A −1μC and a 5μC charge lie on a line, separated by 5cm. What is the electric potential halfway between the two charges?

Alright, guys. Hopefully, you were to figure this one out on your own. If you didn't, if you got stuck, maybe I can help out a little bit. We've got these two charges and they're separated on a line by 5 centimeters. So let me go ahead and just draw a quick little sketch. I've got this negative one microcoulomb charge. Now, I've got a positive 5 microcoulomb charge. And if these things are separated by some distance right here, which is 5 centimeters, then I'm supposed to figure out what the potential is halfway between them. What that means is that these two distances are 2 and a half centimeters respectively, and I just have to be careful about when I, basically convert those into SI units. We know that the potential due to any point charge at any distance, this v here is just equal to k times the producing charge divided by the distance between them. But I want you guys to remember for this video, it's very important is that potentials are scalars, which means we can just add them up assuming that we know all the variables. We don't have to do any composition, like like sines and cosines, things like that we had to do with vectors. These are basically just numbers. We can just go ahead and add them together. So what that means is that the potential, the net potential, if you will, is basically going to be the potential from the negative charge plus the potential from the positive charge. Now, I just have to go ahead and solve those basically, independently. I know that the potential from this negative charge is gonna be 8.99 times 10 to the 9th, then that's gonna be the negative one microcoulomb charge. So negative one times 10 to the minus 6. By the way, that negative sign is actually important there, so don't don't forget to to actually write it. And then this is divided by 0.025 in meters. You don't have to plug that in yet because we're gonna see how this works out. So you also have the potential from the positive point charge. It's gonna be 8.99 times 10 to the 9th. Now we've got the positive 5 times 10 to the minus 6, now divided by that distance, which is 0.025. We know that these basically are gonna add up together. So one of the things that we can do is we sort of see that this 8.99 and the 0.025 appear on both of the equations. So one of the things that you can do sort of as a shortcut, it's almost like using symmetry, is basically, like reduce the amount of calculations that you have to do. It's basically, 8.91 or sorry. 8.99 times 10 to the 9th divided by is 0.025, and and then you can sort of, like, pull this out as a greatest common factor. You just have negative one times 10 to the minus 6 plus 5 times 10 to the minus 6. So you basically can, like, extract this k over r as, like, a greatest common factor, and then just go ahead and add the charges. Now I just wanna point out that this won't always work because it only works because we have the same exact distances involved, but this is a shortcut that you can use. So if you go ahead and plug this in, the potential at this location is gonna be 1.44 times 10 to the 6th, and that's gonna be in volts. So again, a shortcut, but be careful how you use it. It only works out if the k's I mean, the k's are always gonna be the same. It only works if the r's are the same for both of them. Okay? Let me know if you guys have any questions with this.

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A −1μC and a 5μC charge lie on a line, separated by 5cm. What is the electric potential halfway between the two charges?

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