1

concept

## Coulomb's Law

9m

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Hey, guys. So in this video, we're gonna talk about Cool owns law, which is a really, really important law that you need to know for electricity. So basically gives us the electric force between two charges. So go ahead and watch this video as many times you need Thio. We're gonna be covering a lot of examples and practice problems in the videos after this. So basically, electric forces can be attractive or repulsive, and that's a direct consequence of what we talked about with, like, charges, you know, being attracted and or unlike charges attract ing and, like charges repel ing. So, unlike charges, if you have positive, negative or negative, positive will exert attractive forces on each other. Or is if you have to, like charges like two protons or two electrons, things like that. Those things want to fly away from each other so like charges will repel and exert repulsive forces on each other. Now the name for the force is called Columns Law or the Coolum Force, and that gives us the force between two charges. So if you have two charges Q one and Q two and they're separated by some distance, little are thin. The force that exerts between them is gonna be K Q one Q. Two. So both both the charges divided by r squared. It's very similar to how we studied the gravitational force between two masses. So you have some constant times, the two masses divided by the distance between them. Well, columns law, the electric force just as some constant times the two charges divided by the distance between them. So this K constant right here is called columns constant, and that has a number 8.99 times 10 to the ninth. It's really easy to remember, because it's like 899 10 9. So this is something you absolutely should commit to memory. It's very important a lot of times on tests you won't be given this number exactly. So go ahead and commit that to memory, and there's some units associated with that. It's Newtons meters squared per Coolum squared, although that's a lot less important. You probably won't need to know that. So basically, this equation right here gives us the force that exists between two charges and also acts on both of those charges because of action reaction. That's the magnitude of that force. As for the direction that force always points along ah line that connects the two charges. So line connecting to charges. And basically what I mean by that is if you have these two charges Q one and Q two and you know the distance between them, that's little are then, if it's a repulsive force and from Q one and Q two and that's gonna go in this direction, so that's gonna be a repulsive force. And if it's an attractive force, then it's just gonna point in the opposite direction. So that's gonna be, uh, yeah, yeah, that's Ah, that's attractive. And, um, yeah, so it always just exerts along the line that connects those two things. And again, the attractive or propulsion just has to do with whether the fact there are like or unlike charges like repel unlike attract. So I'm gonna give you guys a pro tip in order to figure out the magnitude and direction. So whenever we are trying to figure out cool OEMs law, we're always gonna find the magnitude of the Coolum force just by using positive numbers, and then we'll worry about the direction later. So find the direction by using the attracting and repelling rules. So whenever you're plugging in to this formula that we've given the Coolum force, you're always just gonna use positive numbers and then worry about the direction later. All right, so let's go ahead and take a look. A quick example. In this problem in this example here, we're gonna be calculating the ratio of the electric to the gravitational forces in a hydrogen atom. So in a hydrogen atom, we just have the proton. So we have the mass of the proton and we've got the mass of the electron. But there's electric forces because thes things also have charges. So we have the charge of a proton and the charge of an electron. Now we know that the charges for each of these things were just related to the elementary charge. So this is plus E and this is minus E. So basically, we're trying to figure out what the ratio of the electric force is to the gravitational force. So let's go ahead and solve each one of those separately. So we've got the electric force is gonna be, Let's see, we got K that constant times the product of the two charges. So we've got Q proton que electron and then divided by the distance between them. So that's gonna be r squared. And actually, I have all of these constants just in this nice little table right here. So we know this K constant is 8.99 times 10 to the ninth. Remember that. Now we've got the elementary charge. That's also something you should know. 1.6 times, 10 to the minus 19. And now that's for the proton. For the electron, it should be negative. But again, we're just gonna worry about the magnitude of the force. So we have to just plug in a positive number, right? Worry about the direction later and really were asked to find the ratio of, like, the magnitudes of these forces anyways, so we're just gonna use positive numbers. All right, so we've got the distance 5.3 times, 10 to the