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 Uf - Ui. 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, Qf, is going to be, let's see, it's going to be 80% loss of charge. That means that Qf is going to be 20% of Qi, 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 Uf - Ui is going to be 1/2 (Qf2 / C) - 1/2 (Qi2 / 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.
26. Capacitors & Dielectrics
Energy Stored by Capacitor
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