Hey, guys, let's get more practice involving electric fields. So we have a charge that's at the end of a pendulum. And this pendulum, we're told, is in equilibrium and suspended inside of a uniform electric field. We are told what the electric field magnitude is and our job is to figure out what is the charge on the end of this pendulum? There are a couple of important things to pick out from this problem. The fact that it's in equilibrium means that the sum of all forces in the X and Y directions is equal to zero. All the forces basically cancel out. We're going to see how we can use the forces to solve this problem. But if we're solving for the charge, Q, there's basically one of two equations that we can use. We can use Coulomb's Law, but we don't have another charge, right? There's no second charge over here. So instead, what we're going to do is we're going to use the electric force is equal to Q times E, the charge times the electric field, which we actually have. So, we're going to use this formula to solve for Q. If we go and rearrange this, the Q is just equal to F divided by the electric field. If I can figure out what this electric force is, then I could basically just plug that in and solve for the charge.
To solve that force, there are a bunch of forces acting on this, so let's go ahead and draw a free body diagram for this. We have an electric field on a point charge, so we know that the force is going to be in the same direction. The reason that it doesn't just fly off is that it's held in place partly by this string that it's attached to, which has some tension. There is also another force due to gravity, because this thing has some mass, mg. Since this object is in equilibrium, which means that all the forces are going to cancel out. In order to set this up first, we have to break up this tension into its components. So we have an X and a Y component, and we get those components Ty and Tx by relating the tension with the cosine or the angle θ that it makes with the X axis. Now, this angle, if you sort of project out this horizontal line, this is a 90-degree angle, this is a 15-degree angle, which means this has to make up the difference between 180 degrees. So this angle is actually 75 degrees. So, the tension in the X direction is going to be T cos(θ) and the T in the Y direction is going to be T sin(θ).
Now we have these components, let's go ahead and set up our equilibrium conditions in the X direction. I have that the sum of all forces in the X direction is equal to zero. So, starting off my forces in the X direction, I have the electric force, which is really what I'm looking for, minus the tension or the X component of the tension. In other words, the electric force here is just equal to the tension times the cosine of the angle. The problem is, I know what the cos(θ) is, I don't know what the tension is. So remember, think back to Newton's laws, whenever we had a situation where we needed one variable in the X direction, we usually went to the Y direction to solve for it. So, we are going to do that. In the Y direction, we have the same exact condition, the sum of all forces in the Y direction is equal to zero. Your Ty, which is going to be your T sin(θ), is equal to mg once you move this to the other side. So now, T is actually equal to mg divided by the sin(θ), and I actually have all these numbers: the mass, g and also sin(θ). So, T equals mg/sin(θ) gives you 0.30 Newtons. Now you can just take this number and then plug it back inside of this formula. The electric force is just equal to 0.30 Newtons times the cosine of the angle, which is 75 degrees. So that means that my electric force is equal to 0.78 Newtons. So now what I can do is I can take this electric force and I just have to plug it back into my initial equation to solve for the charge. Remember, that's my final variable. So, Q is equal to 0.78 divided by 100 Newtons per Coulomb. You should get 7.8 times 10-4 Coulombs. This is, by the way, our final answer.
Let me know what you think about this problem. Let me know if you have any questions. I'll happily explain any steps, and I'll see you guys in the next one.