Energy Stored by Capacitor - Online Tutor, Practice Problems & Exam Prep

Capacitance relates charge and voltage, storing potential energy in capacitors. The energy stored is expressed as ${\frac{1}{2}}^c{v}^2$, with alternative forms $\frac{1}{2}qv$ and $\frac{1}{2}{q}^2/c$. Energy density, $u$, is energy per volume, calculated as $\frac{U}{V}$, leading to $\frac{1}{2}\epsilon E^2$ for parallel plate capacitors.

Energy Stored by Capacitor

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 2 plates in a parallel plate capacitor, and that separation leads to some potential energy that's 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 = 1 2 C V 2

But what we can do is we can use the relationship between charge, capacitance, and voltage to change this equation into 3 common forms that you'll see. So using 12CV2, we can sort of use Q=CV to write this also as 12QV. And we can also write this as 12Q2/C, the capacitance. Now, basically, all these three equations are perfectly valid for any capacitor, not just a parallel plate capacitor. And the choice as to which one you're going to 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 between capacitors with charge, and that potential energy we can use in some energy calculations.

Now another important variable to know is the energy density, which is given by lowercase u. And the energy density is basically just the amount of energy per unit volume. So in other words, we have u=U/volume. Now, I'm not going to use capital V because we use voltage for that, so I'm just going to write here volume instead. Whereas the volume is what? Well, the volume is basically the area of these parallel plates. So for instance, the area of these rectangular plates times the distance between them, which is that capital D. 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 going to be length times width, times the distance between them. And so what we can do is we can take this lowercase u and write it in a different way. So we're just going to write this as 12, and we're just going to use the first form, which is CV2. And now we just have to divide it by the area times the distance.

Now one of the things we can use is we can basically use the relationships between Q=CV, 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 12epsilonnaughtE2. So in other words, if this is a parallel plate capacitor, then that means that the energy density is only related to the electric field that exists between the plates. Alright.

So these are basically the 5 equations that you need to know. So there's all 3 or all 5 of them. And, basically, just to get familiar with these, we're going to go ahead and work out a bunch of example problems. Okay. So let's get started. We've got 2 parallel plates, and we've got the area between them, 50 centimeters squared, with the separation distance. And both we're supposed to find a, which is the energy stored, and also what's the energy density. So in part a, we're going to be finding U. And we know that the three forms for U, big U, are 12CV2 equals 12QV equals 12Q2/C., so you can choose which equation to use based on the given variables. 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 what we're being asked in this question is we're asked for the strength of the electric field in a capacitor. We're going to assume that it's a parallel plate capacitor as well. So this right here, this is a joule, so this is millijoules per cubic centimeter. So in other words, the energy, which is big U, is equal to 2.5, and we have millijoules. So that means that it's_times 10 to the minus 3,_that's joules. And now we have per cubic centimeters, 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 If we're trying to figure out what the electric field is equal to, then we have to use the little u is equal to 12epsilonnaughtE2. So now this is our target variable. Let's go ahead and manipulate this. We're just gonna move the one half over to the other side, the epsilon naught over, and we're gonna get 2×u/epsilonnaught=E2. So that means that E=2×u/epsilonnaught. Okay? So let me know if you guys had any questions with this, and let's go ahead and get some more practice.

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 F
• B. 5×10⁻⁴ F
• C. 2.5×10⁻¹ F
• D. 5×10⁻¹ F
• E. 1.0 F

Energy Released by Flashbulb

All right, guys. Let's take a look at this one carefully. We have a flashlight. We're given the capacitance and the voltage of that flashlight. And then what happens is the flashlight goes off, or the flash goes off, and this bulb loses some amount of charge. We're supposed to figure out how much energy is released when that happens. So in other words, we have energy stored inside 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 you have some initial energy and some final energy, and you're finding the difference between them. So Δ is equal to U_f - U_i. Now we have this potential energy that we can relate using our equations. We know that U = 1/2 C V² = 1/2 Q V = 1/2 Q² / C. Now the question becomes, which one of these things are we going to use? Which form? Well, we're told that this potential energy has to do with some loss of charge. So in other words, one of the equations is going to have to use our Q. We know that the charge, Q_f, is going to be, let's see, it's going to be 80% loss of charge. That means that Q_f is going to be 20% of Q_i, because it loses 80%. So we're going to have to definitely relate this to some charge. Now the other question becomes, what other variable do we know? Well, we know the capacitance and the voltage. So now we're kind of stuck between, let's see, we don't have to use 1/2 C V². We're going to use one of these 2. Now what happens is we know that from Q = C V, if you have the capacitance in a parallel plate capacitor is fixed. It can't change. So for instance, if this flashlight has a capacitance of 1,000 millifarads, 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 V is not constant. So we can't use that 1/2 Q V. Instead, we're going to have to relate this back to the charge which is lost, and then the capacitance which is a fixed value. So this is the actual form of the potential energy we're going to use. So we know that this U_f - U_i is going to be 1/2 (Q_f² / C) - 1/2 (Q_i² / C). Okay? And that's basically what we're going to use. So we have 1/2. Now the charge, how do we find the charge? Well, we find it using Q = C V. So the initial charge is going to be the capacitance, which is 1,000 millifarads, which, by the way, 1,000 millifarads just equals 1 farad times the voltage, which is 500. Now we know that that's equal to 500 coulombs of initial charge. How do we find the final charge? Remember, the final charge is 20% of the initial charge, which means that 20% of 500 is just equal to 100 coulombs. So now that we have our initials and finals, now we 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 500 or sorry, that's 100 coulombs. That's going to be squared, and then divided by the capacitance, which is 1, minus 1 half of the final charge or sorry, initial charge, 500 squared, divided by the capacitance. So if you go ahead and work this out, the amount of energy that's released is going to be negative 120,000, and that's going to be in joules. Or you could have written this as 1.2 x 10⁵ in joules. This makes sense because we got a negative number. So in other words, all of this charge gets released, so that means that the amount of energy, the amount of potential energy, is going to decrease. All that stuff gets converted to light. Alright? So let me know if you guys have any questions with this.