Point Charge Inside Capacitor

by Patrick Ford
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Okay, guys. So this problem, this example? Problem has a bit of parts to it, so we're gonna work it out together. So you've got these two parallel plates and I'm gonna draw them out like that. Now I know what the charge is on either of those plates. Both of these plates are charged to 30 nano columns, and that's 30 times 10 to the minus nine columns. I also know that the plate separation between these to positive and negatively charged plates is equal to 10 millimeters. But I need that in the right units. So I have to convert that. That's gonna be 0.1 in meters. Okay. The first thing was supposed to dio is figure out what the voltage is or the potential difference between the plates. So, for part A, we're gonna be looking for what variable is potential difference in voltage? That's V. So we can use V is equal to let's say we have a couple of different ways that we can use this. The easiest approach is gonna is gonna be using Q equals C. V. Right. So this is our approach. So that means that V is equal to the charge divided by the capacitance. Now we have the charge that's accumulated in both of the plates. The one thing we need to find is the capacitance, so we can use that. The other relationship between, uh, the capacitance in a parallel plate capacitor is epsilon, not times a over D. Now what happens is this capacities right here is in the denominator. So what happens is when you divide everything over, that means that these two things in the numerator are actually gonna go on the bottom Epsilon, not a. And then this q is gonna be multiplied by this D right here. So you're gonna have Q times d over Epsilon, not times a right. So it's just because this thing is the denominator, you can work it out yourself. So you got the potential is equal to the charge, which is 30 times 10 to the minus ninth. Now, the plate separation is 100.1 which, by the way, is just this distance right here between the plates and now we've got the epsilon knots. So we've got that epsilon not constant. 8.85 times 10 to the minus 12. Now we have to figure out what the area is now. The area of these 21 centimeter by one centimeter plates is just going to be multiplying those to distance together. But we need to put them in the right units. First, one centimeter is gonna be 10.1 times 0.1 So that means it's gonna be one times 10 to the minus four. That's gonna be in Meter Square. So that's what we actually have to plug into our calculators. So one times 10 to the minus four and then you plug all of that good stuff into your calculator and you should get a voltage off 3.39 times 10 to the fifth. And that's involves. All right, so that's the answer to the first part. Cool. The second part now says, What is the electric field in between the plates? Now, again, we have a couple of different formulas for that. But the easiest way to figure out the electric field is by relating it back to the voltage. We know that E is equal to the voltage divided by the distance, and we actually have what both of those things are now. if you wanted to use the other equation. So if you wanted to actually use this, um, this Q over Epsilon, not times A That also would have been perfectly valid to do that. So you're more than welcome to do that on your own. So just go ahead and plug in whatever whatever form you want to do this in. So I decided to go with the voltage divided by the distance. So in other words, I have the electric field is the voltage 3 39 times 10 to the fifth, divided by the distance 0.1 And you should get 3.39 times to the seventh. And that's gonna be Newtons per cool. Um, so that's the answer. Hopes. My pen is not working today. So that's the answer to part B. Now, this third part here is the interesting one. How much energy will it take to move a charge? A negative five Coolum charge from the positive plate over to the negative plate. So we're just gonna go ahead and assume that this plate is positive. That's negative or positive. Q. This is gonna be the negatively charged plates, and we know that the potential difference the Delta V is gonna be final minus initial. So what happens if we were to take a negative charge and move it across this distance right here? How much energy? What it takes to do that? Let's think about it. We're taking a charge, which is Q and removing it across the potential difference. So what formula is gonna relate that to the energy? Well, that's gonna be the work formula. So the work is going to be equal to the negative Q times Delta V. We just have to teach. Have to figure out what our final minus initial is Now if our we're going from the positive to the negative plates in this Delta V is actually gonna be minus the potential difference which we know is 3.39 times 10 to the fifth. So this is actually what are potential difference is just in this specific case, we're removing a charge from here to here. If we were told it was going from the negative to the positive, then it would have been a positive potential difference. Okay, so make sure that you have the sign correctly on that So we're just gonna plug that formula down into here. We're gonna get the work that's done. All right, so we've got the work is equal to negative. Now, I've got negative five times. 10 to the minus nine. And that's gonna be Nano columns, right? Yes. Five times 10 to the minus nine. And now times the potential difference. Negative. 3.39 times. 10 to the fifth. So now what happens is two of the negative will cancel out, but one of them will still remain. Which means that the work that's done is equal to negative 1.7 times 10 to the and let's see, I've got minus three, and that's in jewels. So the fact that this number right here is negative means that we're taking a charge which is on a positive plates and removing it to where it doesn't want to go. We're moving a negative charge to a higher to a lower potential, which means that we actually have to give this this'll system some energy. So that means that this work is work that's done on the system. And this work is done on the system by you. So you actually have to do the work in order to move this charge from the positive plate over to the negative plate because you're going against what it will naturally would wanna dio. Okay, so these are the answers to the questions. Let me know if you guys have any questions or concerns and I'll see you guys the next one.