Parallel Plate Capacitors - Video Tutorials & Practice Problems

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1

concept

Parallel Plate Capacitors

Video duration:

8m

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Alright, guys, we're a little bit more familiar with capacitance and capacitors. So in this video, we're gonna check out with the most common type of capacity you'll see called the parallel plate capacitor. Let's go and check it out. So we already have the relationship between the capacities. For any given capacitor, it's just C equals Q over V. But the most common type of capacity you'll see is called the parallel plate capacitor, where you have one surface of charge, and that's gonna be positive. Q. You have another surface of some negative charge that's gonna be minus. Q. And these things are separated by some distance, which will call D now. These things don't necessarily have to be rectangles. They can be circles. There could be some odd shape, but they will have an area, and that area A is gonna be the same for both of them. So if you have this kind of situation where you have a parallel plate capacitor, which you have some area and some distance, then we actually have another formula for finding out what the capacitance is. And it's epsilon, not that vacuum. Primitive ity, constant times. The area divided by the distance between them. Okay, so this is another equation to add to our tool belt. Now, if you're wondering when you're going to use one versus the other, that really all just depends on what variables you're given. But both of these equations are perfectly valid to use for parallel plate capacitors. Okay, so basically, that's going to relate to the capacitance of two plates with area A and distance d separating them. So we have the area, the D, distance and Epsilon, Not which we have this constant right here for All right. We also said before that these capacitors, these parallel plates produce an electric field in between them because the electric field lines wanna go from the positive charge over to the negative charge. Now, hopefully, guys remember this. But just in case you don't, you should remember that the electric field between the plates we said was uniforms. In other words, it was the same always everywhere. No matter where you look inside of this parallel plate capacitor, the electric field is gonna have the same magnitude. Whereas if you go outside, if you go out, if you look anywhere outside of the parallel plates. The electric field very quickly approaches zero. So we talked about them being uniforms and zero. We actually didn't know what the equation for. That is what we can actually see. What the magnitude of the equation is within a capacitor by relating it to one of the relationships that we talked about in a different video, which was that the electric field is equal to the negative change and potential divided by the change in distance. This e equals negative Delta view over Delta X. Now we're talking about the magnitude of this electric field, so that means that the negative sign congest drop. And some of you might start to see this equation written in a different way. So if you've seen in your classes or text books, it's written as e equals V, divided by D. This is just a simpler way of writing this expression where this V just represents the voltage and this distance represents the Delta X, but it's really just the same thing. But now what we can dio is we can actually take this relationship and use our new equation for the voltage which is related to the capacitance and write this in a slightly different way to find out what the electric field is. We know that this voltage is equal to Q divided by C. This is just from this equation right here. So just re arrange for V. And we know that this capacitance right here is equal to actually let me write that somewhere else. We know that this capacitance is equal to Epsilon, not times a divided by D. So if you put all of these things together right, so you multiply all of these things out, then you're gonna get is that the electric field is equal to queue divided by epsilon, not times a. So it means the electric field between a capacitor has nothing to do with the capacitance and the voltage. All it depends on is the charge and the area involved. This is actually, believe it or not, a really interesting consequence of Gaza's law. Anyway, the last thing I want to talk about is the echo potential surfaces between these plates. So I know over there I was using green lines for electric field, So I just labeled that I'm using red lines. We said that the electric field inside of a parallel plate capacitor is uniforms. We also know that the electric fields and the equal potential surfaces have to be perpendicular and that the electric field points in the direction of decreasing potential. So I'm just gonna randomly make up numbers here. Let's say this is eight volts and this is 7654 and so on and so forth. Right now, each one of these potential differences or the eco potential differences, is gonna be constants. In other words, a change from one to the other. The change in potential is the same, and they also happen across the same distance. In other words, all these Delta X