1

concept

## Energy Stored by Capacitor

9m

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Hey, guys. So now that we're a little bit more familiar with capacitance and the relationship between capacitance charge and voltage, we're going to see in this video how that all relates to the potential energy stored because remember the capacitors, separate charges between two plates and a parallel plate capacitor, and that separation leads to some potential energy that stored well, How much energy? Well, I'm not going to show you the derivation, but basically the energy stored by any capacitor, not just a parallel plate capacitor is given as this equation. U equals one half see times V squared. But what we can do is we can use the relationship between charge, capacitance and voltage to change this equation into three common forms that you'll see. So using one half C V squared, we consider use Q equal CV to write this also as one half times Q V. And we could also write this as one half Q squared, divided by C, the capacitance. Now, basically, all these three equations are perfectly valid for any capacitor, not just a parallel plate capacitor, and the choices to which one you're gonna use really just comes down to what variables you're given in the problem and that's basically it. So this is the amount of energy that is stored in between capacitors with charge on the potential energy we can use in some energy calculations. Now, another important variable to know is the energy density which has lived given by lower case you and the energy density is basically just the amount of energy per unit volume. So, in other words, we have u equals big you, which is the potential energy divided by the volume. Now, I'm not gonna use capital V X. We use voltage for that. So I'm just gonna right here, volume instead. Whereas the volume is what will. The volume is basically the area of these parallel plates. So, for instance, the area of these rectangular plates times the distance in between them, which is that capital deep? It's basically the volume of the gap between those parallel plates. So, in other words, the volume is equal to the area, which, if you're given a rectangle, is just gonna be length times with times the distance in between them, and so we can do is we can take this lower case you and write it in a different way. So we're just gonna right? This is one half and we're just gonna use the first form, which is C V squared, and now we just have to divide it by the area times the distance. Now, one of the things we can use is we could basically use the relationships between Q equals C V. And we can use the relationships between capacitance, which is in a parallel plate capacitor epsilon, not a over D. And if you go ahead and manipulate these equations and stick them inside for this equation, we can actually find that the energy density of a parallel plate capacitor is something very important and very simple, which is one half epsilon, not times the electric field square. In other words, that this is a parallel plate capacitor. Then that means that the energy density is Onley related to the electric field that exists between the plates. All right, so these air basically the five equations that you need to know. So there's all three are all five of them and basically just get familiar with these. We're gonna go ahead and work out a bunch of example problems. Okay, So let's get started. We've got two parallel plates and we've got the area between them, 50 centimeters squared with the separation distance, and we're both we're supposed to find a which the energy stored and also what's the energy density. So in part A, we're gonna be finding you. And we know that the three forms for you big you are one half C V squared equals one half Q V equals one half times Q squared oversee. So basically, these equations air just gonna be like manipulating variables and seeing which ones you have which ones you can find out using some information. It's basically just like a puzzle. You just have to figure out which equation to use. Let's sleep. So we've got the area between them. So we've got a Let's go ahead, Mark that in black, actually, So we've got the area between the plates. We also have the separation distance, and we have the voltage. So which equation we're gonna use were not told any information about charge. So we're no, we're given no que. So that means that we don't want to use these last two forms right here. We don't have Q V and won't have Q squared oversee because we're not getting any information about the charge. We could probably figure it out. Equals using Q equals C V. But it's just not create more work for ourselves. And we know that these two variables, A and D will be able to find what the capacitance is using these. So this is gonna be a good way to use one half C V squared because we also know what the voltage is. All right, so let's go ahead and figure that out. We know that the capacitance if these two are parallel plates, is gonna be epsilon, not times the area over deep. So that means that instead of actually just figuring this out, what we can do is just right out this equation we've got you. It goes one half that we've got Epsilon, not a over D. Now we just have to multiply by the voltage square and that's it. That's the whole equation. So we just go ahead and plug everything in. We've got one half now you've got Epsilon not is 8.85 times 10 to the minus 12. Now you've got the area of the plates is 50 centimeters squared. So what I want you to have to remember is that 50 centimeters squared is not equal 2.5 m squared. We have to do the conversion twice. So this is equal to 50 times 10 to the minus four. Or this is just gonna 40.5 no, That's 0.0 Yeah, that's 0.5 There we go. So that's gonna be 0.5 Now we have the voltage, which is 20 volts. We have to square that now. The distance between them is 10 millimeters, which is 100.1 m. So we could go ahead and plug all this stuff in. Just gonna make this a little bit neater. And then we just plug it all that and we get a potential energy that stored between the capacitor, that's equal to 8 85 times 10 to the minus 10. And that's in jewels. That's the answer to party. Now, Part B is asking us for the energy density. So that letter is not big. You, it's little you. So this little you right here is equal to big you, which I'll try toe. Sort of like make a little bit more clear, divided by the volume. And we know that this volume over here, this volume is equal to the area of the plates times the distance rights basically the spacing between in between the plates times the area of each one of those plates that gives us a volume. Okay, so that means that the energy density is just gonna be the potential energy. 8.85 times 10 to the minus 10. And now we just have to multiply or divide it by the volume, which is the area which is 100.5 again. Times the spacing, which is 0.1 So you go and work this out. You should get 1.77 times 10 to the minus five. And the units for this. We're dividing a energy divided by an area or sorry, a volume. So that means we're gonna get Jules her meter cute. Alright. And that's the answer. Alright, guys, let's go ahead and work out the second one here, which is what is the strength of an electric field in a capacitor storing this energy per cubic centimeter. So now we're being asked in this question is were asked for the strength of the electric field in a capacitor. We're gonna assume that it's a parallel plate capacitor as well. So this right here, this is a Jules. So this is Milly Jewels per cubic centimeter. So in other words, the energy, which is big U is equal to 2.5 and we have Mila jewels. So that means that it's times 10 to the minus three. That's jewels. And now we have per cubic centimeter. So we have Centimeters Cube right here. That's the volume so we can figure out what the energy density is. But how do we relate that Back to the electric field? Remember that in the we're trying to figure out what the electric field is equal to, and we have to use the little U is equal to one half times epsilon, not which is just a constant times the electric field squared. So now this is our target variable. Let's go ahead, manipulate this. We're just gonna move the one half over to the other side, the epsilon not over. And we're gonna get to times little you divided by Epsilon not equals e squared. So that means that e is equal to the square root of two. And now we have to figure out what the energy density is. They have to be careful because this 2.5 million joules per cubic centimeter is not given in the right units were given all kinds of prefixes. With these units, we have to convert this so that it's in the right units. So what we have to do is we basically have to say that you little U is equal to 2.5 times 10 to the minus three jewels per one centimeter cubed. Now we have to remember is that this one centimeter is equal 2.1 m. So that means that one centimeter cubed has to be one times 10 to the minus six meters. Because you have to apply this conversion three times. We were given three spatial dimensions here. So just remember that because that often trips up a lot of students, you have to have to basically cube the conversion. You have to do it three times because we have three dimensions. OK, so now we've got let's see. So that means that this you over here is going to be 2.5 times to the minus three, divided by one times 10 to the minus six. So it means that the energy density is gonna be 25 times 10 to the third. Now it's gonna be in jewels per meter cubed. And now that it's in the right units, now we can plug it into our equation. So we have two times 2.5 times, 10 to the third. And now we have to divide by the Epsilon, not constant. 8.85 times seven mice. 12. Then you go ahead and square root the whole thing, and you should get electric field. That's 2. 38 times 10 to the seventh. And the units for that are gonna be in Newtons per cool. Um, okay, so let me know if you guys have any questions with this, and let's go ahead and get more practice

