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- [Instructor] Let's try a hard one.

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This one's a classic.

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Let's say you have two charges,

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positive eight nanoCoulombs and
negative eight nanoCoulombs,

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and instead of asking
what's the electric field

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somewhere in between, which is essentially

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a one-dimensional
problem, we're gonna ask,

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what's the electric field
up here, at this point, P?

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Now, this is a two-dimensional problem

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because if we wanna find
the net electric field

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up here, the magnitude n direction

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of the net electric field at this point,

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we approach it the same way initially.

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We say alright, each charge
is gonna create a field

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up here that goes in a certain direction.

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This positive charge creates a field

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up here that goes radially away from it,

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and radially away from this positive

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at point P is something like this.

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I'll call this electric field blue E

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because it's created by
this blue positive charge,

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and this negative charge
creates its own electric field

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at that point that goes
radially into the negative,

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and radially into the negative

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is gonna look something like this.

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I'll call this electric field yellow E

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because it's created by
the yellow electric field.

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So far so good.

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Same approach, but now
things get a little weird.

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Look, these fields aren't even pointing

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in the same direction.

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They're lying in this
two-dimensional plane,

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and we wanna find the net electric field.

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So what we have to do in these
2D electric field problems

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is break up the electric
fields into their components.

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In other words, the field
created by this positive charge

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is gonna have a horizontal component,

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and that's gonna point to the right.

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And I'll call that blue E x
because it was the horizontal

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component created by the
blue, positive charge.

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And this electric field
is gonna have a vertical

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component, that's gonna point upward.

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I'll call that blue E y.

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And similarly, for the electric
field this negative charge

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creates, it has a horizontal component

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that points to the right.

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We'll call that yellow E x,
and a vertical component,

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but this vertical
component points downward.

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I'll call that yellow E y.

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What do we do with all these components

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to find the net electric field?

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Typically what you do in
these 2D electric problems

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is focus on finding the components

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of the net electric field in
each direction separately.

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We divide and conquer.

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We're gonna ask, what's
the horizontal component

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of the net electric field, and what's

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the vertical component of
the net electric field?

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And then once we know
these, we can combine them

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using the Pythagorean
theorem if we want to,

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to get the magnitude of
that net electric field.

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But we're kind of in luck in this problem.

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There's a certain amount of symmetry

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in this problem, and when
there's a certain amount

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of symmetry, you can save a lot of time.

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What I mean by that is that both

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of these charges have the
same magnitude of charge.

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And because this point, P, lies directly

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in the middle of them, the distance

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from the charge to point
P is gonna be the same

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as the distance from the
negative charge to point P,

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so both of these charges
create an electric field

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at this point of equal magnitude.

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The fields just point
in different directions,

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and what that means is
that this positive charge

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will create an electric field that has

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some vertical component upward
of some positive amount.

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We don't know exactly how much that is,

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but it'll be a positive
number because it points up,

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and this negative charge is gonna create

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an electric field that has a
vertical component downward,

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which is gonna be negative,
but it's gonna have

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the same magnitude as
the vertical component

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of the blue electric field.

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In other words, the field
created by the positive charge

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is just as upward as the field created

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by the negative charge is downward.

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So when you add those up, when you add

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up these two vertical
components to find the vertical

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component of the net electric field,

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you're just gonna get zero.

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They're gonna cancel completely,

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which is nice because
that means we only have

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to worry about the horizontal components.

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These will not cancel.

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How come these don't cancel?

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Because they're both
pointing to the right.

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If one was pointing right
and the other was left,

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then the horizontal
components would cancel,

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but that's not what happens here.

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These components combine
to form a total component

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in the x direction that's
larger than either one of them.

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In fact, it's gonna be twice as big

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because each charge
creates the same amount

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of electric field in this x direction

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because of the symmetry of this problem.

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So we've reduced this
problem to just finding

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the horizontal component
of the net electric field.

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To do that, we need the
horizontal components

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of each of these
individual electric fields.

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If I can find the horizontal component

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of the field created
by the positive charge,

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that's gonna be a positive contribution

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to the total electric
field, since this points

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to the right, and I'd add that

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to the horizontal component
of the yellow electric field

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because it also points to the right,

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even though the charge creating
that field is negative,

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the horizontal component
of that field is positive

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because it points to the right.

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So if I can get both of these,

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I will just add these
up, and I'd get my total

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electric field in the x direction.

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How do I get these?

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How do I determine these
horizontal components?

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Well, to get the horizontal component

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of this blue electric field, I first need

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to find what's the magnitude
of this blue electric field.

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We know the formula for that.

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I'll write it over here.

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The magnitude of the electric field

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is always k Q over r squared.

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So for this blue field, we'll say that E

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is equal to nine times 10 to the ninth,

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and the charge is eight nanoCoulombs.

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Nano means 10 to the negative ninth.

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And then we divide by the r,
but what's the r in this case?

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It's not four or three.

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Remember, the r in that
electric field formula

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is always from the charge
to the point you're trying

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to determine the electric field at.

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So r is this.

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This distance is r.

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How do we figure out what this is?

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Well, we're kind of in luck.

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If you know about three,
four, five triangles,

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look at, this forms a
three, this side is three,

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meters, and this side is four meters.

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That means that this side automatically

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we know is five meters.

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If you're not comfortable with that,

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you can always do the Pythagorean theorem.

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Pythagorean theorem says that a squared

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plus b squared equals c
squared for a right triangle,

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which is what we have here.

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A is this side, three.

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B is the four meter side.

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And then c would be r,
I'll call that r squared.

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And if you solve this for r, nine plus 16,

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square root gives you r is
five meters, just like we said.

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But if you know three,
four, five triangles,

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it's kinda nice because
you could just quote that.

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And that's the r we're gonna use up here.

