magnitude of 1.

So let's think, how do we make that?

Well, we need yet another vector.

We need the vector r 2-1.

Ah, the vector hat 2-1.

The vector 2-1 is literally just, I
start my chalk at 2 and I go to 1.

And I put an arrow head on it.

It's the vector from 2 to 1.

So that's pretty straightforward.

There's no guesswork there.

Positive, negative, you don't
think about that.

You just put your pen
on 2, draw it to 1.

That's vector 2-1.

That will be guaranteed to be in the
direction on the axis from 2 to 1.

Now we just need its magnitude
to be correct.

So what do we divide?

Well, we divide by its length,
by the displacement r 2-1.

So that's actually a mathematical way
to get that unit vector, rather than

just looking at and thinking,
well, negative 1.

You might see in some books, they
might write it like this.

And since this is a magnitude,
they might do it like that.

Go ahead and call it the vector r
2-1 and put the bars around it.

Or what we've been doing is when we see
nothing on here, when it's just

empty on top, we know that's
a magnitude, that's a

displacement in this case.

So now, here's a scary part.

Now, we're going to write Coulomb's
law again for the case 2 to 1.

It's going to be ke.

It's still going to be q2 q1.

But now, it's going to be over r
2-1 1 cubed times r vector 2-1.

So before, we had r hat
2-1 1 sitting here.

So now if I bring that here, that puts
another r 2-1 in the bottom.

And now it's cubed.

But I have an r vector 2-1 there.

So when I write it this
way, is it different?

Is it some different Coulomb's law?

Is it a Coulomb's law that
goes 1 over r cubed?

No.

It still goes as 1 over r squared.

We have an r cubed down here.

But now, we have an r in the top.

And when you cancel those out, it's
still going as 1 over r squared.

It's the same Coulomb's law.

It's just a way to write it where we
don't have to guess what r hat 2-1 is.

So let's use it a few times and
let's see if it actually does

anything for us.

So let's write F2-1 is
9 times 10 to the 9--

again, this is all in KMS units--
times 10 times 10 to the minus 6

squared over the separation cubed,
which is 0.15 cubed.

And now, we just write
the vector r 2-1.

Well, it's 0.15.

And it's pointing down.

So it's negative 0.15 in
its i hat direction.

If you multiply that out,
you get minus 0.15.

I'm sorry j hat.

j hat direction, you get minus--

If you multiply that out,
you get minus 40 j hat.

Sorry.

And then we could also do F4-1.

And it's the same thing.

9 times 10 to the 9 over
0.6 cubed times--

And on the top, we have 10 times
10 to the minus 6 squared.

And then we just write
the vector r 4-1.

Well, that's minus 0.6 because it's
this way, i hat direction.

Minus 0.6 i hat.

And that'll give you the same answer
for that component minus

2.5 in the i hat.

And now, I can tell your
massively unimpressed.

It's the same thing.

All I'm doing is cubing it here
and then putting 1 up here.

And they cancel.

And I still had to look at it to figure
out which way r vector was.

So really, is this any better?

Let's do one more part.

Let's do F3-1.

That was the difficult one anyway.

Let's see.

F3-1, that would be ke 9 times 10 to
the 9, 10 times 10 to the minus 6

squared over the separation.

So remember the separation, it
was the square root of 0.6

squared plus 0.15 squared.

Before, I didn't even bother with the
squares and the square roots.

I just wrote it down that way.

But now, it's not squared.

It's cubed.

So now, to write this correctly, it's
0.6 squared plus 0.15 squared.

And it's to the 3/2.

It's a square root of that to the
1/2, but it's also a cube.

So it's to the 3.

So you multiply those.

It's to the 3/2 in the bottom.

And now, we don't write
the unit vector.

We just literally write
our r 3-1 as a vector.

Well, we look at our diagram.

And you remember, it was minus 0.6
i hat and minus 0.15 j hat.

And that's it.

That is your vector.

If you break that down into components,
solve everything, do this

0.6 squared plus 0.15 squared to the
3/2, you'll get minus 2.28 i hat and

minus 0.57 j hat.

It gives the same thing
we got last time.

And then when you go add them up, F2-1,
F4-1, F3-1, of course, you still

get the same thing.

So you see, we got the same
answer with this formula.

But we never did any trig.

How did we do it without trig?

Well, you don't always need trig.

Trig is really, in this case and in a
lot of cases, just a way to do ratios.

There's other ways to get to ratio
something out, rather than having to

use a tangent function or an inverse
tangent to get theta.

So it's really the same thing.

It's just a way to do it without really
having to think about the trig

and the angle and the 14 degrees.

We never put that in.

All we had to be able to do was
write the vectors r, r 2-1,

r 3-1, and r 4-1.

So this is just another way to do it.

But the most important thing to realize
is this formula, Coulomb's

law, still goes as 1 over r squared.

We have a cubed here.

And we have an r magnitude there.

That's the main thing to keep in mind.

But this can save you a lot of
trouble in some problems.