Hey, guys, let's go ahead and work this one out together. We have these two charges right here, one positive one negative. And we're supposed to find out what the electric field is at this point of interest over here. All right, so let's go ahead and do that. We've got a positive charge over here. A negative charge over here. So at this point of interest, we just need to figure out without any just calculations where the electric fields are going to point. So we know from this, uh, charge right here. We know the electric field is going to point off in this direction because it's going outwards, right? It wants to point away from that charge. Whereas over here, the electric field is going to point in this direction towards the negative charge. And that's going to be e minus.
In order to figure out those electric fields, I need the formula. I've got e=kq/r2. So I need to know the distances between the source charge, or the producing charge, and the point of interest right here. Right. So I've got this little r distance over here and I have a triangle. I've got five centimeters and I've got eight centimeters here for the vertical piece. So I can figure out the hypotenuse of this triangle using the Pythagorean theorem. So you work this out, you're going to get 9.4 centimeters. And if you were to do the same thing over here, you would have to do the same exact thing. You know, five and eight, and you have to figure out the hypotenuse. So this is actually starting to look like we have a lot of symmetry in this problem. So we have five and eight, we basically have this sort of, like isosceles triangle.
So let's go ahead and see if we can use symmetry just like we did in the last video, to sort of reduce the amount of work that we have to do. So we need these same charges. So, in other words, the same magnitude of the charges we got the plus one Coulomb and the minus one Coulomb at the same distances. Which means they're going to produce the same electric field, right? And if these things are symmetrically placed around this, these two charges then that means that when we go ahead and break up these electric fields into their components, so ex and ey, we're going to have the angle θ. That's going to be the angle, θ is gonna help us break up these things into the components. But you're also going to have components in the Y direction and the X direction from the negative Coulomb's, the negative piece of the electric field. And that's going to be at an angle θ as well. So in other words, if these things, these test charges and the point of interest, are symmetrically placed, then we're going to actually have the same exact θ for both of them, and what that allows us to do is to eliminate the use of the vertical components. We don't have to worry about them because they're going to cancel out. And instead, what that tells us is that the net electric field is going to point off in this direction. And it's just going to be two times the electric field of the X components in which the X component of that electric field, or the X component of anything, is the magnitude times the cosine of the angle as long as your angle is relative to the X-axis.
So that means that our Enet is just going to be 2×kq/r2×cos(θ). So I have all of these numbers: k is constant, q is the charge, and r is this distance right here. This 9.4 centimeters, all I have to do is to go ahead and figure out what the cosine of this angle in this triangle is. So I have to go and use the triangle. So instead of using this angle right here, we'll have a whole bunch of stuff written out. We can also say that this angle right here is θ because these lines right here are parallel. So this is the same angle θ in the triangle and in any triangle, the way we figure out cosine of θ is if we have the angle, we can just plug it in. But if not, we can use it by relating the sides of the triangle together so we have cosine is the adjacent over hypotenuse. The adjacent side is five, the hypotenuse is 9.4.
Notice how I don't have to convert it to meters or anything like that; I could just plug it in, as meters or centimeters, because this thing is a ratio anyway. So I've got 0.53 for my cosine. So that means that I can actually just plug in this number inside for the cosine of θ. All right? And so now let's go ahead and just plug in all my numbers. I've got 2×8.99×109,1...r distance right here is going to be 0.094...and now for this cosine of θ right here, all you have to do is just substitute 0.53... If you work this out, the net electric field is going to be 1.8×1012N/C. So that's the final answer, and this net electric field points purely in the X direction.
So this is one way that you guys can use symmetry. Get very familiar with it. Watch this video a couple of times. Just so you understand how I was able to work through this symmetry and I'll see you guys in the next one, let me know if you have any questions.