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We'll use five meters squared,

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which, if you calculate, you get

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that the electric field is
2.88 Newtons per Coulomb.

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This is the magnitude
of the electric field

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created at this point, P,
by the positive charge.

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How do we get the horizontal
component of that field?

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There's a few ways to do it.

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One way to do it is first
just find this angle here.

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If we could find what that angle is,

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we can do trigonometry to get
this horizontal component.

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How do I find this angle?

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Well, you note that that angle's gonna be

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the same as this angle down here.

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These are gonna be similar angles

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because I've got horizontal lines

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and then this diagonal line
just continues right through.

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So this angle is the same as this angle,

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so if I could find this angle here,

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I've found that angle up top.

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How do I get this angle?

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I know each side of this triangle,

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so I can use either
sine, cosine, or tangent.

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I'm just gonna use tangent.

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We'll say that tangent of that angle

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is defined always to be
the opposite over adjacent.

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We know the opposite side to this angle

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is four meters, and the
adjacent side was three meters,

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so tangent theta's gonna equal 4/3.

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How do we get theta?

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We say that theta's going to equal

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the inverse tangent of 4/3.

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We basically take inverse
tangent of both sides.

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We get theta on the left,
and if you plug this

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into your calculator,
you get 53.1 degrees.

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So that's what this angle is right here.

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This is 53.1 degrees, but
that means this angle up here

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is also 53.1 degrees because
these are the same angle.

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This horizontal component is not the same

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as this three meters?

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And this diagonal electric
field is not the same

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as five meters, but the angle
between those components

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are the same as the angle
between these length components.

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So what do I do to get
this horizontal component?

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This is the adjacent side to this angle,

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so this E x is adjacent to that angle.

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We're gonna use cosine.

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We're gonna say that
cosine of 53.1 degrees

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is gonna be equal to the
adjacent side, which is E x.

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We'll write this as E x divided

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by the hypotenuse, and
we found the hypotenuse.

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This is the magnitude of the total

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electric field right here,
which is the hypotenuse

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of this triangle, so that's 2.88.

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And we get that E x is
going to be 2.88 Newtons

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per Coulomb times cosine of 53.1,

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which, if you plug that
into the calculator

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is gonna give you 1.73
Newtons per Coulomb.

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This is how much electric field

222
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the positive charge
creates in the x direction.

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That's what this component up here is.

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This is 1.73 Newtons per Coulomb.

225
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So that's what this is.

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That's what I'm gonna plug in here.

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To get this horizontal component

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of the yellow field created
by the negative charge,

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you could go through the whole thing again

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or you could notice that because

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of symmetry, this horizontal component

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has to be the exact same
as the horizontal component

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created by the positive charge.

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They're both 1.73, and
they're both positive

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because both of these
components point to the right.

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So to get the total electric field

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in the x direction, we'll take 1.73

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from the positive charge
and we'll add that

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to the horizontal component
from the negative charge,

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which is also positive 1.73, to get

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a horizontal component in the x direction

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of the net electric field equal to

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3.46 Newtons per Coulomb.

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This is the horizontal component

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of the net electric field at that point.

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We basically took both of
these values and added them up,

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which, essentially is just
one of them times two.

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And now you might be
worried though, this is just

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the horizontal component
of the net electric field.

250
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How do we get the magnitude of
the total net electric field?

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Well, this is gonna be the same value

252
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because since there was
no vertical component

253
00:09:13,169 --> 00:09:15,620
of the electric field,
the horizontal component

254
00:09:15,620 --> 00:09:17,788
is gonna be equal to the magnitude

255
00:09:17,788 --> 00:09:20,104
of the total electric field at that point.

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00:09:20,104 --> 00:09:21,535
If there was a vertical component

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00:09:21,535 --> 00:09:22,720
of the electric field, we'd have

258
00:09:22,720 --> 00:09:25,579
to do the Pythagorean theorem
to get the total magnitude

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00:09:25,579 --> 00:09:27,981
of the net electric field,
but since there was only

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00:09:27,981 --> 00:09:29,678
a horizontal component, and these

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00:09:29,678 --> 00:09:32,427
vertical components canceled,
the total electric field's

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00:09:32,427 --> 00:09:35,094
just gonna point to the
right, and it will be equal

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00:09:35,094 --> 00:09:38,793
to two times one of these
horizontal components,

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00:09:38,793 --> 00:09:39,878
which, when you add them up,

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00:09:39,878 --> 00:09:43,244
gives you 3.46 Newtons per Coulomb.

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00:09:43,244 --> 00:09:45,286
That's the magnitude of
the net electric field,

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00:09:45,286 --> 00:09:47,943
and the direction would
be straight to the right.

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00:09:47,943 --> 00:09:50,810
So recapping, when you have
a 2D electric field problem,

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00:09:50,810 --> 00:09:52,993
draw the field created by each charge,

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00:09:52,993 --> 00:09:56,128
break those fields up into
their individual components.

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00:09:56,128 --> 00:09:57,543
If there's any symmetry involved,

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00:09:57,543 --> 00:09:59,947
figure out which component cancels,

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00:09:59,947 --> 00:10:01,794
and then to find the net electric field,

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00:10:01,794 --> 00:10:03,692
use the component that doesn't cancel,

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00:10:03,692 --> 00:10:05,009
and determine the contribution

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00:10:05,009 --> 00:10:07,359
from each charge in that direction.

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00:10:07,359 --> 00:10:09,076
Add or subtract them accordingly,

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00:10:09,076 --> 00:10:10,576
based on whether those components point

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00:10:10,576 --> 00:10:13,109
to the right or to the
left, and that will give you

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00:10:13,109 --> 00:10:15,392
your net electric field at that point,

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00:10:15,392 --> 00:00:00,000
created by both charges.