minus 11 and that is gonna be squared. So you go out and work this out. You should get 819 times 10 to the minus eight. That's a Newton's. So that's that one. So right, So actually going to use a different color for that. So that's the electric force. Now we just have to do the same exact thing for the gravitational force. So now, gravitational force. Well, and just in case you have you forgotten the gravitational forces capital G times the mass of both objects divided by little r squared. So we've actually have all of these constants over here. So I've got 6.67 times, 10 to the minus 11. That's the gravitational constant. The mass of the proton, 1.67 times 10 to the minus 27. And then all of that's an S. I, by the way and then 9.11 times 10 to the minus 31. And then we've got the same distance between them 5.3 times, 10 to the minus 11 uh, squared. So we work this out and you should get 3.61 times 10 to the minus 47 which is a huge I'm sorry, which is a very, very, very tiny number s so we're just gonna see how tiny that is in a second. So we've got the ratio of these things. Uh, the gravitational or to the electric and the gravitational force. That's just gonna be 8.19 times 10 to the minus eight. Divided by 3.61 times 10 to the minus 47 which is a very, very time number. If you work this out, you've actually plugging those numbers and divide them. You should get 2.27 Let me write that out. Two points to seven times 10 to the 39. And by the way, this is just a dimensional issue. No number because we're trying to basically figure out how much time, how many times stronger is this force than the gravitational force. So, in other words, this thing is trillions and trillions and trillions of times stronger than the gravitational force. Electric force is a very, very strong force, and this is our final answer. So again, there's no units. Alright, eso Let's go ahead and take a look at another example here. So we've got two identical charges and they're connected by five centimeters water wire. Now what's happening is we have two identical charges that are on the end of a string like this and because they're like charges they want to repel away from each other. But as they start to do that, there's some tension that's created in the wire. And using that we're supposed to figure out what the magnitude of these charges are. So the first thing is that we know that these two charges are identical. What that means is that we have. Q one is equal to Q two, so that means that we can just use Q in our equation, not have to worry about Q. One Q. Two. They're the same exact thing. So if we're trying to figure out what the charges are, then we just start off with cool OEMs law. So in other words columns, laws saying that we've got K Times Q one Q two divided by r squared. But again, if these two things are the same thing, this is just gonna turn into a K Q squared. Divided by R squared, right, because Q Times Q. If they're the same exact thing is just Q squared. So all we have to do is just figure out what this Q squared is. All right. So let's see. We've got these, like charges and they're trying to exert repulsive forces on each other like that, right? So cute one is trying toe push away Q two and Q two is trying to push away Q. One action reaction, and the reason that they're not flying apart is because there's some tension that exists between the wire that basically keeps these things together. So we know that the tension here is equal to 10 Newtons, and so if everything is in equilibrium, then the tension the tension is equal to the electric forces. So we've got the electric forces right here and here. And I know I just didn't draw them equally sort of to scale. But these things should be equal to each other. So in other words, the tension is equal to the electric force. That's why nothing is actually flying apart. All right, so let's see, we've got, um let's set up the equation If I wanted to Q squared. So I just have to do Q squared over r squared. Now I'm just gonna isolate Q squared by moving this over and then moving the K to the bottom. So we're gonna get that f e the electric force Times R squared, divided by K is equal to Q squared, All right. And if you go ahead and look through our numbers I have with the electric forces because I know it's just the tension. Now I've got r squared. Now, I just have to I just have to realize that this is in five centimeters, so I have to convert it to s I first. So I mean, I have that are, uh this distance right here is equal to 0. m. So I've got that's and then K is just the constant, right? So I'm ready to go. So I've got 10 Newtons times the R squared, which is 0.5 I've got a square that divided by 8.99 times 10 to the ninth. If you go ahead and work this out, you should get, um Let's see. I got some number. Oh, that's right. We have to square root this thing first. So all you have to do is for this number, you have to take the square roots of that. And that should equal Q. So I'm gonna write that you're gonna do that. And if you work this out and I'll take the square root. You should get a charge that is equal to 1.67 times 10 to the minus six. And that's in cool OEMs right there. So that is the answer. Let me know if you guys have any questions with this.

