 Okay, so chapter one's really about the electric field. So if I have a point charge like this, the electric field due to that goes outward in all directions, and if it's negative it's going in, and I can describe the electric field at any location a vector r away as e equals 1 over 4 pi epsilon not q over r squared r hat. Okay, so 1 over 4 pi epsilon not just a constant, q is the value of the charge, r is the magnitude of that distance, and r hats that unit vector outwards. Okay, so that's that's that. Now what's really important is you really just have one charge if multiple charges. So the electric field obeys the superposition principle. So if I have two charges, say there's a negative charge right there, then the electric field at that location is the electric field due to this charge plus a charge. So you can add them together. I need to find a better marker. Let's see if this works. Okay, that one's better. Okay, so let me go ahead and just go through a derivation that's in the book, but I think it'll be useful anyway. This is the suppose that we have a positive charge and a negative charge separated by distance s. This is called a die, and they have the same charge q and plus and minus q. This is called a dipole. What if I want to find the electric field anywhere? Well I would just add up the electric fields due to the two things, but let's say I want to find it as it moves up and along this y-axis, so a distance y away from the center of that dipole, I want to find the electric field. Okay, well that's the case. I have two electric fields that I need to add together. Let me put this y over here. So I have the electric field due to the positive charge is going to be, let me put the point down a little bit lower just so it looks, it's easier to see. Let's find the electric field right there, just for simplicity. So I have this distance away and that electric field I'll call E plus, and then the negative charge has this distance and it has electric field going towards it E minus. So now I just need to find the values of the E plus and E minus vectors and add them together. Okay, so let's do E plus first. So let's first find, let's say right here, E plus 1 over 4 pi epsilon knot q over r squared r hat, r hat plus, and I just left r squared since this r is the same as that r, it doesn't matter which one it is, they have the same value. Let's find out what r is. If I look at this triangle right here, this is s over 2, this is y, and this is r. So that's a right triangle. So r squared is going to be y squared plus s squared over 4, because it's a Pythagorean theorem. This squared is that square plus that squared. What about r hat? r hat is going to be, well I could write r hat as r hat is the vector r divided by the magnitude of the vector r, but maybe it might be better just to write this as 1 over 4 pi epsilon knot q the vector r over r cubed. I can do that, right, because if r hat is r over the magnitude of r, then I can turn in r hat and put another r down the bottom. So in that case I should write r cubed. So if this is r, r cubed, r is going to be the square root of that, and r cubed would be to the 3 half power. And then what about r? Now that's going to be r plus. So now I can just write the vector going from here to there. It's going to go over s over 2 in the x direction and up y in the y direction. So r plus is going to be s over 2 y zero. Okay, now what's next? I could put that in there for r, and I will just for completeness. E plus 1 over 4 pi epsilon knot q. Now let me put in for r cubed this. It's going to be y squared plus s squared over 4 to the 3 halfs. Then I have s over 2 y zero. Okay, now I need to do the same thing for e minus. But this is easy because it's the same. The r is the same, right? It has the same positive value, and r cubed is still even the same because it's the magnitude. The only thing that's different is r minus. r minus is going to be the vector from, it's still from here to there, right? So it's going to be negative s over 2, negative s over 2, y, zero. And the other thing that's going to be different is q. Now I have a negative charge, so q is going to be negative. I'm going to put a negative sign in for that. 1 over 4 pi epsilon knot, negative q, that's a g. Why would I draw a g? That's just dumb. It's a q. It's a q. Okay, y squared plus s squared over 4 to the 3 halfs, and then I have negative s over 2, y, zero. Now just for simplicity, to add these things, let me get a common denominator by factoring out a negative sign out of here, which will cancel that, and this will become positive and that will become negative. So now I see these are the same exact terms in front of vectors, and I can add the vectors together. So I get s over 2, I get two of those, but the y's cancel. So then my total electric field is going to be 1 over 4 pi epsilon knot. Now I have, I'll just write it as, I'll write it as 2. Well, yeah, I'll try to write that as s, okay. So q, y squared plus s squared over 4 to the 3 halves, and then I'm going to have s over 2 plus s over 2 is s, zero, zero. And then let me factor out the s and make that a unit vector. So now I know the unit vector and I can get the magnitude. And that is the answer, okay. But a lot of times this dipole, the separation is very small compared to the distance away, so y is much greater than s. And if y is much greater than s, then y squared is going to be way bigger than s squared over 4. So it's not a terrible approximation to say that this is just going to be, and I'll just write the magnitude, 1 over 4 pi epsilon knot qs over y squared to the 3 halves, okay. But y squared to the 3 halves is going to be y cubed. And then I have the unit vector, but I didn't include that in there. Okay, so that's the answer. But we need to check two things that we can always check when we have an answer like this. First, the units. Does this have the same units as the electric field due to a point charge which I erased? Well, it's got the 1 over 4 pi epsilon knot and the q just like the point charge. And the point charge had 1 over distance squared. This has a distance divided by distance cubed, which is 1 over distance squared. The other thing to check is what happens as I get really, really far away. If I get really far away from this, the electric field, the magnitude, should go to zero. So as y goes to zero, I mean, as y goes to infinity, e does go to zero. So I mean, I admit that this, doing something like this may look just totally crazy, but you know, this is something that you definitely need to know how to do. If this is something that's causing you problems with the vectors, then that's what you need to work on.