 Ok, so I said I would work out this problem for you, hopefully I'll do it for the battery and the camera dies, but it might. But you'll never know because then I won't finish, so why am I wasting my time telling you this? I don't know. Ok, so here's the situation, here's the X and Y axis and here's a ring, a C is for cookie shaped rod of negative charge Q spread over that and this has a radius of R and we want to find the electric field at that point right there. So the way we do this is we always say, ok, I'm going to break this into a little piece, I'll call that DQ, I'm going to find the electric field due to that piece and then I'm going to add up all the electric fields due to all the pieces. So let's locate this piece by the angle theta right there, which isn't the normal use of theta but it doesn't matter, you can do whatever you want. That's going to make an electric field going this way, DE. And if I wanted to, I could write DE equals 1 over 4 pi epsilon not DQ over R squared because this is a distance R away right, so that's the magnitude of that and then I can write that as a vector, I can write that as negative, I want to be in that direction, negative sine theta, cosine theta, zero. That gives me the unit vector in that direction if I have that angle. So that's my piece, I mean really if I wanted to I could say add up all the pieces set up in the integral going around the whole thing but I want to be a little bit smarter than that. I want to look at another piece straight down here that's symmetrical to that piece, it's going to make an electric field going this way of the same magnitude because of the same distance and it's going to have the same, it'll have, this will be, this calls 1, DE2 is 1 over 4 pi epsilon not DQ over R squared and this one's going to have negative sine theta, negative cosine theta, zero because it's going down. So if I add those together the Y components cancel and I get two of the X components. That's if I treat these in pairs, that's what I'm going to do. So I can do that and I can find out this whole thing but the problem is, is anything changing? Yes, something's actually changing right? As I move over here theta's going to change so I need to integrate over theta and find out what happens. In fact I'm going to, since I'm including this piece down here I'm going to integrate from zero to pi over two. But my problem is that I have DQ and I don't have D theta in the thing. So let's just call this angle in there D theta, right? It's theta to D theta tells me how big of a p-set is. And then I need to get a relationship between DQ and D theta so I can say Q over pi equals DQ over D theta. If the charge evenly distributed over this angle then I can say the total charge over the total angle, which is pi, would be that little piece divided by that little angle. So that means DQ is going to be D theta Q over pi. So now I have an integral here. Now I can say E1x, I only have the x component that survives, 1 over 4 pi, epsilon knot, the Q comes out, Q, the R squared comes out, pi, R squared. All I have left is, and there's a two because I have two of those, all I have left is the integral from zero to pi over two. I have D theta, and then I have negative sine theta, let me move the negative over here. So why, wait, negative sine theta, that's right, if I'm getting another negative, because when I integrate that I get, where does that negative come from? That's right there, sine, okay, let's just proceed. Negative two over 4 pi, epsilon knot, Q over pi, R squared, and it's going to be negative cosine theta from zero to pi over two. Okay, it's going to work. Okay so this is going to be cosine of pi over two is zero minus cosine of zero, which is one. So that's going to be this whole thing is one. So the answer is equal to E, I can write it as a vector even, negative two over 4 pi, epsilon knot, Q over pi, R squared, and then I have negative one, zero, one, zero, Newton's per coulomb or volt per meter. And then we can check, we can check some things, does it have the right units, does it have units of one over four pi, epsilon knot, Q over R squared? This pi doesn't have any units, so it does, okay. I have the right direction, what about as, what about, well we can't do as you get further way because we did it at that point. So there's nothing else we can really, we can really check. I mean you could do this, you could say what if I put in the other half, then I would get the exact same answer but the electric field would be going that way and then it would cancel that up to zero which agrees with the value for the ring. So, okay, that's it, I think I made it through the battery, let's see, barely, barely.