2

Problem

If the force between two charges is F when the distance is d, what will the force between the two charges be if they were moved to a distance of 2d?

A

2F

B

F/2

C

F/4

D

4F

3

example

## Charges In A Line (Find Zero Force)

7m

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Hey, guys, how's it going? So I wanted to work this problem out together. We're supposed to figure out where we need to put a one Coolum charge so that the force on it is equal to zero. So what does that mean? We're just gonna assume that the net force So what's asking for is that the net force needs to be equal to zero, because we have to cool OEMs, or we're sorry. We have to charges on the outsides. So what this problem is saying is that we need to put a one Coolum charge somewhere around here in order for the forces on it to cancel out. So before I actually start plugging anything in, I just want to figure out qualitatively meaning no numbers where this thing needs to be in order for those horses to cancel out. So let's take a look at the left side. So imagine I dropped a one Coolum charge here and I want to figure out if those forces would cancel. So all I have to do is just figure out the directions of the forces on it from the two Coolum charge. Remember, all of these things are positive. So they're all going to repel. They're gonna repulsive forces on each other. This thing has to move to the left. Right? So this is the force from the two Coolum charge that's gonna point to the left. But the force from the three Coolum charge also has to point to the left. It might be stronger or weaker, we don't know, but it's going to point to the left. So in other words, there's never going to be a situation where these forces are going to cancel out. They're always going to add together and point to the left so it can't be here, right? And so I'm right here that these things never cancel. There's never gonna be a situation in which one will point in one way one will point in the other. So it can't be in this area in this region right here. So let's look at the same. Let's look at the situation on the rights. Who got this one cool in charge here and you don't do the same thing from the three Coolum charge. You have a four set points in this direction. The too cool own points points in this direction. So for the same exact logic this thing will never cancel out. These forces will never cancel out. So that means it has to be somewhere in the middle. And the reason for that is if you work out the forces here, the force from the too cool on charges gonna point in this direction and the fourth from the three column charge needs to point in this direction. So that means that there is some magical distance in which these things will cancel out. Now we have to be careful because it won't exactly be in the center. If this thing were in the center than this stronger charge here will produce a stronger force. So this thing actually needs to be slightly closer to the left so these things will balance out. So that means that the condition that we're trying to solve for the rest of the problem is that we need the magnitude of the two Coolum force to be equal to the magnitude of that three Coolum force or the forces from those two charges. In other words, this is the condition that we need to solve, right? They're gonna be pointing in opposite directions. But if their magnitudes of the same, they're going to cancel out. So let's go ahead and write out the expressions for those two. So the two Coolum charge right here I need a distance between these two charges. By the way, this is a one Coolum test charge that we're gonna drop right here. So I'm gonna call this distance here, X. And that means that if this distance between the two and the three Coolum charge is are so this is our equals 10 centimeters, then this is just our minus X rights. That's the distance between eyes just the whole entire distance minus that little chunk of X R minus X. So I've got the K constant times the two charges that are involved. In other words, this is gonna be the too cool in charge and the one Coolum charge divided by the distance between them. That's X squared. Now I set up the equation for F three. So that's this guy right here between the one Coolum charge. So okay, and then I've got the three and one. So that's three columns one cool arms divided by the distance between them squared R minus X right and then squared, so I don't know what those distances are. Now again, what we're supposed to solve is that these two things are supposed to be equal to each other. So in other words, these expressions right here that I've just figured out have to be equal to each other. So let's just go ahead and actually said those things equal. So we've got K to one divided by X squared equals que three one divided by R minus X squared. So this thing is an equal sign, so we can actually cancel out anything that appears on both sides of the equation. We're gonna get the case will cancel, and also the ones will cancel. I mean, one doesn't really do anything, so I mean it. Just cancel it anyways. And so we come up with an easier expression. Now, what I'm trying to solve is where I need to put this one cool in charge. So that's a distance. So in other words, I'm gonna go ahead and solve for X. So X is my target variable here. So just because of this, the easier one to work out like I could actually go ahead and solve for what ar minus X is and refined. But it's just that X is an easier variable to solve. So let's go ahead and do that. If I wanted to solve for X, I had you to get all the things involving X to the one side. So I'm gonna move this ar minus X over to the other side, cross multiply, and then this to needs to come down and basically trade places with it. Right, So you just have to cross multiply. Okay, so we get our minus X squared, divided by X squared is equal to three over to right. We have the three and the two on the outside. So all you have to do now is just take the square roots. So you might think you have to foil this thing our our might X squared, and you're gonna get these all these ugly terms. But the thing is that both of the terms in this expression R squared so you can actually just take the square root of both of the whole entire thing. So if you take the square root of both sides where you're gonna get his ar minus X just one term and over X is equal to the square root of three over to. So now how do we get How do we start isolating this X? I just have to move this stuff to the other side. So I have to bring this X over. And this becomes ar minus X equals the square root of 3/ X. And now all I have to do is this a subtraction? So I could move this X to the other side and and add it right? So I got r equals the square root of 3/2 X plus X, right? So and then I have this X as the greatest common factor inside of both of these terms. So God's r equals that about X parentheses, the square root of 3/2 plus one. Right. So if you distribute this X back inside this expression, you should get back to what the thing on the top is. All right, so now all we have to do Is this our distance? Remember, it was the 10 centimeters. So that means if I wanted to figure out in meters, that's just gonna be 0.1 So I got 0.1, and now I always to do is just divide this over to the other side in order to get rid of it. So I've got this expression here on the bottom is gonna be square root of 3/2 plus one, and that's gonna equal X. If you plug all of this stuff into your calculator carefully making sure that you have a parentheses here in the denominator, you should get an X distance of 0.45 That's in meters. So in other words, that's 4.5 centimeters. And that's our answer for X. So, in other words, going back to the diagram, this thing, this one Coolum charge has to be 4.5 centimeters away from the left. Coolum are from the left charge, and that means that this remaining distance over here is gonna be 5.5 centimeters. So in other words, it confirms we were saying before, this charge has to be slightly closer to the weaker charge so that smaller distance can balance out the larger charge on the right. Alright, guys, that's basically it. Let me know if you have any questions with this

