Coulomb's Law (Electric Force) - Online Tutor, Practice Problems & Exam Prep

Coulomb's law describes the electric force between two charges, which can be attractive or repulsive based on their nature. The formula is given by $F^e=K{Q}^1{Q}^2/r^2$, where K is Coulomb's constant (8.99 x 10⁹ N m²/C²). The direction of the force is along the line connecting the charges, with like charges repelling and unlike charges attracting. Understanding this principle is crucial for analyzing electric interactions in various contexts.

Coulomb's Law

Hey, guys. So in this video, we're going to talk about Coulomb's law, which is a really, really important law that you need to know for electricity. It basically gives us the electric force between two charges. So go ahead and watch this video as many times as you need to. We're going to be covering a lot of examples and practice problems in the videos after this. Basically, electric forces can be attractive or repulsive, and that's a direct consequence of what we talked about with charges: unlike charges attract and like charges repel. So, unlike charges, if you have positive, negative, or negative, positive, will exert attractive forces on each other. Or if you have two like charges, like two protons or two electrons, 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 Coulomb's Law or the Coulomb force, and that gives us the force between two charges. So if you have two charges Q1 and Q2, and they're separated by some distance, little r, then the force that exerts between them is going to be:

K Q1Q2 r 2

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, in Coulomb's law, the electric force is just some constant times the two charges divided by the distance between them. So this K constant right here is called Coulomb's constant, and that has a number: 8.99 × 10⁹. It's really easy to remember, because it's like 899 times 10⁹. 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 are some units associated with that. It's Newtons meters squared per Coulomb 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 a line that connects the two charges. And basically, what I mean by that is if you have these two charges Q1 and Q2, and you know the distance between them, that's little r, then if it's a repulsive force from Q1 to Q2, that's going to go in this direction, so that's going to be a repulsive force. And if it's an attractive force, then it's just going to point in the opposite direction. So that's going to be, uh, yeah, that's attractive. And, um, yeah, so it always just exerts along the line that connects those two things. And again, the attractive or repulsion just has to do with whether the fact they are like or unlike charges; like repel, unlike attract.

So I'm going to give you guys a pro tip in order to figure out the magnitude and direction. So, whenever we are trying to figure out Coulomb's law, we're always going to find the magnitude of the Coulomb 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 into this formula that we've given for the Coulomb force, you're always just going to use positive numbers and then worry about the direction later.

Alright, so let's go ahead and take a look at a quick example. In this problem, we're going to 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 are electric forces because these 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 are just related to the elementary charge. So this is +e and this is -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 going to be, let's see, we've got K, that constant, times the product of the two charges. So we've got:

K QprotonQelectron r 2

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 × 10⁹. Remember that. Now we've got the elementary charge. That's also something you should know, 1.6 × 10^-19. And now that's for the proton. For the electron, it should be negative, but again, we're just going to 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 we are asked to find the ratio of like, the magnitudes of these forces anyway, so we're just going to use positive numbers.

Alright, so we've got the distance 5.3 × 10^-11, and that is going to be squared. So you go ahead and work this out, you should get 8.19 × 10^-8. That's in Newtons. So that's the electric force.

Now we just have to do the same exact thing for the gravitational force. So now, the gravitational force, well, just in case you have forgotten the gravitational forces:

G mprotonmelectron r 2

So we've actually have all of these constants over here. So I've got 6.67 × 10^-11. That's the gravitational constant. The mass of the proton, 1.67 × 10^-27. And then all of that's in SI, by the way, then 9.11 × 10^-31. And then we've got the same distance between them, 5.3 × 10^-11, squared. So we work this out, and you should get 3.61 × 10^-47, which is a very, very, very tiny number. So we're just going to see how tiny that is in a second.

So we've got the ratio of these things. The gravitational or to the electric and the gravitational force. That's just going to be:

8.19 × 10 ^ −8 / 3.61 × 10 ^ −47

If you work this out, you've actually plugging those numbers and divide them. You should get 2.27 × 10³⁹. And by the way, this is just a dimension issue. No number because we're trying to basically figure out how much stronger 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. The electric force is a very, very strong force, and this is our final answer. So again, there are no units.

Alright, let's go ahead and take a look at another example here. So we've got two identical charges, and they're connected by a five-centimeter 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 Q1 is equal to Q2, so that means that we can just use Q in our equation and not have to worry about Q1 Q2. They're the same exact thing. So if we're trying to figure out what the charges are, then we just start off with Coulomb's law. So in other words, Coulomb's law is saying that we've got K times Q1 Q2 divided by r squared. But again, if these two things are the same thing, this is just going to 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. Alright. 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 Q1 is trying to push away Q2, and Q2 is trying to push away Q1, 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 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.