is between all of these spacings of these eco potential surfaces are the same. Now we know that the relationship between the electric field is again negative. Delta V over Delta X. So what I'm saying here is that the changes in potential are constant and they happen over the same constant amount of distance. So you have the same delta over the same distance. Then that means that your electric field is going to be constant. So basically all I'm doing here is I'm just reinforcing the idea that the electric field is uniforms in between the plates and then anywhere outside of it, it's equal to zero. All right, so let's go ahead and check out this example right here off a parallel plate capacitor and go ahead and solve some problems. So you've got this parallel plate capacitor we're told with the area of this thing is in centimeters squared the separation distance, and we're told that the voltage is so our first step is basically I'm just gonna go ahead and draw a quick little sketch of what's going on. So I've got these charges right here. These plates gonna do my best and I've got the separation distance. In other words, D is equal to 10 millimeters, but that's actually equal to 0.1 m. We have to change the everything into meters, so we have the right around units. Right now, we're told that the area of each one of these plates is equal to five centimeters squared. Now we have to do is we have to convert this 2 m squared. So we have to do is we actually have to time multiplied by 10 to the minus four because what happens is we have to do the conversion from centimeters, centimeters twice. All right, so that's the area. And we know that the potential difference between these two things is equal to 100 volts. So in this point, in this part right here, we're supposed to figure out what is the charge on the capacitor. Well, let's go and use our charge formula. We could use the capacitance times voltage. So we have with the voltage is now we just need to figure out what the capacitance is. But this is a parallel plate capacitor, so that means I can use my other formula for C, which is absolutely not a divided by D. Some of the words the capacitance is 8.85 times 10 to the minus 12 at the Epsilon, the area is five times 10 to the minus four. And now that separation distance is 0.1 So that means that the capacities here is equal to 4.43 times 10 to the minus 13. That's in ferrets. That's the unit for that. So that's the capacitance. I'm not quite done yet because now I actually have to plug it back into that formula. So that means the charge, which is gonna be my final answer. Let's write that down. The charge is equal to 4. 43 times 10 to the minus 13. Now, in times the voltage, which is 100. Yep. So that means that the charge is equal to 4.43 times 10 to the minus 11. That's gonna be in columns. So that's our final answer for part A. So now what we can dio is for part B. We can figure out what the magnitude of the electric field between the plates is, and we can actually use a couple of different formulas to find this out. So if we're trying to figure out what the electric field is, we can use that equation, which is that Q divided by epsilon, not times A is equal to the electric field. Remember, this Onley works for parallel plate capacitors, so you have to be careful when you use this. But basically we could just go ahead and solve because we know that the hue is equal to 4. times 10 to the minus 11 and we have the Epsilon knots, which is 8.85 times 10 to the minus 12. And now we could just multiply by the area in the area. It was equal to five times 10 to the minus four. We go ahead and move, and if you work this out, the electric field is equal to 10,000 newtons per cool. Um so that's one way you could get to this, by the way, another way that you possibly could have done this. So here's the alternative approach. So you also could have done this by relating this to e whoops equals the voltage divided by the distance. That probably would have been simpler. Approach the vultures Eyes 100 the distance is 1000.1 That's the separation distance. And if you work this out, you'll also get 10,000 Newtons per cool. Um so this is just another way. And we just got the same answer using two different approaches. Alright, let me know if you guys have any questions with this and I'll see you guys the next one

2

Problem

Problem

Two circular plates of radius 2cm are brought together so their separation is 5mm. What is the capacitance of these plates?

A

2.23×10^{−8}

B

2.23×10^{−10}

C

2.23×10^{−11}

D

2.23×10^{−12}

3

example

Point Charge Inside Capacitor

Video duration:

6m

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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.

4

Problem

Problem

A 3 F capacitor is given a potential difference across its plates of 10 V. What is the charge built up on its plates? If the source of the potential difference across the plates is removed, but the plates maintain their charge, what is the new potential difference across the capacitor if the distance between the plates is doubled?

A

V_{new} = 5 V

B

V_{new} = 10 V (no change)

C

V_{new} = 15 V

D

V_{new} = 20 V

E

V_{new} = 30 V

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