Is this idea that force is a vector. Force is a vector and therefore Coulomb's law we have to worry about adding those forces as vectors. F equals k q-1 q-2 all over R squared and then there is this direction associated with it. So let's try an example and let's see if we can make maybe a simpler example than we had talked about earlier. Let's do the following. Let's say we draw a coordinate system here x and y and let's do this let's put a charge Q right there. We'll put another charge Q right there and then we'll put a third charge out here equal distant away and let's see if we can find the force on this charge due to the other two, okay? Very straightforward vector problem. Now let's give some dimensions here we'll say that this distance here is L and we will call that distance L and that distance L, so everything is L away from the origin and now let's figure out what the force is on this little guy out here. Okay, we wrote down the force law but what we had in there was an n hat and the N hat says the direction is along the line between the two charges. So if this is the line then the force on this guy has to be in that direction. We'll call that f1. The other guy is coming up like that and so the force from it has to be along that line f2. Let's just pretend they're all positive charges okay? Three Q's, all of them greater than zero. All right. So now we just have to add up f1 and f2 and you know what's going to happen they're going to add up to some f3 that is off in that direction, okay. Let's see if we can calculate what f3 is. All right, first off let's take a look at how about f2? F2 is the following, it is K Q Q over R squared and then there's going to be some direction associated with it. Okay. And we're not exactly sure how to write the direction yet so let's make this the magnitude of f2 okay? That's what the magnitude of f2 is it's k QQ over R squared. F3 is of course going to have the exact same magnitude and that R squared is what? It's this, that's our R, but this is a right triangle so R squared is in fact l squared plus l squared which is 2 l squared. And the same on f3, k QQ over to l squared. Alright so those are the strengths of the two forces due to those guys okay? We're looking at the force on this one right here but now we have to worry about the direction. Okay? Let's think about the direction here, f2 is going up like that and f2 therefore has some component in that direction and some component in that direction and the components are the following. The horizontal is f2 cosine theta the vertical is f2 sine theta and likewise for f3 we're going to get something similar, f3 is just--alright sorry, f1 is just going down and so for the two sides of that triangle we would get f1 cosine theta and f1 sine theta. All right, and we know what theta is right if this is L and that's L then we know that that's 45 degrees right there, that has to also be 45 degrees and so now we're almost there because we have the magnitude we're almost done with the direction and we can figure out how to add this up. Okay, f2 as a vector is what? It is the magnitude of f2 cosine theta in the X direction which we call i-hat, plus f2 sine theta in the Y direction which we call j-hat. And we know exactly what f2 is, it's kqq over 2 l squared and then we have cosine of theta i-hat plus KQQ over 2 l squared times sine theta j-hat. F3--oh I'm sorry f1. F1 which we said is pointing down at a angle that's going to be KQQ over 2 l squared cosine theta i-hat. And then we have a sign on the other one but it's pointing down and so we need a minus sign. Minus KQQ over 2 l squared sine theta j-hat. And now what we're looking for is f3. F3 is equal to f1 plus f2. Well here's f2, here's f1, so f3 is f2 plus f1 and look what happens if I add these up. These two are the same, these two are equal and opposite. And so the sine components in fact drop out entirely and we just end up with this and since we have two of them we can double it. KQQ over l squared, the 2 in the bottom goes away, cosine theta and the direction here is i-hat. Okay? And if you want the magnitude of f3 then it's just the stuff that's out in front. Now in our example we picked L and L and so we also know that theta is going to be 45 degrees here, and 45 degrees, cosine of 45 degrees gets us a root 2 in the denominator. Okay. F3, that's your remaining force, it's this, kqq over root 2 l squared and it's in the x-direction i-hat. If you want to calculate the magnitude of it, it's just that stuff out in front, okay? So you kind of knew the answer right? You knew that f3 had to be pointing to the right because those other two factors were going to cancel in the vertical direction and they would contribute in the x direction, in the horizontal direction. Alright, questions about this stuff? Feeling okay with it? Alright.