Potential Difference Between Two Charges

Alright, guys. So you've got these two charges they're lying on this line and we're supposed to figure out what the potential difference is between point A and point B. So, in other words, what is the potential between these two? Sorry, the potential difference between these two points. Another thing that could have asked us is instead of this, they could have asked us for the voltage because those things meet in the same exact thing. In other words, or in any case, we know that the potential difference between or from point A to point B is just gonna be VB minus V a. So, basically, we know that at these at these points right here, we're gonna have to contributions at the from the potential we're gonna have the negative three Q charge and acute charge producing potentials at both of these locations. Basically, which need to do is we need to figure out what these things are Now. I just want to warn you, the math here in the algebra is gonna get really messy. But the procedure is fairly straightforward. We know that at point B, we're gonna have to contributions the potential from Q at this point, plus the potential from the negative three Q at this point, So basically, we just need to figure out what those things are. We know our equation. We know that V is just equal to K. Big Q over a little are. In fact, I'm just gonna I'm just gonna scoop this up over here, So that means the potential from Q at point B is just going to be. Let's see, we've got K. Then we've got to queue. Now we've got divided by the distance involved here. Now, where this gets a little tricky is that we have this s distance, that the complete distance between these two charges and then we have these X is right here. So that means that the distance between Q and B is actually s minus X. Is this whole entire distance minus this little piece right here? So that means it's just gonna be s minus X now from the negative 33 charge. We just have this distance Little X over here, So this is gonna be K times negative three Q divided by X Now, one of the things that we can actually do in this case is we have these denominators that aren't the same. So one of the things we could do with fractions that don't have the same denominators is essentially, like cross multiply. So they do have the same denominator. So basically, what we have to do is just move these things over and multiply. So what? This ends up being and this is where the algebra gets a little messy is this is gonna be cake you times x from this guy over here divided by and then the denominators become the same s x s minus x. And then this guy over here this You have to remember that this is a negative three. Q charge is gonna be minus three K. Uh, this is gonna be cake. You This is gonna be s minus X divided by X s minus X. All right. So you can already see how the math is gonna start to get a little tricky over here. Um, so let's see, I've got how can I simplify this? Okay, well, I've got this is gonna basically gonna expand out to K Q X over S X s minus X and then I have minus three k Q s over, X s minus X. And then we've got this minus and the minus over here are actually gonna sort of turn into a plus. And this is gonna be another three K Q X over X s minus X. So one of things that happens is that this guy over here well, actually group up with this term in the front, and all you're gonna get is you're just gonna get, uh, four k Q x over X s minus X minus three K Q s over. X s minus X. All right, so this is what the potential looks like at point B. Now all we have to do is basically the same exact thing. So I'm just gonna separate this the same exact thing for the potential at point A. So we've got at a This is gonna be Let's see, we've got a k q X over our Sorry Que que? Let's see cake divided by X. That's right. Yeah. So we've got this distance right here, and then we've got plus K negative three Q divided by And now we have this, uh, this piece right here, which is s minus X again. So this is just gonna be s minus X. And now we basically have to do the exact same procedure. Eso We're basically just gonna take these terms up here and then on, then cross multiply them. So I know this is gonna get a little tricky, Guys. Eso we have cake you and then we have s minus X over X s minus X and then we're gonna have, uh, Plus, Or I guess, actually, this is gonna turn into a minus sign because of this negative right here. This is gonna turn into three K Q x over X s minus X. Okay, so one of the things that we can do is we can sort of distribute this this cake, you It's kind of like the same way how we pulled out this k this three k Q x term over here, we just have to distribute this cake you into this minus X, and then this cake you X that you get from this term and this minus three cake X are actually going to be the same thing. So, in other words, we're just gonna get cake us over X s minus X minus four k Q x over X s minus X So I just wanna point out. So just give myself some room. Just wanna point out that when you distribute this cake, this cake s term comes from the first and this cake you into the negative X term actually goes inside of this and groups up, and that's why it becomes a negative for So I just want to point out, just in case you got lost a little bit. So basically, let's see what is the sense of being is Yeah, this is the actual final results. So that means that we can get back to the voltage calculation, which is basically going to be the subtraction of these two things. So we have this term, which is VB, and then this term, which is va and we all you have to do is just subtract them. Eso we have this. Let's see, we have, um, Delta V is gonna be Let's see, I've got this four k Q x term over here, and then I'm gonna subtract this entire term over here. When I do that, the subtraction of this term is actually going to turn into a positive. So this and this actually group up to form eight K Q x over X s minus X. And then what happens is this term over here, this negative three cake us when I subtract v A, which is cake us. It's just gonna be minus four cake us over X s minus X, and that is the answer. So that is the potential. And then all you need to do to actually solve numbers for this is you just need to know exactly what the distances involved are. So I know again, the algebra is kind of messy. But if you guys could do this, then you should have no problem with working with these potential differences between point charges. All right, let me know in the comments that any one of these steps sort of like tripped you up or got you confused. I'd be happy to explain it. Just let me know, guys. All right,