INT FIGURE EX23.25 shows a 1.5 g ball hanging from a string insid...

Question

INT FIGURE EX23.25 shows a 1.5 g ball hanging from a string inside a parallel-plate capacitor made with 12 cm×12 cm electrodes. The electrodes are charged to ±75 nC . What is the charge on the ball in nC?

Accepted Answer

Hello everyone. Let's go through this practice problem. A parallel plate capacitor has plates with dimensions 15 centimeters by 15 centimeters and a charge of 60 Nano Coomes as depicted in the figure, a 2.5 g cube hangs freely within the capacitor and is attracted to one of the plates. Determine the charge on the cube. We are given a diagram showing the system and four multiple choice options. Option A 1.9 multiplied by 10 to the power of negative four columns. Option B 1.9 multiplied by 10 to the power of negative seven columns. Option C three multiplied by 10 to the power of negative eight columns and option D 2.9 multiplied by 10 to the power of negative nine columns. So the first thing you might want to notice is that this diagram clearly shows that the cube is being pulled towards the negative plate of the capacitor due to the law of charges that opposites attract. That must mean that the cube has a positive charge. So whatever our answer is, it's going to be positive though this doesn't actually narrow down our options all that much since all four of our multiple choice options are written as positive in order to find the actual magnitude of this charge, we're going to need to analyze in more depth all the different forces acting on the cube. So as you might have guessed, we're going to start this problem off by writing a free body diagram for the cube. So I'm going to draw a crude little diagram over here. And so some of the forces acting on the cube. First off there is the tension from the string that is acting in a kind of diagonal direction. So I'm going to draw it like this where we have the tension force acting up into the right. That is simply how I'm drawing the diagram. And by the way, I'm going to define the coordinate system in the usual fashion. And because this tension forces at an angle, the diagram gives us that angle as kind of looking like this. It is going to be helpful for us when we're writing our force equations to have it broken up into its components. So I'm going to, I'm gonna add to this drawing notes of the vertical and horizontal components. So this is the horizontal component which this angle here is going to be the same value theta due to the fact that they are interior opposite angles. And so just due to the geometry of the situation, this vertical component is going to have a magnitude of t multiplied by the cosine of theta. I'm going to do a similar thing for the horizontal component. It's going to be directed toward the right, of course, and is going to have a magnitude of T multiplied by the sign of theta. Again, this is just based on the geometry of the situation. That's not the tension from the string isn't the only force acting on the cube. Of course, there's also going to be a weight that is always acting downwards and it always has a magnitude of M multiplied by G, the mass of the cube multiplied by the gravitational acceleration. And lastly, there is a force acting towards the left and it's the electrical force from the capacitor. It's the force pulling the cube to the side in the first place. So I'm calling it F sub E. And before we move on with our force equations, I want to establish real quick what the electrical force is because there isn't there, there is unlike for the weight, the electrical force isn't something that always has a super simple formula for especially in a case like this where we aren't given things like the magnitude of the electric field strength. So generally speaking, an electric force due to an electric field is equal to the magnitude of the charge of the object within that field. In this case, the charge on the block multiplied by the magnitude of the electric field strength. So Q multiplied by E that said however, nothing in the problem gives us E. Now, when we have a parallel plate capacitor, there are a few different formulas for the electric field strength that we can use. And the one that seems most relevant here is the formula that tells us that the electric field strength is equal to the electric charge density on either of the capacitors divided by the electric constant epsilon knot. So we can expand this formula into something that works better for us as Q multiplied by the electric charge density sigma divided by the electro constant epsilon knot. We're not given the electric charge density of the, of the uh capacitor plates, but we are given the total amount of charge as well as their dimensions. So we can use the fact that an electric charge density is defined as an amount of charge using a capital Q to distinguish it from the charge on the cube divided by the area of the capacitor plate. So divided by the area of one plate or alternatively divided by the square of the side length. Since in our case, the side lengths are the same 15 centimeters. So really the most complete way we can write out the electric field strength is as the charge multiply the charge on the cube multiplied by the