Hey, guys, let's tackle this problem together. So we've got these two charges which lie on the X-axis, and we're supposed to figure out at what distance or at what 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 going to be using the fact that outward or positive charges produce outward-going electric field lines, they always want to go away. Whereas, negative charges produce inward-going electric field lines.
This is going to be useful because we actually have a positive charge right here and a negative charge right here. So we're going to need to remember which directions those electric fields go in. We've done these kinds of problems before when we were looking at Coulomb's law. So, in order to figure out, basically before we start plugging in numbers, we have to figure out in which region it is 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 at the middle, what's going to happen at this point of interest, this point right here, the electric field from this positive Coulomb charge and the electric field lines are going to point away. So, that means that at this position right here, the electric field from that positive charge is going to be towards the right. You see, it 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 going to be a situation in which these things will cancel out. They're always going to add together, no matter where I am 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. We have the inward-going electric field lines, so that's going to be E-. And from this positive charge, we have an outward-going electric field line. Now, you might be tempted to say we might be able to find a place where these things cancel out. But I want you to notice that the electric fields for any charges are just k times big Q divided by little r2. So, that means it's the magnitude of the charge divided by the distance between them squared. So this three Coulomb charge is always going to be at a smaller distance from this point of interest relative to this entire distance to the weaker charge. What this means is that at this region here, the magnitude 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, the magnitude of this electric field is always going to be bigger than the magnitude of that electric field. So it's never going to cancel, so it can't be this, and instead, we're going to have to use our last scenario, which is on the left. The magnitude of the electric force goes in this direction, the direction of the electric field.
Wow, and we've got the inward-going electric field from the negative three Coulomb 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 at a larger distance so there might be a way 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 going to do is I'm just going to set the origin over here to be zero. So this is just going to be X equals zero, which means I'm probably going to get a negative number, or I'm just going to have to specify 'to the left of the Coulomb charge' or something like that. 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, the magnitude 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 magnitudes of the electric fields are for This guy over here, it's going to be K. We've got the producing charge, which is the two Coulomb charge. And this is going to be at a distance of X squared. Now, that has to equal the inward-going electric field. That's going to be K times the three Coulombs. Notice how I didn't put the negative sign there because I'm just working with the magnitudes of them. 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 going to be X plus 7 centimeters squared, so I could just write that. So, I've got X plus 0.07 squared. That's 7 centimeters. Just written in SI.
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 on 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 going to just continue writing all this stuff in black because it's going to be a pain if I try to switch colors. So, I've got X plus 0.07 squared, divided by X squared equals 3/2.
Great, now, rather than starting to foil all of this 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 are squared, which means that we can take the square root of each side. So, we're going to take the square root, take the square root, and then it just becomes X plus 0.07 over X equals the square root of 3/2. Okay, so now all we have to do is start isolating X. I'm just going to move it over to the other side. This becomes X plus 0.07 equals square root of 3/2 times X. And now all I have to do is subtract this X to the other side. So, by subtracting X, I have 0.07 equals square root of 3/2 X minus X. Now what I can do is pull this X out as a common factor because it appears in both of these terms, so basically, I have 0.07 equaling X times (square root of 3/2 minus 1). Notice how if you distribute this X back inside of both of these terms, you're just going to get back to the expression that you had right above, so it's a good shortcut to 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 0.07 divided by the square root of 3/2 minus one, and if you just plug that stuff in, you should get the X distance, and you should get approximately 0.31 meters or that's just 31 centimeters. But remember we said that it's going to be to the left of that two Coulomb charge. So we just have to specify it's to the left of the two Coulomb charge. And this is the answer. So if there were 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 going to keep going with some more examples.