All right. So, uh, but this problem, we have this, uh, this set up so to threat, connected with two balls and, uh, angle between the thread and the vertical is data. So the mass of the boys and and the charges, Q one. So this is also an on the q two. The information we know Ah, about the state of it is that we know. Ah, the mess. Em is eight grand, and the CDA is 20 degrees and l l which is the lens of the threat, which is 200.5 meters. Ah. So, uh, part, eh? We want to show the free body diagram off the off the board, and we want to label the forces that down the hall. So we're just four simplicity. We just look at the right one. We just look at this one. A CZ. You can see that the club is bowl. We have the gravity queer de and we have the coon Falls because these two bowls, they carry the same, uh, listen, can this same kind of charge, So the colon forces upstate to each other, so it's pointing to right direction. And this is the force on the threat. So let's say this is ft. This is F C. And this is empty. Okay. And the four party, we want to find out the colon force. So if you're looking at this diagram, we see that the f t. It's basically office. It is opposite to the combination off F C plus F G. In that case, we know that I've see over FT Is tension pita So weird. This is the data. Okay, so we know that we can find out f c quote f g times tending beta and ft is simply the gravity. So this is m g times tended data. We already know where the mass and we know that data. So just plugging the two values and see that FC for their 20.28 54 nude is okay. And, uh, we also want to know on the fourth on the threat of tea, uh, again, we just look at this free body diagram. We know that f a g over ap ti equal course on data. Okay, so just the plug in the video full f g and the data second, see that t a quote point on 0834 new. It's okay. And for Parsi for here. A Parsi. Ah, we want to get some information about the Cuban. Que tu uh, I'll say that at this moment we can know determine the value for Kyu won And que tu However, we can steal some information, so we already have the magnitude of the cool and force over here. Right? So we know the expression for cool imposes. Have see, they call one over four pi Absolute knots times you want you to be. Bye bye. The separation. Here's the separation. So just utilize the strangler knowledge and you could see that separation. The call to L. A times son Peter with bread. Okay, so this value equal 0.28 54 as we found previously. Okay, So, uh, we can solve for Q one times Que tu by using this expression So we see that is equal to 3.71 times 10 to the negative. The thing Call him squared again Because this is a Q one dance, you two. So this is the only information we camped in for the time being and for party. Uh, so the thing is. We suppose we connected the two bowls with the with the metal with the metal wire so that the charge on the two bowls Ah, uh, now they're equal. Ah, and the we get in this case, we will obtain. Ah, I mean, if it were up to a new angle beta So they had a prime April 30 degrees. Yeah. So now that we have a new angle leader prime, however, the gravity stays as constant. So in that case, we can turn the new cool in Force FC Prime in this case, equal and G times 10. Jane's fleet of prime. Okay, so this sequel 0.0 fourty five 26 notice. And this equal to the expression for the colon force over here. One over four pi astronauts. You want you to worry it's not. Q on Q two. In the case, it's simply a Q quit because the two bulls in this kid's carried the same kind of charge the same amount of charge and the separation in this case is too. L sign made a prime. Quit. Okay, so in this expression Ah, the only all new variable is this Q. We know l we know Peter Prime. So just plugging the vetting for Aaron of Prime, we can find out Cube. So I saw I got q equal, Thio one point 12 times 10 to the negative six quotes. Okay, so remember that the charge is conserved on the two bowls. So in that case, Q, we go to half off Q one o'clock. You, too. Right. So, uh, we know that the committee we know where we know that some that some charge of Q and Q two by using this compression. And we also know that the product of Q one Q too, so we can solve for Q on q two. So the final solution I got is, uh, I like beer. So cute boy is 1.8 times 10 to the negative seven cools. And the cue to is, uh, it's the rested er, er I don't have much room here. I just square to label This is the final solution. And the other one, Q two is, uh, 2.6 times 10 to the native six homes. Okay, there's a six. So these or the funding solutions okay,