Anderson Video - Electric Field Example with Four Charges

Professor Anderson
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And let's do an electric field example, and let's do the following. We're going to take four charges, q1 q2 q3 and q4, and let's calculate the force right at the centre of this square. So this side is L, this side is L, and we're going to figure out--not the force I'm sorry, the electric field right at the center of the square. And let's give you some values here for the Q's. Q1 is going to be the same as q2 and q3, which is two point four times ten to the minus twelve coulombs. Q4 is going to be negative two point four times ten to the minus twelve coulombs. Okay, and let's see if we can calculate e right at the centre of this configuration. And we'll say that the length of each side is four centimeters. All right. So, how can we do this? Well, electric fields are vectors. Vectors obey the principle of superposition which means we can add them up according to the vector rules. So, if we want to calculate e at the center, we could calculate e due to number one, e due to number two, e due to number three, and e due to number four and just add them up. Okay, so let's draw it again right here. Here's our square, here we are at the center. This was q1, this was q2 q3 and q4. Let's think about q1 for a second. At this position right here, what direction is the electric field from q1? It has to be along the line between the two and q1 was a positive charge, right? We said that was positive. Positive means the electric field is going to point away from it, so that is e1, that's the electric field from q1. Q2 is also positive and the electric field from q2 is going to point that way, e2. Q3 is also positive and so the electric field from it is going to point radially away from it, so it's gonna be right there. And then finally q4 is a negative, and so the electric field from q4 is going to point towards it in the same direction as e1. So look what happens, e1 and e3-- I'm sorry not e1 and e3, e2 and e3 are equal and opposite, they cancel. Right? E2 is pointing up to the right, e3 is down to the left and so those two exactly cancel out. E1 and e4 are in the same direction with the exact same strength in fact, and so they add up. Okay, and so now the problem is really simple. All we have to do is calculate the electric field due to one of them and double it. All right, let's do it. What is e total? It's just twice e1 and we know what e1 is. It's K q1 divided by this distance R squared. Okay, and we know K. K is 9 times 10 to the 9. We know q1, it's two point four times ten to the minus 12. That's saying on the board? Yeah, it's alright. And then we got to divide by R squared. Now we have to be a little bit careful about R here, because this side of the triangle is in fact L over 2, that side of the triangle is also L over 2. So R squared is in fact L over 2 squared plus L over 2 squared, and so we get 2 times L over 2 squared. All right. L over 2 squared plus L over 2 squared equals R squared, and we know that number now. So now we can plug all this in to your calculator, once you try it, tell me what you get. Okay we've got L we said was four centimeters, so four centimeters there is 0.04 but we're going to divide it by two so that's 0.02. We're gonna square that and then we have a two out in front of it. That's convenient because that 2 is gonna cancel with that 2 and so I will approximate it here and you guys can tell me what you get. Nine times two point four is 18-23 ish. 23 times ten to the minus three and then we've got it two times ten to the minus two squared, which is four times ten to the minus four, and that I'm gonna say is 23 over four is pretty close to six times ten to the one. So we're gonna say it is 60 newtons. Is that close to what people got, or not even close? >> [inaudible] You got 54? All right. So we're pretty close. 54 Newtons--I'm sorry, it's not Newtons right we're calculating electric field. Electric field is volts per meter, that's the SI unit, volts per meter. Okay, that's the value of the electric field right in the center. Okay, and we took advantage of some real important things here, we took advantage of this notion of superposition, adding up electric fields according to the vector rules and that greatly simplified the problem because two of them cancelled and the other two doubled. Okay, and then the rest was just plugging in some of the numbers. All right. Any other questions about that one? You guys feeling okay? All right, hopefully that's clear if not definitely come see me in officers. Cheers!
And let's do an electric field example, and let's do the following. We're going to take four charges, q1 q2 q3 and q4, and let's calculate the force right at the centre of this square. So this side is L, this side is L, and we're going to figure out--not the force I'm sorry, the electric field right at the center of the square. And let's give you some values here for the Q's. Q1 is going to be the same as q2 and q3, which is two point four times ten to the minus twelve coulombs. Q4 is going to be negative two point four times ten to the minus twelve coulombs. Okay, and let's see if we can calculate e right at the centre of this configuration. And we'll say that the length of each side is four centimeters. All right. So, how can we do this? Well, electric fields are vectors. Vectors obey the principle of superposition which means we can add them up according to the vector rules. So, if we want to calculate e at the center, we could calculate e due to number one, e due to number two, e due to number three, and e due to number four and just add them up. Okay, so let's draw it again right here. Here's our square, here we are at the center. This was q1, this was q2 q3 and q4. Let's think about q1 for a second. At this position right here, what direction is the electric field from q1? It has to be along the line between the two and q1 was a positive charge, right? We said that was positive. Positive means the electric field is going to point away from it, so that is e1, that's the electric field from q1. Q2 is also positive and the electric field from q2 is going to point that way, e2. Q3 is also positive and so the electric field from it is going to point radially away from it, so it's gonna be right there. And then finally q4 is a negative, and so the electric field from q4 is going to point towards it in the same direction as e1. So look what happens, e1 and e3-- I'm sorry not e1 and e3, e2 and e3 are equal and opposite, they cancel. Right? E2 is pointing up to the right, e3 is down to the left and so those two exactly cancel out. E1 and e4 are in the same direction with the exact same strength in fact, and so they add up. Okay, and so now the problem is really simple. All we have to do is calculate the electric field due to one of them and double it. All right, let's do it. What is e total? It's just twice e1 and we know what e1 is. It's K q1 divided by this distance R squared. Okay, and we know K. K is 9 times 10 to the 9. We know q1, it's two point four times ten to the minus 12. That's saying on the board? Yeah, it's alright. And then we got to divide by R squared. Now we have to be a little bit careful about R here, because this side of the triangle is in fact L over 2, that side of the triangle is also L over 2. So R squared is in fact L over 2 squared plus L over 2 squared, and so we get 2 times L over 2 squared. All right. L over 2 squared plus L over 2 squared equals R squared, and we know that number now. So now we can plug all this in to your calculator, once you try it, tell me what you get. Okay we've got L we said was four centimeters, so four centimeters there is 0.04 but we're going to divide it by two so that's 0.02. We're gonna square that and then we have a two out in front of it. That's convenient because that 2 is gonna cancel with that 2 and so I will approximate it here and you guys can tell me what you get. Nine times two point four is 18-23 ish. 23 times ten to the minus three and then we've got it two times ten to the minus two squared, which is four times ten to the minus four, and that I'm gonna say is 23 over four is pretty close to six times ten to the one. So we're gonna say it is 60 newtons. Is that close to what people got, or not even close? >> [inaudible] You got 54? All right. So we're pretty close. 54 Newtons--I'm sorry, it's not Newtons right we're calculating electric field. Electric field is volts per meter, that's the SI unit, volts per meter. Okay, that's the value of the electric field right in the center. Okay, and we took advantage of some real important things here, we took advantage of this notion of superposition, adding up electric fields according to the vector rules and that greatly simplified the problem because two of them cancelled and the other two doubled. Okay, and then the rest was just plugging in some of the numbers. All right. Any other questions about that one? You guys feeling okay? All right, hopefully that's clear if not definitely come see me in officers. Cheers!