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All right, why don't we try another charge example but let's just do one dimension. Okay, let's say this is our xy coordinate system, let's put one charge right here, q1 right at the origin, we'll put q3 right there, and let's put q2 halfway in between. Okay, but let's give you some values. Let's say that q1 is positive 5 times 10 to the minus 19 coulombs, and we'll say that q3 is equal to negative 5 times 10 to the minus 19 coulombs, and let's ask the question, what is the force on number two? Okay, what is the force on two and let's say that these distances are equal, and we will let l equal 10 microns, which is 10 to the minus 10, 10 times 10 to the minus 6 meters. Okay, that's 10 microns. All right, how do we do this? Well we write down our force equation. Force equation is f equals k q1 q2 over r squared and we remember that it's a vector and so there is a direction associated with it, and now let's put our three charges right here. We've got q1, we've got q2, and we've got q3. So there is a force on two, due to one and we need one more piece of information here, right? We need to know what the charge is at q2. So let's say that q2 is equal to also 5 times 10 to the minus 19 coulombs. If I think about these charges now, this one is positive, q1 we said was positive, q3 was negative. And we want the force on q2 due to the other two. So the force on q2 due to q1, is it attractive or repulsive? Repulsive because they are like charges. So q1 is trying to push away q2. But between q2 and q3, that's attractive because q3 we said was negative and so there's another force in that direction as well. So there's two forces acting on q2, both of them are to the right. And so what is the net force on f2? It's going to be f on 2 due to 1 plus f on 2 due to 3. And now we know exactly what those are. It's k q2 q1 divided by l squared, right? We said the distance we called l. And then the other one is going to be minus k q2 q3 divided by l squared and the direction on both of those is to the right which we're calling i-hat. Now why did i put a minus sign here, right? We just said that both those forces are to the right. I put a minus sign there because q3 is negative. So when I plug in a negative number here, it's going to cancel with that negative and give me a positive. All right, so let's try it with some of these numbers and see what we get. Okay, so let's calculate f2. We said that it is--well looks like we have some common factors here, right? We have a k q2 over l squared and then the first term is q1 and the second term is negative q3, all in the i-hat direction. And now we can put in all those numbers. K we said was 9 times 10 to the 9. Q2 we said was 5 times 10 to the minus 19 coulombs. L we said was 10 microns, 10 times 10 to the minus 6 meters and we got to square that, and then we have this quantity which is q1, 5 times 10 to the minus 19 minus q3 which is negative 5 times 10 to the minus 19. Okay, so we have 5 minus a negative 5 and so that thing just doubles and so what do we get? We get 9 times 10 to the 9. 5 times 10 to the minus 19. We have a 10 times 10 to the minus 6 which is a 10 to the minus 5. If I square that I get a 10 to the minus 10 down there, and over here I have a 5 plus a 5, so I have 10 times 10 to the minus 19. And now we can do this by hand but you can double check on your calculator if you want. We've got 45 times 10 to the minus 10. That's a 10. And then we've got 10 to the minus 18 and we're going to divide by 10 to the minus 10. So the 10 to minus 10s drop out and we get 4.5 times 10 to the minus 17. Yeah? >> [inaudible] >> So the question was about the 10 to the minus 19 here, what happened to that, and it became, let's see, the 9 times 5 got us a 45 and then we had a 10 to the 9 and a 10 to the minus 19 so that became a 10 to the minus 10. And then this one right here, 10 times 10 to the minus 19, the 10 means I have to add 1 to the exponent, and so adding 1 to minus 19 becomes minus 18. Right? Are we okay? All right good, I think that's right. Good so we've got 4.5 times 10 to the minus 17 newtons. Okay, that would be the force on q2 due to those other two charges. All right, which sounds like a small number but it's because we'd use charges that were sort of five electrons or five protons big and yet the distances were large compared to atomic distances, right? 10 microns is much much bigger than 10 angstroms. Okay any questions about that one?

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