24. Electric Force & Field; Gauss' Law

Electric Field

# Zero Electric Field due to Two Charges

Patrick Ford

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Hey, guys, lets you got this problem together. So we've got these two, these two charges which lie on the X axis, and we're supposed to figure out what distance or a point on the X axis Is the electric field going to be zero? But I want you to remember that for this problem, we're gonna be using the fact that again outward or positive charges produce outward going electric field lines, they always wanna go away. Whereas negative charges produced in we're going electric field lines. This is gonna be useful because we actually have a positive charge right here and a negative charge right here. So we're gonna need to remember which directions those electric fields go in. And we've done these kinds of problems before when we were looking with columns law. So in order to figure out basically before we start plugging in numbers, we have to figure out which region is this possible for the electric fields to cancel out. So I've got three basic choices I've got on the left I've got in the middle and I've got on the right. Now, if you look the middle, what's gonna happen is at this point of interest. Some of the words this p right here, the electric field from this positive to cool, um, charge all the electric field lines, we're gonna point away. So that means that at this position right here, the electric field from that positive charge is gonna be towards the right. You see, You see, that always wants to go away. Whereas the electric field from the negative charge always wants to go towards that charge. So that means that at this point, it wants to go to the right as well. So, basically, this is the electric field from the negative charge. And there's never gonna be a situation which these things will cancel out. They're always going to add together, no matter where I am along along this line between them. So that means that it can't be in this area because they never cancel out. So let's take a look at the right region now. So we have the inward going electric field lines, so that's gonna be e minus. And from this positive charge, we have a an outward going electric field line. Now, you might be saying you might be tempted to say Hell we actually might be ableto you know, find a place where these things cancel out. What I want you to notice is that the electric fields for any charges is just k times big Q divided by little r squared. So the words it's the magnitude of the charge divided by the distance between them squared. So this three Coolum charge is always gonna be at a smaller distance from this point of interest remembers to some random point here than this entire distance to the weaker charge. Now, what this means is that at this region here, the magnitude of the net of the negative electric field is always going to be bigger than this positive electric field because it's a larger magnitude of a charge, and it's always going to be a smaller distance because this distance right here is always going to be larger, and it's a weaker charge. So, in other words, the magnitude of this is always gonna be bigger than the magnitude of the electric field. So it's never going to cancel, so it can't be this, and instead we're gonna have to use our last scenario, which is on the right. The magnitude of the electric force. It goes in this direction, their direction of the electric field. Wow. And I've got the inward going electric field from the negative three Coolum charge. Now there is possibly a scenario where I could find a distance in which these will cancel out the weaker charges at a smaller distance. The larger charges that a larger distance so there might be a way than for them to balance out the variable I'm looking for is this distance right here, which is X at what point on the X axis is the electric field equal to zero? And what I'm gonna do is I'm just gonna set the origin over here to be zero. So this is just gonna be X equals zero, which means I'm probably gonna get a negative number or I'm just gonna have to specify to the left of the cuticle, Um, charge something like that. And the condition that I'm trying to solve for is that these things should be equal but opposite in direction to cancel out. So in other words, I should have that the magnitude of the e of the electric field from the positive charge should be equal and opposite to the magnitude of the negative electric field. All right, so we know what the magnitude the electric fields are for This guy over here, it's gonna be K. We've got the producing charge, which is the two Coolum charge. And this is gonna be at a distance of X squared now that has to equal the inward going electric field. That's gonna be K times the three. Cool. OEMs. Notice how I didn't put the negative sign there because I'm just working with the magnitudes of them, right? So just set their magnitudes equal to each other. And now this distance right here is actually not going to be X. But it's gonna be X plus the seven centimeters square, so actually could just write that. So I've got X plus 0.7 square. That's seven centimeters. Just written in s I Right. So now what I can do is I can try to solve for this x variable over here. So I got this equation and I could go ahead and cancel some stuff out. I got the K's. And now, in order to get this X over one side, I have to move this thing over and I have to move the two down, so basically, they just have to trade places. And if I do that, I'm gonna just continue writing all this stuff in black because it's gonna be, ah, pain if I try to switch colors. So I've got X plus 0.7 squared, divided by X squared equals And then I've got three halves. Great. Now, rather than starting to foil all of this this whole, this whole mess out and getting a whole bunch of quadratic terms, things like that, what you can do is realize that both of these terms on the top and bottom r squared, which means that we can take the square root of each side. So we're gonna take the square root, take the square root, and then it just becomes X plus 0.7 over X equals the square root of three halves. Okay, so now all we have to dio is to start isolating for X. I'm just gonna move it over to the other side. This becomes X plus 0.7 equals square root of 3/2 times X. And now all I have to dio is I have to subtract this X to the other side. So by up to minus X, and I have this 0.7 equals the square root of three halves X minus X. Now I can do is I can pull this X out as a common factor because it's like, you know, it appears in both of these terms, so basically have 30.7 is equal to x times the square root of three halves minus one. Notice how if you distribute this X back inside of both of these terms, you're just gonna get back to the expression that you had right above right? So it's just it's a good shortcuts is basically just pull out that greatest common factor. Okay, and now what we could do is we could basically just move that over and divide. So we have 2.7 divided by the square root of three halves minus one, and if you just plug that stuff in, you should get the X distance, and you should get 0 31 meters or that's just 31 centimeters. But remember we said that it's going to be to the left of that too cool in charge. So we just have to specify its to the left off the too cool, um, charge. And this is the answer. So if there was, there was, like, an actual coordinate system here, this would be, like minus 31 centimeters anyway, but this is actually perfectly fine. If you just left it like that, let me know if you guys have any questions. And if not, we're gonna keep going with some more examples.

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