Alright, so let's see, we've got, let's set up the equation If I wanted to Q squared. So I just have to do:

Q2/r2

Now I'm just going to isolate Q squared by moving this over and then moving the K to the bottom. So we're going to get that FE, the electric force times r squared, divided by K is equal to Q squared, alright. 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 realize that this is in five centimeters, so I have to convert it to SI first. So I mean, I have that r, this distance right here, is equal to 0.05 m. So I've got that, 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 0.05, I've got to square that divided by 8.99 × 10⁹. 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 take the square root of this thing first. So all you have to do is for this number, you have to take the square root of that. And that should equal Q. So I'm going to write that you're going to 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 × 10^-6. And that's in Coulombs right there. So that is the answer. Let me know if you guys have any questions with this.

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

Charges In A Line (Find Zero Force)

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 Coulomb charge so that the force on it is equal to zero. So what does that mean? We're just going to assume that the net force we're asking for needs to be equal to zero, because we have two Coulombs, or, sorry, we have two charges on the outsides. So what this problem is saying is that we need to put a one Coulomb charge somewhere around here in order for the forces on it to cancel out. 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 forces to cancel out.

Let's take a look at the left side. So imagine I dropped a one Coulomb 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 Coulomb charge. Remember, all of these things are positive, so they're all going to repel. They're going to have repulsive forces on each other. This thing has to move to the left, right? So this is the force from the two Coulomb charge that's going to point to the left. But the force from the three Coulomb 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 writing here that these things never cancel. There's never going to be a situation in which one will point in one way and 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 situation on the right. Who got this one Coulomb charge here, and you do the same thing from the three Coulomb charge. You have a force pointing in this direction. The two Coulomb 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.

The reason for that is if you work out the forces here, the force from the two Coulomb charges is going to point in this direction, and the force from the three Coulomb 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, then this stronger charge here would produce a stronger force. So this thing actually needs to be slightly closer to the left so that 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 Coulomb force to be equal to the magnitude of that three Coulomb force or the forces from those two charges. In other words, this is the condition that we need to solve, right? They're going to be pointing in opposite directions. But if their magnitudes are the same, they're going to cancel out.

So let's go ahead and write out the expressions for those two. F 2 for the two Coulomb charge, where I need a distance between these two charges. By the way, this is a one Coulomb test charge that we're going to drop right here. So I'm going to call this distance here, \( X \). And that means that if this distance between the two and the three Coulomb charge is, \( R \) so this is \( R = 10 \) centimeters, then this is just \( R - X \), that's the distance between \( R \) minus that little chunk of \( X \), \( R - X \). So I've got the \( K \) constant times the two charges that are involved. In other words, this is going to be the two-column charge and the one-column charge divided by the distance between them. That's \( X^2 \). Now I set up the equation for \( F3 \). So that's this guy right here between the one Coulomb charge. So \( K \) and then I've got the three and one. So that's three Coulombs one Coulomb divided by the distance between them squared \( R - X \) 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 set those things equal. K 2 1 / X 2 = K 3 1 / ( R - X ) 2 So this thing is an equal sign, so we can actually cancel out anything that appears on both sides of the equation. We're going to get the \( K \)s will cancel, and also the ones will cancel. I mean, one doesn't really do anything, so I mean just cancel it anyway. And so we come up with an easier expression. Now, what I'm trying to solve is where I need to put this one Coulomb charge. So that's a distance. So in other words, I'm going to go ahead and solve for \( X \). \( X \) 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 \( R - X \) is and refine. 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 to get all the things involving \( X \) to the one side. So, I'm going to move this \( R - X \) over to the other side, cross multiply, and then this two needs

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; 3.8×10¹⁴ m/s²
• B. left; 6.2×10¹³ m/s²
• C. left; 6.2×10⁹ m/s²
• D. right; 1.1×10¹⁵ m/s²

Charges In A Triangle (Rank Force Pairs)

