This one's a classic.

Let's say you have two charges,

positive eight nanoCoulombs and negative eight nanoCoulombs,

and instead of asking what's the electric field

somewhere in between, which is essentially

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

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

Now, this is a two-dimensional problem

because if we wanna find the net electric field

up here, the magnitude n direction

of the net electric field at this point,

we approach it the same way initially.

We say alright, each charge is gonna create a field

up here that goes in a certain direction.

This positive charge creates a field

up here that goes radially away from it,

and radially away from this positive

at point P is something like this.

I'll call this electric field blue E

because it's created by this blue positive charge,

and this negative charge creates its own electric field

at that point that goes radially into the negative,

and radially into the negative

is gonna look something like this.

I'll call this electric field yellow E

because it's created by the yellow electric field.

So far so good.

Same approach, but now things get a little weird.

Look, these fields aren't even pointing

in the same direction.

They're lying in this two-dimensional plane,

and we wanna find the net electric field.

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

is break up the electric fields into their components.

In other words, the field created by this positive charge

is gonna have a horizontal component,

and that's gonna point to the right.

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

component created by the blue, positive charge.

And this electric field is gonna have a vertical

component, that's gonna point upward.

I'll call that blue E y.

And similarly, for the electric field this negative charge

creates, it has a horizontal component

that points to the right.

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

but this vertical component points downward.

I'll call that yellow E y.

What do we do with all these components

to find the net electric field?

Typically what you do in these 2D electric problems

is focus on finding the components

of the net electric field in each direction separately.

We divide and conquer.

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

of the net electric field, and what's

the vertical component of the net electric field?

And then once we know these, we can combine them

using the Pythagorean theorem if we want to,

to get the magnitude of that net electric field.

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

There's a certain amount of symmetry

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

of symmetry, you can save a lot of time.

What I mean by that is that both

of these charges have the same magnitude of charge.

And because this point, P, lies directly

in the middle of them, the distance

from the charge to point P is gonna be the same

as the distance from the negative charge to point P,

so both of these charges create an electric field

at this point of equal magnitude.

The fields just point in different directions,

and what that means is that this positive charge

will create an electric field that has

some vertical component upward of some positive amount.

We don't know exactly how much that is,

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

and this negative charge is gonna create

an electric field that has a vertical component downward,

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

the same magnitude as the vertical component

of the blue electric field.

In other words, the field created by the positive charge

is just as upward as the field created

by the negative charge is downward.

So when you add those up, when you add

up these two vertical components to find the vertical

component of the net electric field,

you're just gonna get zero.

They're gonna cancel completely,

which is nice because that means we only have

to worry about the horizontal components.

These will not cancel.

How come these don't cancel?

Because they're both pointing to the right.

If one was pointing right and the other was left,

then the horizontal components would cancel,

but that's not what happens here.

These components combine to form a total component

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

In fact, it's gonna be twice as big

because each charge creates the same amount

of electric field in this x direction

because of the symmetry of this problem.

So we've reduced this problem to just finding

the horizontal component of the net electric field.

To do that, we need the horizontal components

of each of these individual electric fields.

If I can find the horizontal component

of the field created by the positive charge,

that's gonna be a positive contribution

to the total electric field, since this points

to the right, and I'd add that

to the horizontal component of the yellow electric field

because it also points to the right,

even though the charge creating that field is negative,

the horizontal component of that field is positive

because it points to the right.

So if I can get both of these,

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

electric field in the x direction.

How do I get these?

How do I determine these horizontal components?

Well, to get the horizontal component

of this blue electric field, I first need

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

We know the formula for that.

I'll write it over here.

The magnitude of the electric field

is always k Q over r squared.

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

is equal to nine times 10 to the ninth,

and the charge is eight nanoCoulombs.

Nano means 10 to the negative ninth.

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

It's not four or three.

Remember, the r in that electric field formula

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

to determine the electric field at.

So r is this.

This distance is r.

How do we figure out what this is?

Well, we're kind of in luck.

If you know about three, four, five triangles,

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

meters, and this side is four meters.

That means that this side automatically

we know is five meters.

If you're not comfortable with that,

you can always do the Pythagorean theorem.

Pythagorean theorem says that a squared

plus b squared equals c squared for a right triangle,

which is what we have here.

A is this side, three.

B is the four meter side.

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

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

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

But if you know three, four, five triangles,

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

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

We'll use five meters squared,

which, if you calculate, you get

that the electric field is 2.88 Newtons per Coulomb.

This is the magnitude of the electric field

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

How do we get the horizontal component of that field?

There's a few ways to do it.

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

If we could find what that angle is,

we can do trigonometry to get this horizontal component.

How do I find this angle?

Well, you note that that angle's gonna be

the same as this angle down here.

These are gonna be similar angles

because I've got horizontal lines

and then this diagonal line just continues right through.

So this angle is the same as this angle,

so if I could find this angle here,

I've found that angle up top.

How do I get this angle?

I know each side of this triangle,

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

I'm just gonna use tangent.

We'll say that tangent of that angle

is defined always to be the opposite over adjacent.

We know the opposite side to this angle

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

so tangent theta's gonna equal 4/3.

How do we get theta?

We say that theta's going to equal

the inverse tangent of 4/3.

We basically take inverse tangent of both sides.

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

into your calculator, you get 53.1 degrees.

So that's what this angle is right here.

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

is also 53.1 degrees because these are the same angle.

This horizontal component is not the same

as this three meters?

And this diagonal electric field is not the same

as five meters, but the angle between those components

are the same as the angle between these length components.

So what do I do to get this horizontal component?

This is the adjacent side to this angle,

so this E x is adjacent to that angle.

We're gonna use cosine.

We're gonna say that cosine of 53.1 degrees

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

We'll write this as E x divided

by the hypotenuse, and we found the hypotenuse.

This is the magnitude of the total

electric field right here, which is the hypotenuse

of this triangle, so that's 2.88.

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

per Coulomb times cosine of 53.1,

which, if you plug that into the calculator

is gonna give you 1.73 Newtons per Coulomb.

This is how much electric field

the positive charge creates in the x direction.

That's what this component up here is.

This is 1.73 Newtons per Coulomb.

So that's what this is.

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

To get this horizontal component

of the yellow field created by the negative charge,

you could go through the whole thing again

or you could notice that because

of symmetry, this horizontal component

has to be the exact same as the horizontal component

created by the positive charge.

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

because both of these components point to the right.

So to get the total electric field

in the x direction, we'll take 1.73

from the positive charge and we'll add that

to the horizontal component from the negative charge,

which is also positive 1.73, to get

a horizontal component in the x direction

of the net electric field equal to

3.46 Newtons per Coulomb.

This is the horizontal component

of the net electric field at that point.

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

which, essentially is just one of them times two.

And now you might be worried though, this is just