2

Problem

A cardiac defibrillator can be modeled as a parallel plate capacitor. When it is charged to a voltage of 2 kV, it has a stored energy of 1 kJ. What is the capacitance of the defibrillator?

A

2.5×10

^{−4}FB

5×10

^{−4}FC

2.5×10

^{−1}FD

5×10

^{−1}FE

1.0 F

3

example

## Energy Released by Flashbulb

3m

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Alright, guys, let's take a look at this one carefully. We have a flashlight were given the capacitance and the voltage of that flashlight. And then what happens is the flashlight goes off, the flash goes off and this bulb loses a some amount of charge, and we're supposed to figure out how much energy is released when that happens. So in other words, we have energy stored inside of this flashlight right here, some initial energy. And then what happens is it releases some of that charge. 80% of it. We're supposed to figure out how much energy is released by that. So in other words, we're figuring out what happens when when you have some initial energy and some final energy and you're finding the difference between them. So Delta U is equal to u F minus you. I Now we have this potential energy that we can relate. Using our equations, we know that you is equal to one half c v. Squared equals one half Q V and one half Q squared oversee. Now the question becomes which one of these things where we're gonna use which which form we're told that this charged or sorry. This potential energy has to do with some loss of charge. So in other words, one of the equations is gonna have to use our Q. We know that the charge que final is gonna be Well, let's see, it's gonna be 80% loss of charge. That means that cute final is going to be 20% of Q initial because it loses 80%. So we're gonna have to definitely relate this to some charge. Now the other question becomes, what other variable do we know? We know the capacitance and the voltage. So now we're kind of stuck between Let's say we don't have to use 11 half C V squared. We're going to use one of these two now What happens is we know that from Q equals C V. If you have a capacitance that capacities in a parallel plate capacitor is fixed, it can't change. So, for instance, if this flashlight has a capacity of 1000 military IDs, then it's always that value, and it has to do with the charge and the voltage. What happens is as the charge gets dropped, Then the voltage also decreases. But this capacitance stays the same. So that means that we can't use Q V because because I'm not right It right here. So can't use this one because V is not constant. So we can't use that one half Q V Instead, we're gonna have to relate this back to the charge which is lost and then the capacity to which is a fixed value. So this is the actual form of the potential energy we're gonna use. So we know that this you final minus you initial is gonna be one half. Now we've got two Q final squared, divided by the capacitance, minus one half que initial squared, divided by the capacitance. Okay. And that's basically what we're gonna use. So we have one half now the charge. How do we find the charge when we find it using Q equal CV? So the initial charge is going to be the capacitance, which is 1000 militaries, which, by the way, 1000 military it's just equals one Ferid times the voltage, which is 500. Now we know that that's equal to 500 cool OEMs of initial charge. How do we find the final charge? Remember the final charges 20% of the initial charge, which means that 20% of 500 is just equal to 100 cool homes. So now that we have our initials and finals now you can just plug these values inside of these equations right here and figure out what the change in energy is the amount of energy that's released. So you got one half of five. Sorry. That's 100 cool homes that's gonna be squared and then divided by the capacitance, which is one minus one half of the final charge or sorry, initial charge 500 squared, divided by the capacitance. So you go ahead and work this out. Thea's want of energy that's released is going to be negative on 120,000, and that's gonna be in jewels. Or you could have written. This is 1.2 times 10 to the fifth in jewels. This makes sense because we got a negative number. Some of the words all of this charge gets released, So that means that the amount of energy that amount of potential energy is going to decrease. All that stuff gets converted toe light. All right, so let me know if you guys have any questions with this

Additional resources for Energy Stored by Capacitor

PRACTICE PROBLEMS AND ACTIVITIES (3)

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