Okay, So in this problem, you have to particles. They each have a charge of five times 10 to the minus six. Cool. OEMs. Right down and their distance between them. The distance between them is 24 centimeters. Let's convert that two meters 20.24 meters, and we want to get the force exerted. 1/3 particle. So I'll call this. I don't know. Thank you. And that has a charge minus 2.5. Micro cool ums. Um, and that's 13 centimeters from each of them. Um, so I'll call that. I don't know. Let's see Big D, I guess to keep along with the theme. Not that That's Ah, convention. Okay, so, um, our goal is to get that the forest that they exert on this particle. So, um, actually, I would need to know how it's configure it, so let me really the problem. I guess it had It almost has to be a triangle. Right? So if the two cues are like, um so this they're this far from each other, right? This is our d. Here's our two cues in order for them for, uh, this que to be 0.13 from both of them. It kind of has to be up here or down here, um, sort of. And so So, Yeah, that's probably important to acknowledge that this is the picture we need tohave based on the description. So, um so here we go, And then our goal is to get the force on here, so I'm just gonna draw free body diagram on cue. So it's gonna feel a attractive for us to both of the charges. So something like this, And, um uh, but the horizontal components are gonna cancel, so we're just gonna be left with the vertical, some of these vertical components. So, basically like this, that would be the vector sum of these two. Probably a little bit shorter. Um, but, um yeah, so it's just and then these air the same magnitude vectors. So we're just gonna take the vertical component and double it. So you basically, this problem boils down to finding the vertical component of the force of, um, one of these charges on one of these little cues on the big Q. So let's write that down, then. Um, that's right down after you Q first. So it's just gonna be a k q q over, um, Big D squared. Um, and then now we need to get the components, and so we need to figure out the ankle. Um, so it's gonna be pointing along this line that connects the two. Um, this line has a magnitude of d. And then, um, we know that the full line is off is, um de so this side's gonna be d over two. Alright, so I've just been trying to get this angle, so I'm drawling this triangle to represent with crayon. So, um, and then this is going to be our data that we and we can take the vertical component by, um, looking at the length of of this poor sorry by, um, let me just remind a little bit to make sure I'm making full sense. Yeah, I just I drew the triangle that was connecting all of these three charges and to in order to get the angle that this vector is putting down because I know the vector is gonna be pointing along the side of the triangle and so that I can identify this data and then, um, I can if I multiply this forest by the coastline of this data that I'm getting the vertical part of the forest. So it's gonna be so I'm and need to do coastline data and then to get the net forest, I need to double this. Right? As I was saying before, so then I get that, um f net equals this times two. So now it's I just need to plug in numbers co sign of data. It's adjacent over high pot news. Um, and so I could if I wanted, I could find that I using tangent here. I could say tan data is your little do or two, um divided by big D But I can also use Pythagorean two to determine this side, and then I could just to directly do this side over D so the site is gonna be my Pythagorean theorem D squared minus B squared over four. So then the coastline, if data is just gonna be this divided by d. So actually, I'll just go ahead and said that in there. And, um, I'll put my two in the front so I don't forget about it. Is that, too, that I raced? And yet now I'm gonna do a Jason over iPod news. So it's d squared minus d squared over four, divided by D. So that's gonna make this d cubed. Okay, great. Now I can start plugging in numbers. Um, you pull up a tab. Okay, So we're gonna do two times eight point. How many Sig figs? I think it was to about 32.92 times 8.99 times 10 of the nine. And then I'm just gonna use magnitudes of Q. Um, because I already took into account for direction. Um, in my drawing, which sign is the only time would affect the direction. And I already took into account the direction of drawing. Um, that's personally how I like to approach these problems. So then I multiplied the two charges, and then I'm gonna look a climber. The square root of big B squared big. These points 313 squared minus little D is 0.24 square divided by four square. That whole thing you're not going to arrive at by 40.13 to the third power. So let's see what I got. Um, I got 50 and ice. It comes out to just know ah, times 10 to the anything. So 5.11 Newtons is, um the what I get for the net force.