[Music] So, I'm going to take you through one of the kind of standard examples of calculating the electric field due to some point charges. And that's uh what's known as a dipole. It's just two charges equal size but opposite magnitude spaced by a distance d. And I'm going to line these things up along the x-axis. And I'm going to ask what the field is if I stand somewhere off to the side midway between them. So I'm going to go right to the midpoint and go out a distance s from there and ask what is the field due to the blue charge first. Then we'll calculate the field due to the red charge. Then we'll put those two together. So that blue charge we just go grab our definition what the electric field should be. Kq R 2 in the R hat direction. And the R is the vector that points from Q to the field point to where I'm standing in that that white circle there. Well, r is found pretty easily from the pagorean theorem. Here I have s^2 + t /2^2 root gives me the length of r. And the direction of r I can say just points that way. And I can identify that better by looking at the angle formed down there. Theta, which I'll be able to use some trig identities to uh to nail down in a second here. Basically, what's going to happen here is the cosine of theta is going to be the xhat component of the the unit vector and the s of theta is going to be the yhat component. So let's go ahead and start plugging some of this stuff in. I have E is equal to KQ R^2. The direction, as I just said, is cossine theta Xhat and sin theta Yhat. And I can clear this up a bit by noting that the cosine and s of theta I can find by looking at that triangle I got there. The cosine is going to be d / 2 / r and the s is going to be s / r. And I'm just going to hold this in this format for a second because I want to flip around and look at the red guy first and then we'll put the two together and and the algebra will clean up nicely. So if I flip around and look at the red guy, the main thing to notice is my r is the same length but points in a different direction. Now it's pointing from the red one to the white spot. Uh so my unit vector now is negative cosine theta xhat plus sin theta yhat. But the negative sign on that q means this thing's really going to flip around. So the field is going to point back toward the negative charge. And when I plug in the values for cosine theta and sin theta, now what I get is that E is equal to K q R 2 D /2 RXhat minus S / R Yhat. When I put these guys together, I'm going to add that blue unit vector and the red unit vector. The the fields due to both those two charges I just calculated. And just doing a vector addition, I can see what I'm going to get here. Something that points straight out in the X direction. And you can actually look up top and see that happening in the algebra. The y components are equal and opposite to each other. the x components add together. So the d over2 just becomes d and I get k qd / r cubed pointing in the xhat direction. Now this is a a sort of fundamental charge distribution that gets used a lot in electricity magnetism. It's known as a dipole. And since the geometry of it is just the charges and that distance d, we go ahead and define something known as a dipole moment defined with the letter p. and it's equal to Q * D where Q is the charge that's separated and D is the distance that it is separated. Um the dipole moment literally is a vector quantity but we aren't going to worry about that for now. We're just going to use this P to plug back into our field equation and show that it is KP over R cubed. Now I'm going to move off to the side here though and look at it from a different point of view. I'm going to move a distance X away from the center line where I'm lined up with the dipole. I'm looking end on if you will. And again, calculate the field due to the blue charge and the field due to the red charge. Things are a little simpler here though because all I have is X component to worry about. And the distances are just X + D /2 for the red charge and X minus D /2 for the red charge. I'm sorry, I said red a second ago. I meant blue. Uh, and I can add these guys together so that what I get is KQX + D2 D over2 minus KQ because it's a negative charge over X - D over2. Uh, now I'm going to do a little algebra here just to clean this up. And there's a couple ways I could do this. I'm going to take you through kind of the the brute force method, which is to cross multiply those x + d over2 and x - d over2. So bring each one up to the opposite term and what I get is x - d /2 ^2 minus x + d over2 in the numerator and then the multiple of the two the common denominator. I've gone ahead and pulled the kq out front because that's just going to be riding along the whole way that we do this. And now I'm going to go ahead and complete the squares up top. So I have x^2 - xd + d over 2 over 4 and then x^2 + xd + d^2 over 4. And since I'm subtracting those, the first thing I'm going to notice is that the x^2 and the d^2 over 4 terms are just going to cancel each other out. Down in the bottom, what I've done is I've pulled the square out and considered it x + d /2 * x -2. That whole quantity squared because when I do that, that's a kind of common algebra identity written down there that I have x^2 minus d ^2 over 4 all squared. And up top, all I have left over is the xd term. So I have -2 * xd. Now the next step of this is to to look at the situation. What happens if I make x pretty large? If I if I get a far way away from it so that x is big compared to d then what happens is the d over two d over2 well the d^2 over 4 term down there gets really small in comparison to the x^2. So I can just kind of ignore it as time goes by if I get far enough away. And the denominator just becomes x to the 4th. If it's x to the 4th, then the x up top will cancel one of those. And what I get is an x cubed in the bottom. And I'll notice that the qd up there is just my dipole moment again. So I have min -2 kp overx cubed in the xhat direction. So once again, I have something that depends on the distance cubed, which is common for a dipole field. Now, what we've done so far, we found that the field off to the side points horizontally. The field down there at the end points in toward the negative charge. If I wrapped around to the other side, I could prove to you that the field points out from the positive charge the other way. And if I sort of map all these together with a whole bunch of others, I can get this classic picture of what a dipole field looks like. The field lines coming out of the positive charge and wrapping back around into the negative charge. And this is a just a a common picture of how to to represent the the dipole field. And it just pulls together all those things we just got done calculating plus any other place I might go in space which for for some place off to the side and and up that the math would be a little bit more confusing. But the there's more to this and it's it's one of the most fundamental problems in electricity and magnetism and it's sort of the leaping off point in physics 350. But for now let's just use this as sort of an example on how to go about solving a problem involving two-point charges.