Three point charges are arranged on a line. Charge q3 = +5.00 nC and is at the origin. Charge q2 = -3.00 nC and is at x = +4.00 cm. Charge q1 is at x = +2.00 cm. What is q1 (magnitude and sign) if the net force on q3 is zero?

Video transcript

Hey, everybody today, we're dealing with the problem about Coghlan's law. So we're being told that in an electrostatic slab activity, there are three charged fears that can be modeled this point charges that are being used to study electric forces. Now, spheres, S three is placed at the origin as to is that X is equal to four cm and S one is placed at X is equal to eight cm. Spheres, S three and S one have charges of negative foreign nano columns and negatives or, and six nano columns respectively. So with all this in mind, we're being asked to find what is the magnitude and sign of the charge S two, if the net force from the other charges, S one or the net force of the other charges on S one is zero. So let's go ahead and take a look at this, right? So there's a few things I'd like to know before we get into the nitty gritty is the first is that we're going to be utilizing cool owns law which states that the force being exerted between two charges will be equal to the coolants constant multiplied by the magnitude of the two forces in question, the two charges Divided by the square of the distance between the two or few other things to keep note of equilibrium means that the sum of all forces acting within a direction or equal to zero. And we have two conversions. We want to remember that one nano column is equal to 10 to the negative nine columns. And that one centimeter is equal to 10 to the negative second meters. So this is mine. Let's go ahead and sort of draw out what's going on here. So if we assume we have three balls, three point charges, that's a horrible point charge. We have this ball and I'll just copy and paste it and copy it. And then we can so important. There we go, I can paste that right and then we can do it again, paste. So with that little, there we go. So let's label these. We can say that this is X is equal to zero, Which means this is fear S3, this is Fear S two, which is said to be located at X is equal to four centimeters. And we'll say this is S one which is located X is equal to eight point oh centimeters. No, we're, we're being told that the sum or the net force of the other charges on S one is zero, right? So on S one, the net charges Between S three and S two R 0. So let's write that out the sum, The charges on one are equal to zero. Now noting that the force between S one or S three and S one is attractive, right? Because S three has in charge of and alright. Charges here it has a charge of Negative for Times 10 to the -9 columns. Oh, sorry. It's S one has that church, right? No, my bad. Yes. S three has a charge of negative four times 10 to the 99th columns. While S one has a charge of six times 10 to the NATO ninth columns. So since they're opposite in direction or the opposite sign, it's an attractive force which means they will be attracted to each other and directed towards the origin. The force between S two and S one must therefore be repulsive, right? Because if S one is attracted towards the origin, but we're having a state of equilibrium where the sum of the forces acting upon S one or zero, Right? Then that must mean that the force of S two on S one must also be repulsive enough so that it maintains equilibrium there and directed away from the origin, this being the origin. So if we write that out, we can say that the force of one on two minus the force of one on three B zero. And this is because the force of one on 2, excuse me will be a negative value. Well, it'll be a positive value will be in positive X, but it'll be counter active to the force that one exits on three, which is attractive. Remember F one on two is a repulsive force. One or two is a repulsive force. While F on F of one on three is an attractive force. But this therefore means that the magnitude of the forces must be the same. So if we expand this using coolant flow, we get that this is Q1 on Q two over R 21 or 12 squared, which will be equal to. And I'll let me write this side and read KFQ one on cue three, Oops Uber or one on three squared. Now a few terms will cancel out the K and Q ones will cancel out and we will be left with Q two over R12 Squared is equal to Q three. The charge of S three, The magnitude of the charge of S three over the distance between S one and S three squared that too in meters. Now let me move this to the side, move this over here to have a little more room to work. We want to find out what is the magnitude and sign of the charge as to correct. So let's excuse me, let's rewrite our equation And solve for Q two. So rewriting that's us Overseitude of Q two will be equal to um Charge of Q three, The magnitude of the charge of Q three multiplied by, I just read it in black R 12 oops Divided by R13 sq substituting in our values we get Q three has a charge of four times 10 To the -9. Cool homes. Again, this is ah absolute values to magnitude. It's very like so multiplied by four times 10 to the oops. That's not a four, that's a nine, but it will be goodbye four times 10 to the negative 2nd m meters Because it's four cm away from the origin as to one on two divided by The distance from 1 to 3 which is eight times 10 to the 92nd m equating. This gives us a final answer of one point oh times 10 to the negative ninth columns or converting back 1.0 nano columns And the magnitude of discharge on S one is positive, right? The magnitude will be positive because it is going in the positive X direction. It's going towards us one away from the origin in the positive X direction. So it will be a positive uh force giving us answer choice. Hey, I hope this helps and I look forward to seeing you all in the next one.