4

Problem

In which direction will the −1 C charge move? If it has a mass of 10 g, what will its initial acceleration be?

A

left; 3.8×10

^{14}m/s^{2}B

left; 6.2×10

^{13}m/s^{2}C

left; 6.2×10

^{9}m/s^{2}D

right; 1.1×10

^{15}m/s^{2}5

example

## Charges In A Triangle (Rank Force Pairs)

1m

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Hey, guys. So in this problem, we have to rank although the possible pairs of charges in this diagram by which one has the greatest possible electric force. So we're talking about cool owns law and specifically for point charges. So we've got to use columns. Law K Q one Q two divided by the distance between them square. That's little are in this diagram, though we actually have the distance as D, but it's the same exact distance through all of the points in this triangle, which means that this D isn't gonna be a factor in determining which one is the strongest in which one is the weakest. Instead, what we have to dio is we have to just look at the product of the charges themselves, and we have three pairs that we're gonna discuss. We're going to discuss the pair between two and three e to e and sorry to an e two and three and then e and three. So I've got these little loops right here that represent those pears, and all we have to do is just compare the magnitude of the charges involved. Now, all these air like charges that all exert repulsive forces on each other, so we just have to figure what Q one Q two is Now. I've got to e and E and remember that e stands for the elementary charge, so I've got to eat times E. So, in other words, that's just gonna be two e squared. Remember that E is just 1.6 times 10 to the minus 19 just stands for, like, the electron or the country charge. And if I do the same exact thing over here, I've got Q one. Q two is equal to two e times three. And then that's just gonna equal six e squared. And then I've got Zoo Wannabe, the minus sign and then for the last one. I've got e times three. So that's just gonna be three e squared, right? That's Q one Q. Two. All right, so we're just gonna basically plug in these things if we're actually trying to figure out what the forces are inside for Q one. Q. Two, Which one is the greatest? Well, the six e squared is first, and then the three e squared is next. And then, finally, this two e squared is third, so that's That's the rank in terms off, sort of descending, greatest electric force. Alright, guys, let me know if you guys have any questions with that.

