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Anderson Video - Electric Field from Two Charges

Professor Anderson
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 Say we have a charge. Positive charge sitting right there, and a positive charge sitting right there. And let's say these have values of capital Q, what is the electric field halfway in between? I'm asking you guys, what do you think? Zero, right? We know that, it's zero. Why is it zero? Because the top charge is giving us an electric field that looks like that, pointing down, the bottom charge gives us an electric field like that, pointing up. If that point is halfway in between the two, they're equal and opposite, it cancels out, everybody's happy. But let's make it a little more complicated now, let's bring out our point to the side. So let's put our point of interest right there and let's say that this distance is l, that distance is l, that distance is l, and let's calculate the electric field right there. How do we do it? Well, we know that we have an electric field from this one, it must be pointing in that direction. Let's call that e1. We know we have an electric field from this one that's pointing in that direction, call that e2. Those are positive charges so e-field lines leave the positive charge. And if we want to calculate the net field at this point, how do we do it? E3 is just going to be the sum of those two but they're vectors, right? We have to remember to put the little arrows on top. And now we have to be a little bit careful about determining what those vector components are. E1 is at some angle, it has some horizontal component and it has a vertical component. E2 is also at an angle, so it has some horizontal component and some vertical component. And if we call this angle theta, which means this is also angle theta, now we can determine what some of those components are. E1 is just going to be some x component, e1 x in the i hat direction. Remember i hat is a unit vector along x, j hat would be a unit vector along y. So we're going to have some x component i hat, there's going to be some y component along j hat. We need to figure out what those things are. Well, this e1 is a triangle. The hypotenuse is e1, the sides are e1x and e1y, and we said that is angle theta. So this just becomes e1 cosine of theta i hat, that's this side of the triangle, minus, since it's pointing down we have to put a minus sign right there, e1 sine theta j hat, and now we can do the same thing for e2. E2 looks like this, it has an x component to it, it has a y component to it, this is angle theta. And so we can write down e2 very simply, it is going to be e2 cosine of theta i hat, it's going to the right, plus e to sine theta j hat, it's going up. Now we have e1 and we have e2 and so we can add them up to get our resultant E. E3 is going to be what? Well let's just look at this, I have e1 cosine theta, e2 cosine theta, but by symmetry those have to have the exact same strength. Right? Same charge, same distance, everything is symmetric, so e1 has to be equal to e2. E1 equals e2, and so we can rewrite this with ones here. And now when I add them up look what happens, this one doubles, this one goes away, and so we get 2 e1 cosine theta i hat plus 0 j hat. It's all in the horizontal direction. And now we're nearly done, the only thing we need is the strength of e1 and that angle theta. So let's put that together now. So let's redraw the picture just for one of them. We've got a q and it was l on the top, l on the side, there's our point, that was our e1. Okay, and we need to calculate the strength of e1. It doesn't really matter if that was e1 or e2, since it's symmetric it's going to have the same strength. What is the strength of e1? Well, we just learned that it's the field of a point charge, kq over r squared. Okay, that's the magnitude of e1. We've already taken into account the directions here with the i hats and the j hats, and so we just have k q over r squared for the magnitude. Do we know r? Well, r is the hypotenuse of that triangle, that's a right angle. If that's a right triangle, then we know that r squared equals l squared plus l squared, which is just 2 l squared, and so I can put that in right here. This is kq over 2 l squared. And now we're almost there, we just need one little piece of information which is theta. Theta we said was this angle but that angle is exactly the same as this angle and since this side is l and that side is l, it's a 1 1 square root of 2 triangle which means theta is 45 degrees, and cosine of 45 degrees is 1 over root 2. And so now we have everything we need. E3, my three got a little funky, e3 is equal to 2 times e1 k q over 2 l squared times the cosine of theta one over root two and we have an i hat. Okay, and you can simplify that because the two goes away and we get k q divided by root 2 l squared, all of that times i hat. That is our resultant field e3 due to those two charges. Questions about that? If you are confused about the unit vector stuff, the i hat the j hat the k hat, click to my youtube channel and there is a whole lecture about unit vectors, okay? And it'll walk you through it. The idea is that i had and j hat have unit magnitude but they have direction associated with the x-axis for I, and the y-axis for j. That's all it means, are you pointing in the x or are you pointing in the y? Good. Everybody okay with that?