charge on the plate divided by epsilon knot multiplied by L squared. Or again, just to be clear here, I am defining L as the side length of the cube of the, of the plate. Now moving on, let's discuss, let's write some force equations for the free body diagram that we established since that will allow us to have a relationship between all the different forces. And it should be noted that the problem specifically states that the cube is hanging freely, which means that it is currently in its hanging position at rest, which means the net four senate right now is zero. First, let's discuss the horizontal forces or net force or the net of the X forces. Again, this is going to be equal to zero because of the fact that the block is hanging freely with all the forces on it being balanced. And now let's consult our diagram to see what these forces are well toward the left here. In the negative X direction, we have the electric force, this one right here and in the positive direction, we have t multiplied by sine theta. So that means that in the X direction we have, I'm gonna say I'm gonna write negative of the electric force since it's in the negative direction. And then plus the X component of the tension force or T sine theta. Now, because this force equation is equal to zero, we can add the electric force to both sides to make it positive. So that we'll have a formula just equating two forces with each other. And let's also plug in the large formula we found earlier. In for F sub E. So what we find is that the charge in the cube multiplied by the charge of the plate divided by epsilon, not L squared is equal to T sine theta. Now let's move on and write out a force equation for the Y axis. So the net force in the Y direction, once again, the net force is going to be zero because it's at rest. And if we look at our diagram, we can see that we have the weight of the cube pointing downwards so negative MG and then the positive direction, we have the Y component of the tension pulling upwards which is T coast data. So once again, if we set these forces equal to each other, add MG to both sides of the equation, we find that MG is equal to T cos theta. Now we do not know the value of T, the tension in the string. But we want to solve for Q and equate these equations together in order to do so. So the way I'm going to do this is I'm going to divide this top equation here by the lower equation to combine them effectively. And so doing that algebraically, the left hand side of our new equation becomes small, Q multiplied by the capital Q divided by the epsilon knot, MGL squared. And then this is equal to T sine theta divided by T Cota. The T is the tensions will of course cancel out and then sine theta divided by cosine theta is of course equal to the tangent of theta. So what we're left with now is a single equation in which everything is known, except for small Q, the charge on the cube that we're trying to find. So all that's left for us to do is multiply both sides of the equation by epsilon dot mgl squared and divide both sides of the equation by big Q that we get a formula for a little Q all by itself. And what we find is a formula saying that small Q is equal to epsilon knot MGL squared multiplied by 10 G of theta all divided by big queue, the charge on the capacitor plate. And this is really it, all that's up for us to do is take all the knowns given to us and plug them into this equation. So let's do that now. So small Q is equal to epsilon knot 8. multiplied by 10. It raises the power of negative 12 plum squared per Newton meters squared. This is gonna get pretty big. Actually, I'm gonna have to kind of cram these in there multiplied by the mass of the cube which is given to us as 2.5 g or 2.5 multiplied by 10 to the power of negative 3 kg multiplied by the gravitational constant 9.81 m per second squared multiplied by the square of the side length of the capacitor or 0.15 m squared multiplied by the tangent of the angle, which is 20 degrees. So make sure your calculator is in degrees mode for that part. And we divide all of this by the charge on one of the capacitor plates which is given as 60 Nano Coomes or 60 multiplied by 10 rate the power of negative nine columns. So once you manage to type all of that into your calculator, we find a charge on the cube of approximately 3.0 multiplied by 10 to the power of negative eight columns. And if you plug that and that's what you get. And so this is the answer to a problem. And if you look at the multiple choice options, we can see that this agrees with option C so option C is the answer to this problem and that is really it for this problem. I hope this video helped you out if it did and you need more practice. Please consider checking out some of our other tutoring videos which will give you more experience with these types of problems and that's all for now. And I hope you all have a lovely day. Bye bye.

INT FIGURE EX23.25 shows a 1.5 g ball hanging from a string inside a parallel-plate capacitor made with 12 cm×12 cm electrodes. The electrodes are charged to ±75 nC . What is the charge on the ball in nC?

Key Concepts

Electric Field

Coulomb's Law

Charge Conservation

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