6

example

## Charges in a Plane

7m

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Welcome back, guys. Let's try to solve this one together. So we've got these three point charges arranged in this triangle right here, and we're trying to find out what the net force on this three Coolum charge is. So let's go ahead and first draw. What? The forces that are acting on this three cologne charge. So we got We're focusing on this guy right here. And the first is that he feels a attractive force from this negative too cool in charge. Why is it attractive? Because these things are unlike charges, ones positive ones. Negative. So this is gonna point in this direction, and I'm gonna call this F two because it comes from the too cool in charge. Now, there's also this cool, this one cool in charge here on the bottom left corner, that is some distance are away from this other charge, and it's gonna exert a repulsive force on the three cool in charge because these things are like charges. They're both positive. I'm going to call that F one, and what we're supposed to do is take thes forces, which is the only force acting on it and figure out what the net force is on this object. So in order to do that, we have these forces there pointing in different directions. We have to use vector addition. So with vector addition, we always have these steps right here. First we get a label and calculate these forces, which we've labeled already. So I'm gonna take that over here, and then we have to decompose them, pick our pick our directions and add them in this all for the, uh, net force. All right, so let's go ahead and calculate the forces. So I got f two and I'm gonna be using columns. Law write columns. Law is just k times the Q one Q two divided by the distance between them squared. So that means that F two, which is the force between the negative two columns and the three columns, is gonna be K which is that constant? 8.99 times 10 to the ninth. And then I've got the two charges involved. So I've got to two and three divided by the distance between them square 0.6 square. Now, some of you might be wondering why I haven't chosen that negative sign and it's because remember, whatever we're finding, the Coolum force were always just going to plug in positive numbers. We're gonna worry about the direction later. So if you work this out, you should get The F two is equal to 1.5 times 10 to the 13th. And that's a Newton right here. Right? So you don't need the units for that if we try to work out F one. So in order to work out F one, we need 8.99 times 10 to the ninth. We've got the two charges, right? You've got this three Coolum and the one cool. Um, but we don't have the center of mass or sorry. We don't have the distance between them. That little are. So hopefully you guys realize that this is a 68 10 triangle. Right? So this should be 10. But if you didn't, you could always just use the Pythagorean theorem. So you've got six squared plus eight squared, and that equals 10 centimeters. All right, so we want this in s eyes. This actually just be 100.1 m. So first we have to go. We have to do three columns, One column and then divided by zero. We got 0.1 square. And if you work this out, you should get 2.7 times 10 to the 12th. Alright, I'm trying, right? That's small enough just so we don't run into this little thing right here. But anyways, now we've labeled and calculated both of these forces. So we're done with that step that first step. Now we just have to go ahead and decompose this thing. Right? So you've got these forces that are pointing in opposite directions at once pointing to the left one is pointing upwards, so we got to decompose it. So I've got to decompose this F one vector by basically moving it and projecting it down to the X and y axis. And I do this by needing by using my trig and my co signs and things like that. Right? So, in other words, I need my f one X components and my F one. Why components. Now I've got this. Um, I know if I wanted to break up this F one into its X and Y components, I know I relate that using the sine and cosine. So F one is gonna be Kasey in F one X is gonna be a F one co sign of data and F one. Why is gonna be f one of sign of data? So I know what these forces are I'll have to do is just figure out how to get Thetas. So I use that by going ahead, looking at my triangle, and usually I'm told what the angles of these things are so I can go ahead and figure what this angle is. Or I could just use the relationship