Hey, guys. So in this problem, we have to rank all the possible pairs of charges in this diagram by which one has the greatest possible electric force. So we're talking about Coulomb's Law and specifically for point charges. So we've got to use Coulomb's Law, KQ1Q2/d2, where d represents the distance between the charges. 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 going to be a factor in determining which one is the strongest and which one is the weakest. Instead, what we have to do is just look at the product of the charges themselves, and we have three pairs that we're going to discuss. We're going to discuss the pairs between charges 2 and 3, 2 and e, and 3 and e. So I've got these little loops right here that represent those pairs, and all we have to do is just compare the magnitude of the charges involved. Now, all these are like charges that all exert repulsive forces on each other, so we just have to figure what Q1Q2 is. Now, I've got 2e and e, and remember that e stands for the elementary charge, so I've got 2e∙e. So, in other words, that's just going to be 2e2. Remember that e is just 1.6×10-19, which represents the electron charge. And if I do the same exact thing over here, I've got Q1Q2 equal to 2e∙3, and that's just going to equal 6e2. And for the last one, I've got e∙3, so that's just going to be 3e2, calculating Q1Q2. If we're actually trying to figure out what the forces are and considering Q1Q2, the ranking from greatest to least is: first is 6e2, followed by 3e2, and finally 2e2. That's the ranking in terms of descending greatest electric force. Alright, guys, let me know if you have any questions with that.

Charges in a Plane

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 Coulomb charge is. So let's go ahead and first draw. What? The forces that are acting on this three Coulomb charge. So we got We're focusing on this guy right here. And the first is that he feels an attractive force from this negative two Coulomb charge. Why is it attractive? Because these things are unlike charges, one's positive, one's negative. So this is going to point in this direction, and I'm going to call this F2 because it comes from the two Coulomb charge.

Now, there's also this one Coulomb charge here on the bottom left corner, that is some distance R away from this other charge, and it's going to exert a repulsive force on the three Coulomb charge because these things are like charges. They're both positive. I'm going to call that F1, and what we're supposed to do is take these 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 label and calculate these forces, which we've labeled already. So I'm going to take that over here, and then we have to decompose them, pick our directions, and add them in this all for the net force.

Alright, so let's go ahead and calculate the forces. So I got F2 and I'm going to be using Coulomb's Law. Coulomb's Law is just k times the Q1 Q2 divided by the distance between them squared. So that means that F2, which is the force between the negative two Coulombs and the three Coulombs, is going to be k which is that constant 8.99 × 10⁹. 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, whenever we're finding the Coulomb force, we're always just going to plug in positive numbers. We're going to worry about the direction later. So if you work this out, you should get the F2 is equal to 1.5 × 10¹³. And that's in Newtons right here. Right?

So you don't need the units for that if we try to work out F1. So in order to work out F1, we need 8.99 × 10⁹. We've got the two charges, right? You've got this three Coulomb and the one Coulomb. but we don't have the center of mass or sorry. We don't have the distance between them. That little R. So hopefully you guys realize that this is a 6-8-10 triangle. Right? So this should be 10. But if you didn't, you could always just use the Pythagorean theorem. So you've got 6 squared plus 8 squared, and that equals 10 centimeters. Alright, so we want this in SI this actually should be 0.1 m. So first we have to go. We have to do three Coulombs, one Coulomb, and then divided by 0.1 square. And if you work this out, you should get 2.7 × 10¹². 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 F1 vector by basically moving it and projecting it down to the X and Y axis. And I do this by needing by using my trigonometry and my cosines and things like that. Right? So, in other words, I need my F1 X components and my F1. Why components. Now I've got this. Um, I know if I wanted to break up this F1 into its X and Y components, I know I relate that using the sine and cosine. So F1 is going to be cosine F1 X is going to be F1 cosine of theta and F1. Why is going to be F1 sine of theta? So I know what these forces are all I 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 theta right here, which, by the way, is the same as this angle theta right. These things are parallel lines. This thing is a parallel horizontal lines, so could basically complete this little triangle. And I know that this is 8 centimeters. So I've got 6-8 and I know the hypotenuse of this triangle is 10. So that means Well, let's use our sine and cosine rules. Right? The cosine of anything is just adjacent over hypotenuse. So, in other words, cosine of this angle right here is the adjacent side over the hypotenuse. That's just 6/10. And if I did the same thing for sine using so gotta I've got opposite over hypotenuse. 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 these cosines and sines with these fractions right here that we figured out it's the same exact thing. So let's keep going. We've got F1 X is just going to be F1, which is 2.7 × 10¹², and I have to multiply by cosine of theta, 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 × 10¹². Now, the Y component is going to be very similar to 0.7 × 10 of the 12 Now, instead of 6/10 you do 8/10 for a sign. So if you work this out, you're going to get 2.16 × 10¹².

