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### Course: Electromagnetism (Essentials) - Class 12th > Unit 3

Lesson 2: Electric field due to multiple charges - Superposition!# Net electric field from multiple charges in 2D

In this video David solves an example 2D electric field problem to find the net electric field at a point above two charges. Created by David SantoPietro.

## Want to join the conversation?

- An example without symmetry would be amazing. Would we just use the vectors to determine the direction and magnitude?(64 votes)
- It would be like how we sound the horizontal component in the example, but we would also use the method to find the vertical componant(15 votes)

- at6:00, david solved this example taking into account the angle between the electric field and its x-component, isn't there any alternative to solve this problem other than finding the angles and then solving it?(9 votes)
- The alternative way is looking at the original triangle and observing that the horizontal component is 3/5 of the hypotenuse. So the electric field in the horizontal direction is 3/5 of the original electric field (2.88 N/C in our case).

(3/5)x(2.88 N/C)=1.728 N/C

Dave rounded that off to 1.73 N/C(23 votes)

- Are the words charge density and charge distribution refers to the same thing ?(4 votes)
- At9:21, If we did have a vertical component...how do we get the final value i.e how do you apply the pythagorean formula?(5 votes)
- Just as in pytagorean theoren we have h=√(x^2+y^2) we can plug Ex(net) and Ey(net) in there place

Here you have to be careful about Ex(net) and Ey(net) they are the resolved components that is Ey(net)=Ey(from 1st charge)±Ey(from 2nd charge).The sign will depend on directon.If you only have one charge then Ey(net)=Ey(of the charge). In the video Ey=(sin58.1)2.88=2.30N/C but it will get cancelled out by other charge(2 votes)

- Do negative charges always have their vectors pointing towards them, and vice versa?(3 votes)
- Yes, electric field "points" away from a positive charge and toward a negative charge.(3 votes)

- if the negative charge was not -8.0 C. Let's say it is -10.0C

Would the y-components cancel out?(3 votes)- of course not, it would be -2 net downards so then you would have to go through the same process as the E_X but for E_Y which is a substraction since oposite directions, you would then have to use sin to solve for those components rather than cos for the X(2 votes)

- DUDE. This is the first subject that is actually AWESOME. If I could learn this in school, I would wake up EVERY DAY excited to go to school.(2 votes)
- how did david got the horizontal and vertical component at1:19?(1 vote)
- By resolving the two electric field vectors into horizontal and vertical components.

here is a Khan academy article that will you understand how to break a vector into two perpendicular components:

https://tinyurl.com/zo4fgwe

this article uses the example of velocity but the concept is the same. You can also search online for more tutorials on the resolution of vectors into horizontal and vertical components.(3 votes)

- at6:00, to calculate the electric field in the x-direction created by the blue (positive) charge, can we not simply use 3 m as the distance? so: E = k (8 nC / 9 m^2)

finding the magnitude of the net electric field and then using trig to calculate its magnitude in the x-direction seems like going about the problem the long way around.(2 votes) - At3:14, how do the y-components of both the electric fields cancel out?(1 vote)
- Because they have the same magnitude and have opposite directions.(2 votes)

## Video transcript

- [Instructor] Let's try a hard one. 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 the horizontal component
of the net electric field. How do we get the magnitude of
the total net electric field? Well, this is gonna be the same value because since there was
no vertical component of the electric field,
the horizontal component is gonna be equal to the magnitude of the total electric field at that point. If there was a vertical component of the electric field, we'd have to do the Pythagorean theorem
to get the total magnitude of the net electric field,
but since there was only a horizontal component, and these vertical components canceled,
the total electric field's just gonna point to the
right, and it will be equal to two times one of these
horizontal components, which, when you add them up, gives you 3.46 Newtons per Coulomb. That's the magnitude of
the net electric field, and the direction would
be straight to the right. So recapping, when you have
a 2D electric field problem, draw the field created by each charge, break those fields up into
their individual components. If there's any symmetry involved, figure out which component cancels, and then to find the net electric field, use the component that doesn't cancel, and determine the contribution from each charge in that direction. Add or subtract them accordingly, based on whether those components point to the right or to the
left, and that will give you your net electric field at that point, created by both charges.