of sine and cosine in the triangle itself. What I mean by that is that I've got this angle Fada right here, which, by the way, is the same as this angle theta right. These things are parallel lines. This thing is a parallel parallel horizontal lines, so could basically complete this little triangle. And I know that this is eight centimeters. So I've got 68 and I know the hypothesis of this triangle is 10. So that means Well, let's use our sine and cosine rules. Right? Co sign of anything is just adjacent over high pot news. So, in other words, co sign of this angle right here is the adjacent side over there. Hi, pot news. That's just 6 10 6/10. And if I did the same thing for sign using so gotta I've got opposite over iPod news. So I've got 8/10. So, basically, instead of having to figure out the angle theta and then having to use that, which is normally what we do here, all we have to do is just replace thes co signs and signs with these fractions right here that we figured out it's the same exact thing. So let's keep going. We've got f one X is just gonna be F one, which is 2.7 times 10 to the 12th, and I have to multiply by co sign of fatal, which is just 6/10. So if you do that and you work it out, you should get Let's see, I've got 1.62 times 10 to the 12th. Now, the white component is gonna be very similar to 0.7 times 10 of the 12 Now, instead of 8/10 we just do 10. Uh, sorry. Instead of 6/10 you do 8/10 for a sign. So if you work this out, you're gonna get 2.16 times 10 to the 12th. All right, so now you've got the components. So now you have to basically just pick our X and Y sorry Are positive or negative directions. So we've been working with up into the right. Usually that's what we picked to be positive. So let's go ahead and just stick with that up into the right is gonna be positive. Which means left and downwards gonna be negative. And now all we have to do is at all of our components together. So I'm gonna go down here and make some room for myself. So I got Let's see, that should be enough room, That little box right here. I'm gonna have the F two and F one, and I add them together to figure out what the Net Force is thes air my X and Y components. Now, if you remember, F two pointed purely along the X direction and to the left. So that means that X component is going to be negative. So that's negative. 1.5 times 10 to the 13th and a white component of zero. And now the F one pointed in this direction, which means both of the components gonna be positive up into the right. So we got positive components for that. We've got 1.62 times 10 to the 12th and then 2.16 times 10 to the 12th. So, again, this is why we don't wanna concern ourselves with the negative directions by plugging in the charges first. Because all these things are gonna change anyways, So we're gonna go ahead and pick our directions after all of that stuff. That's why this is that protest is really important. And if you go ahead and add everything straight down, we're gonna get that the F net components, there's gonna be negative 1.34 times 10 to the 13 because it's actually one power higher. Let me just go ahead and make sure you can, As you can see, that that's gonna be 13 and then this is just gonna be 2.16 times 10 to the 12th. So if I actually go back into the diagram, I can actually draw out what that vector is going to look like. It's gonna look like this is gonna have a high negative x component, and then it's gonna have a positive Why components gonna look something like that? Because this F two is actually really, really powerful compared to the white components or start compared to the X component of this guy. This one's actually relatively weak. Um, okay, so I've got my components all added up, So that means that the magnitude of the net Force is just going to be using the pie, the Pythagorean theorem. So I've got negative 1.34 times 10 to the squared. Plus, Now I've got 2.16 times 10 to the 12th, and then I've got to square that Let me just make sure that's written properly. So 10 of the 12 square that when you get the magnitude of F net is equal to what I got was 1.36 times tend to the 13th, because again, this is like one power higher. So it's actually really strong. And that's Newton's. Now the question didn't ask us to find out what the direction is, but when you know, you could do that using tangents and things like that. All right, so that's how you find the net force of these kinds of problems. Let me know if you guys have any questions

