Alright, guys, in this example, we're upping the ante. Now we're gonna figure out what the potential difference is not just a one charge, but due to two charges. So we've got this five Nano Coolum charge and a negative Nano three Coolum charge on the same line. So I'm just gonna draw a quick little sketch right here. I've got this five Nano Coolum charge Negative three Nano Coolum charge. And I know that they're on the same line and that distance is six millimeters now, in part A. What I'm asked for is a mask for the potential directly between them. So in other words, this is gonna be the potential at V A. So let me do a better job. It's sort of highlighting that. So that is gonna be the potential at V A. All right, So let's go ahead and do work that out. We've got the potential at V. A. Is basically going to be the potential due to both of these charges at this location. But one of the things we know is that if this is half of the distance between these two charges, then that means that this distance right here this three millimeters is actually going to be the same for both of them. So in other words, we've got the potential at point A is gonna be the potential due to the five Nano Coolum charge, plus the potential due to the negative three Nano column charge. We're just gonna add those things up together because their scale er's so we we have that the potential difference over here is just gonna be K Times Q one over R A plus K Times Q two over are a but one of things we notice is that we have the same exact case and raise for both of these things. So, in other words, since both of these are the same, we can actually use a little shortcut that we've used before in reducing these potentials down to, like even simpler forms we have. The potential is basically just equal to K. Over are a times, and then we're just gonna add the charges up together, and it's only because we have symmetry. It's only because thes things are the same distances that we can actually do. You use this rule so, in other words, that the potential appoint a is going to be 8.99 times 10 to the ninth, and then this is gonna be divided by the distance. We have to be careful because this is three millimeters, which means it's 0.3 and then we're just gonna add up the charges. We have five Nano columns minus the three Nano columns. So that means that the results here so both of these things together are just gonna add up to two Nano Coolum charges, right? So you're just gonna add in both of these charges and you get to nano columns, which, by the way, can be represented by two times 10 to the negative nine. That's gonna be in columns, right? So that means the potential at point A is just going to be, Let's see, I got six times 10 to the third. That's in volts. Alright. So we can use that shortcut for this case right here. So in part B, now in part B, we're supposed to figure out what is the potential at this point B, which is now halfway between the charges. But now it's, um, extra distance above that line. So, in other words, point B is somewhere over here we're supposed to be figuring What is the potential at this point? So basically we need to Dio is now that we have a different point, we actually have to calculate what the distance is to both of these charges. Now, fortunately, we've got here, we've got a three and we know that this line right here is four millimeters above so we can recognize this is a 345 triangle. If you wanted to figure out this our distance, if you didn't know it was 345 you could always just use the Pythagorean theorem. But in any case, we've got five millimeters right here for this distance. And this R B is actually gonna be the same for both of these things, right? Because it's symmetrically placed around around this, uh, this access so you can kind of use the same shortcut that we did here, but now we're just gonna sort of like, do it a little bit quicker. So this VB is just gonna be k over RB and then times again, the addition of the two charges which we already know, is to nano columns. So in other words, just gonna be five nano colognes minus three Nanako loans. And if you write it all out, if we write it all that we get that the potential at point B is 8.99 times 10 to the ninth. And now we've got the distance, which is 0.5 And now we've got two times 10 to the negative nine. And if you work that out, you should get the potential at this point is equal to 3.6 times 10 to the third, and that's in volts. So that is part A and part B. And now the last part over here, which I'm gonna do Let's say over here is I need to figure out what the work that's done on a one nano Coolum charge. And let's see, we've got from the first point to the second point. In other words, we're going from point A over two point B, and we know that a work that's done on a feeling charge through a potential difference is negative. Q. And such the feeling charge times the potential difference at this point, and this is the potential from a to be So, in other words, what happens is the work that's done is equal to negative Q times. The difference in final minus initial potentials. VB minus v a. If it had said the second point to the first that we would actually reverse that, so it's very important that you do the right step here. All right, so we've got the work that's done. It's gonna be negative. We've got one times 10 to the minus nine. That's the feeling charge that we have this negative one Nano Coolum charge is this feeling charge over here? And then we've got the potential difference. So VB was equal to 3.6 times 10 of the third V A was six times 10 to the third. And if you work all that out on your calculator, by the way, you could have actually just figure out what the potential difference is just, like actually gotten the subtraction. And then you would just plug that in here. But I'm just doing it sort of like the expanded way. And if you work this all out, you should get a work that's done, which is equal to 2.4 on That's times 10 to the negative six, and that's in jewels. So that's the work done in moving in charge from that first point over to the second. All right, let me know if you guys have any questions with this and I'll see you guys the next one.