Alright, 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 going to have the F2 and F1, and I add them together to figure out what the Net Force is these are my X and Y components. Now, if you remember, F2 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 × 10¹³ and a Y component of zero. And now the F1 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 × 10¹² and then 2.16 × 10¹². 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 process 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 × 10¹³ because it's actually one power higher. Let me just go ahead and make sure you can, As you can see, that that's going to be 13 and then this is just gonna be 2.16 × 10¹². So if I actually go back into the diagram, I can actually draw out what that vector is going to look like. It's going to look like this is going to have a high negative x component, and then it's going to have a positive Why components gonna look something like that? Because this F2 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 Pythagorean theorem. So I've got negative 1.34 × 10 to the squared. Plus, Now I've got 2.16 × 10¹², 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 × 10¹³ 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

Exploiting Symmetry

Hey, guys. So in this video, we're going to cover something really, really important that's going to help you in a lot of problems called exploiting symmetry. And it happens when you have these sorts of arrangements of charges. Let's go ahead and check out two examples. So in this example, we've got two Coulomb charges stationed away from this one Coulomb charge. And we want to figure out just the direction, not the magnitude of the net force on this thing.

Alright, let's go ahead and work out what the forces acting on the one Coulomb charge are. There's a repulsive force coming from the two Coulomb charge, and there's also a repulsive force due to the other two Coulomb 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 want to exploit something called symmetry. Notice how these two Coulomb charges are both positioned the same distance away, so you usually have symmetry. Whenever you have the same cues, the same charges positioned the same distance away. So you have distance or R; 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 Fs here are also the same. So, in order to figure out the net force, we'd have to decompose them into their components. Right? So we have an x-component and we have a y-component over here, and we do that by using our trig—our sines and cosines 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 x-component, there is an angle theta, and there is a y-component. So what happens is if you have the same Fs and these things are symmetrically placed, then you're going to have the same exact angle theta. And in this situation, what's going to happen is that the opposite components here are going to 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 y-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 on 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 y-components here.

Alright, so that's one way you can use symmetry. Let's see another example. So in this situation, it's going to be very similar, except instead of both of them being positive charges, we have one positive and one negative. Okay, so we know there's a repulsive force from this positive two Coulomb charge. But now this negative two Coulomb charge over here is going to exert an attractive force because these things are unlike charges, so we're going to have an attractive force that points off in this direction. Now, these are going to have these are going to be F and F over here. We've got the same exact distance and the same exact charges. So that means these forces are going to be the same. So we've got the same cues. We've got the same distance, so that means the forces are going to be the same. And these things are positioned sort of in a symmetrical way, which means we're going to have the same exact theta. So, if we start breaking these things up into their components, right, we've got the y-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 going to have the same exact angle theta. So it's going to 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 going to cancel out. And now the net force over here is going to be in the purely X directions. The net force is going to be two times the x-component of both of those forces together. Alright, so this is just the direction. So this is a really, really important thing that we're going to use. See if you can exploit some symmetry in your problems. Sometimes you'll get like squares with a whole bunch of charges. You'll have to figure out what the directions of the net force. It's a very important shortcut. You guys need to know. Let me know if you have any questions.

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

Electroscope (Find Charge)

Welcome back, guys. We've got a fun example involving an electroscope. If you haven't seen them before, maybe those of you who are taking labs in physics will 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, 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 an electroscope is. We're told the mass of each leaf in this electroscope, and we're also told the deflection angles to 30 degrees. We're supposed to figure out what the charge on the end of each leaf is.

So what I want to point out first is that we're dealing with two identical charges, which means Q₁ and Q₂ are going to be the same. That actually helps us simplify things because we know that if Q₁ is equal to Q₂, then we could just replace them both with just Q. And we also know that these things are going to repel from each other because they're like charges. So that means that the force right here is going to be out in this direction. That's the electric force and these things are going to basically push off from each other, so they're just going to point off in opposite directions. Now, the reason they don't fly off is because they're connected sort of by the strings here. So we're going to see that in a little bit, how that plays out.

So, basically, if we wanted to know what the charge of each leaf is, we have to use Coulomb's law, right? We have to use the force between two point charges. That's going to be Q₁Q₂ over r². Remember, no Q₁Q₂ because these things are the same thing. So basically, this just turns into KQ2/r2, and we're going to be solving for these Q² over here. So basically, all I have to do is move this r² up to the top, move the K down to the bottom, and I'm just going to raise that. Basically, what this becomes is RFe/K=Q2. So all we have to do is just take the square roots. So we get Q is equal to the square roots of RFe/K.