7

example

## Exploiting Symmetry

3m

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Hey, guys. So in this video, we're gonna cover something really, really important That's gonna help you in a lot of problems called exploiting symmetry. And it happens when you have these sort of arrangements of charges. Let's go ahead and check out two examples. So in this example, we've got to to cool, um, charges, stationed away from this one Coolum charge. And we want to figure out just the direction, not the magnitude off the net force on this thing. All right, so let's go ahead and work out what the forces acting on that are on the one column charge. There's a repulsive force coming from the two column charge and there's also repulsive force due to the other two Coolum charge. Now, if you wanted to figure out what the net force is, we have to calculate those forces and then decompose them. Right, So we've got f here and then we've got f over here. Alright, so we wanna exploit something called symmetry. Notice how these two Coolum charges are both position the same distance away, so usually have symmetry. Whenever you have the same cues, the same charges positioned in the same distance away So you have distance or are it could be whatever letter they choose to be the distance. Now, if you have these two things, the same charge and you have the same distance that means that the force acting on these forces, both of these efs here are also the same. So in order to figure out the net force would have to decompose them into their components. Right? So we have an ex components and we have a white components over here and we do that by using our trig are signs and co signs and things like that if you could figure out what that angle is, but we also have to do the same thing over here there's an ex components, there is a angle theta and there is a white component. So what happens is if you have the same efs and these things are symmetrically placed, then you're gonna have the same exact angle feta. And in this situation, what's gonna happen is that the opposite components here are gonna have the same exact magnitude. So these things are always going to cancel. You'll never have to worry about these X components and the white components are basically going to add together, and they'll be the exact same thing. Whoa! So they're always going to add together. And so that means the net force of this object is not just going to be somewhere pointing upwards. It's going to be exactly pointing upwards. The net force is just going to be the sum of both of the white components here. Alright, so that's one week where you can use symmetry. Let's see another example. So in this situation, it's gonna be very similar, except instead of both of them being positive charges, we have one positive and one negative. Okay, so we know that there's a repulsive force from this positive to cool, um, charge. But now this negative to cool. Um, charge over here is gonna exert a attractive force because these things are unlike charges, so we're gonna have a attractive force that points off in this direction. Now, these are gonna have these are gonna be f and f over here. We've got the same exact distance and the same exact charges. So that means these forces were gonna be the same. So we've got same cues. We've got the same distance so that means the forces were gonna be the same. And these things are position, sort of like in a symmetrical way, which means we're gonna have the same exact data. So means when we start breaking these things up into their components, right, we've got the white components, the X components by using that trig well, the attractive force this red one can also be broken up into its components. And we're gonna have the same exact angle theta. So it's gonna end up happening is we've satisfied all the conditions for symmetry. So that means that the components in the Y direction, the ones that point in opposite directions are always gonna cancel out. And now the Net force over here is going to be in the purely X directions things. Net force is going to be two times three x component off both of those forces together. All right, so this is just the direction. So this is a really, really important thing they were gonna use. See if you could exploit some symmetry in your problems. Sometimes you'll give him like squares with a whole bunch of charges. You'll have to figure what the directions of the Net force. It's very important. Shortcut. You guys need to know. Let me know if you have any questions.

8

Problem

What is the direction of the net force on the charge at the center of the square in the following figure?