So if we look at this equation, this is really all we need to solve. But we don't know what the distance between these two charges is. So in other words, that's this distance right over here, which I'll call little r. 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. How do I figure out r? 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 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, then this whole entire thing is just an equilateral triangle. All the sides have to be equal to each other. So that means that this little r is going to be 0.5.

Let me know if that makes sense for you guys. So basically, we now know what this little r distance is. Now we have to just figure out what the electric force is. How do we do that? Remember, there are multiple forces involved here, so let's go ahead and draw a free body diagram for each object. 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 going to 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 simplicity, I'm just going to work with this free body diagram here. This one's going to be the exact opposite. Just everything's going to be reversed.

Okay, that's the free body diagram. And we've got the r 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 in 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 are going to cancel out with mg 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.

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 binary intelligence 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 cosine of theta, if our angle is relative to the X-axis, and T_Y is going to be t sine of theta 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 cosine theta 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 forces equal to zero in the Y direction. Now I've got the tension of the Y direction is t sine of theta and that's going to be minus mg is equal to zero. So if we move this sine of theta over to the other side and also move the mg, what we're going to get is that t is equal to mg divided by sine of theta. Right? So we have t equals that, now we can actually go ahead and solve for this. We've got the mg, which we're told that's equal to 0.0 kg. Remember, everything has to be in SI 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 in, you get a tension of 0.057 Newtons.

So now we can go ahead and do it. 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 0.057 times the cosine of 60 degrees, and that's just equal to 0.0285. Great. So now that we have this electric force right here, we can go back to the original expression that we had at the 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 Q is equal to the square roots of this r distance, remember here is just 0.5. So you've got 0.5 squared, and then you've got the electric force, which is 0.0285 divided by this K constant, 8.99 x 10⁹. If you go ahead and work this out, you're going to get Q is equal to 2.84 x 10⁻⁶ Coulombs or this 10⁻⁶ just has a symbol, micro, or that's that Greek letter mu. So you could also say this is 2.84 micro Coulombs, 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 have any questions with this.

Here’s what students ask on this topic:

What is Coulomb's Law and how is it used to calculate the electric force between two charges?

Coulomb's Law describes the electric force between two point charges. The formula is given by:

$F_e=K{Q}_1{Q}_2/r^2$

where $K$ is Coulomb's constant (8.99 x 10⁹ N m²/C²), $Q_1$ and $Q_2$ are the magnitudes of the charges, and $r$ is the distance between them. The force can be attractive or repulsive depending on the nature of the charges: like charges repel, unlike charges attract. This law is crucial for understanding electric interactions in various contexts, such as in atomic structures and electrical circuits.

How does the electric force between two charges compare to the gravitational force between two masses?

The electric force between two charges, described by Coulomb's Law, is generally much stronger than the gravitational force between two masses, described by Newton's Law of Gravitation. The formula for the electric force is:

$F_e=K{Q}_1{Q}_2/r^2$

while the gravitational force is given by:

$F_g=G{m}_1{m}_2/r^2$

where $G$ is the gravitational constant (6.67 x 10^-11 N m²/kg²). In a hydrogen atom, for example, the electric force is approximately 10³⁹ times stronger than the gravitational force, highlighting the dominance of electric interactions at the atomic scale.

What is Coulomb's constant and why is it important?

Coulomb's constant, denoted as $K$, is a proportionality factor in Coulomb's Law that quantifies the electric force between two point charges. Its value is approximately 8.99 x 10⁹ N m²/C². This constant is crucial because it allows us to calculate the magnitude of the electric force between charges accurately. Understanding and memorizing this constant is important for solving problems related to electric forces in physics, as it often appears in equations and calculations involving electric interactions.

How do you determine the direction of the electric force between two charges?

The direction of the electric force between two charges depends on whether the charges are like or unlike. Like charges (both positive or both negative) repel each other, so the force vectors point away from each other. Unlike charges (one positive and one negative) attract each other, so the force vectors point towards each other. The force always acts along the line connecting the two charges. To determine the direction, first calculate the magnitude of the force using Coulomb's Law, then apply the rules of attraction and repulsion to find the direction.

What is the relationship between the distance between two charges and the electric force between them?

The electric force between two charges is inversely proportional to the square of the distance between them, as described by Coulomb's Law:

$F_e=K{Q}_1{Q}_2/r^2$

As the distance $r$ between the charges increases, the force decreases rapidly. For example, if the distance is doubled, the force is reduced to one-fourth of its original value. This inverse-square relationship is crucial for understanding how electric forces behave over different distances.