A

Right

B

Left

C

Down

D

The net force is 0 so its direction is undefined

9

example

## Electroscope (Find Charge)

7m

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Welcome back, guys. We've got a fund example involving an Electra scope. If you haven't seen them before, maybe you you who are taking labs and physics, we'll see one. But basically, it's just a little device with these two conducting leaves on it. And what happens is you turn on the machine and these conducting leaves pick up charge and they because they're like charges they want to repel from each other. So the leaves start to separate, and that's why they don't fly off. So that's what Electra scope is. We're told the mass of each leaf in this Elektra scope, and we're also told, uh, the deflection angles to 30 degrees. We're supposed to be figuring what the charge on the end of each leaf is. So what I wanna point out first is that we're dealing with two identical charges, which means this Q one and Q two are going to be the same. That actually helps us simplify things because we know that Q. One If Q one is equal to Q two, then we could just replace them both with just queues. And we also know that these things are going to repel from each other because they're like charges. So that means that this force right here is gonna be out in this direction. That's the electric force and these things. I want to basically push off from each other, so they're just gonna point off in opposite directions now, the reason they don't fly off is because they're connected sort of bythe strings here. So we're gonna see that in a little bit how that plays out. So, basically, if we wanted to know what the queue of each leaf is, we have to use cool owns law, right? We have to use the force between two point charges. That's gonna be que que que over r squared, Remember? No. No Q one Q two because these things are the same thing. So basically, this just turns into K Q squared over r squared, and we're gonna be solving for these Q squared over here. So basically all have to do is move this r squared up to the top, move the k down to the bottom, and I'm just gonna raise that. Basically, what this becomes is our f e uh, sorry. R squared Effie, divided by K, is equal to Q squared. So all we have to do is just take the square roots. So we get Q is equal to the square roots of our sorry r squared F E divided by K. All right, so if we take a look at this equation, this is really all we need to solve. But if we look at this thing, we don't know what the center, We don't know what the distance between these two charges are. So in other words, that's this distance right over here which I'll call Little are. And we also don't know the force involved. So let's go ahead and take this step by step and solve each one of those separately. So how do I figure out our well, hopefully you guys realize that this is a 60 degree angle between these two sides and that these two sides are the same. So one way to think about this is that if this is 60 or this is sorry, if this is 30 and this is a right angle, this also has to be 60 degrees, right? And if this is 60 degrees and this is 60 degrees and this whole entire thing is just unequal lateral triangle. All the sides have to be equal to each other. So that means that this little are is gonna be 0.5. All right, let me know if that makes sense for you guys. So basically, we know now what this little our distance is. Now we have to just figure out what the electric force is now. How do we do that? Remember, there's multiple forces involved here, so let's go ahead and draw a free body diagram for each object. So the reason that these things don't just simply fly away from each other is because there's a tension in the string that connects these things. And there's a tension here, and there's a tension here, basically the free body diagram. So each of these things, they're just gonna be mirror opposites of each other. We also have a weight that pulls them straight down mg, and we're told the masses of each of these objects Now, just for sort of simplicity. I'm just gonna work with this free body diagram here. This one's gonna be the exact opposite. Just everything's gonna be reversed. Okay, so that's the free body diagram. And we've got the our distance. So how do we go about finding what this electric force is? Well, the reason that this electric force doesn't just simply push this thing off is because there is a component in the tension force that basically keeps it and perfectly balances it. This thing is an equilibrium. So remember that equilibrium condition When things are in equilibrium, it means that the sum of all forces in X and Y are equal to zero. All the forces basically cancel out. So you've got these components of these forces the tension in the X direction and the tension in the Y direction. And those things were going to cancel out with M G and the electric force. So let's go ahead and set up our conditions for equilibrium. Let me go ahead and make some room down here. Okay, So I've got to the in the X direction. Actually, we go ahead and do that in black. So I've got in the X direction I've got the sum of all forces should equal zero. What are the two forces acting on it? Well, if I just go ahead and pick this direction to be positive, then I've got the electric force minus the tension that in the story the tension component or the X component of the tension force which, if you guys remember, that's going to be tension times the cosine of the angle. Remember that T X is going to be t co sign of data. If our angle is relative to the X axis and t y is going to be t sign of data if it's relative to the X axis, Right, So we've got t cosine. Theta is equal to zero. Sorry, f e minus T coastline data is equal to zero. So if I figured out, basically the electric force is just going to be equal to the tension divided or sorry times the cosine of the angle theta. Now the problem is, I don't know what this tension force is. And whenever I have an unknown in the X direction, I have to go to the Y direction and solve for that. So in the Y direction Now I have to set all the force is equal to zero in the Y direction. Now I've got the tension of the Y direction is t sign of data and that's going to be minus M G is equal to zero. So if we move this sign of data over to the other side and also move the MG, what we're gonna get is that T is equal to m G divided by sine of theta. Right? So we have t equals that now he could actually go ahead and sulfur this. We've got the 50 g which we're told that's equal to 0. kg. Remember, Everything has to be an s I times 9.8 divided by the sine of the angle. And the angle here is just 60 degrees. So go ahead and plug that stuff and you shoot attention of zero point. Uh, let's see, I got 57 Newton's so now we can go ahead and Dio is we can plug this tension in that we figured out the reason we came to the Y direction is to solve for what? This tension was over here. So now we can just keep on going and figure out what the electric force is. So we've got the electric force is just equal to 0. times the cosine of 60 degrees and that's just equal to 0. 29. Great. So now that we have this theoretic force right here, we can go back to the original expression that we had up top and basically solve for what that charge is. And I could basically plug that in for the electric force. So finishing off and plugging everything in we've got to Q is equal to the square roots. This our distance, remember here is just 05 So you've got 05 squared and then you've got the electric force, which is 0.29 divided by this K constant 8.99 times 10 to the ninth. If you go ahead and work this out, you're gonna get Q is equal to 2.84 times 10 to the minus six columns where this 10 to the minus six just has a symbol, micro. Or that's that Greek letter mu. So you could also say this is to 84 micro columns and both of those things would be perfectly fine. You get full credit for both of those answers. Alright, guys, let me know if you guys have any questions with this

Additional resources for Coulomb's Law (Electric Force)

PRACTICE PROBLEMS AND ACTIVITIES (8)

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