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Ch 24: Capacitance and Dielectrics

Chapter 24, Problem 24

A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 3.90 uC. (a) What is the potential difference between the plates? (b) If the charge is kept constant, what will be the potential difference if the plate separation is doubled? (c) How much work is required to double the separation?

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Welcome back everybody. We are making observations about a device that consists of two identical capacitors. We are told that the capacitance on each one is 430 PICO fair ads or 430 times 10 to the negative 12 fair ads. We're told that the charge on each of the plates is 1.45 micro columns or 1.45 times 10 to the negative six columns. And we are tasked with finding a couple different things here. A We are tasked with finding what is the voltage difference across the plates. Be? What will be the new voltage difference when we triple the distance that they are currently at. Which by the way, we're not told that. So we're just going to say their current distance is just equal to D. And then see we are tasked with finding what is the work done by tripling the separation here. So let's go ahead and start out with part a part is pretty easy. We're just gonna use this formula here that the voltage difference is simply just the maximum charge divided by the capacitance. So we have that this is 1.45 times to the negative six, divided by 430 times 10 to the negative 12. This is equal to my apologies. This is equal to 3.37 times 10 to the 3rd. 4. Our voltage difference. No, we are now going to take the metal plates here and separate them by triple the distance at what they were at. And now we need to find what the new voltage difference is. Well, the charge, the maximum charge is gonna stay the same. But now there's going to be a new capacitance and here's why we have that are capacitance is equal to epsilon. Not times the surface area divided by the distance. But now we are going to triple the distance. Right? So now we are going to go to epsilon not times the area uh divided by three times the original distance, which really, if you think about it, is just the original capacitance divided by three. This is our new capacitance. So let's plug this into our equation right here. And I'm actually just going to scroll down just a little bit. So we have that our new voltage distance is equal to you divided by C. Divided by three A K A three times Q over C. Well, we already calculated our, our old voltage here. So it's just going to be three times 3.37 times 10 to the third. Giving us a new voltage difference of 1. times 10 to the fourth vaults. Great. So now let's go ahead and move on to part C. Here. We need to figure out the work done by tripling the separation between the two plates. Well, the work done is just going to be the change in the internal energy stored. A K. You're going to have your final energy stored minus your initial energy stored. But what exactly is that going to be? Well let's look at the formulas just for the stored energy here, our initial stored energy will just be the maximum charge squared divided by two times the capacitance. Our final stored energy is going to be Q squared divided by two times the new capacitance. Well we know that the new capacitance is just C divided by three. So we're gonna have the same thing. We're gonna have three times U squared divided by two C. Which is just three times the initial stored energy. So now just scrolling down just a little bit more here we have that the work done is equal to three U. I minus you. I this is equal to two times our initial energy stored. So let's go ahead and calculate that we have two times two Q squared all over to see and we can plug in all of these numbers. So we have two times 1.45 times 10 to the negative six. Where'd all over? Two times 430 times 10 to the negative 12. Giving us a final answer of 4.89 times 10 to the negative third jewels. So now we have found the original voltage. The new voltage difference in the work done by making that separation which corresponds to our final answer choice of the Thank you all so much for watching. Hope this video helped. We will see you all in the next one
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Textbook Question
For the system of capacitors shown in Fig. E24.16

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Textbook Question
In Fig. E24.20

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Textbook Question
A 5.80-uF, parallel-plate, air capacitor has a plate separation of 5.00 mm and is charged to a potential difference of 400 V. Calculate the energy density in the region between the plates, in units of J/m^3.
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You have two identical capacitors and an external potential source. (a) Compare the total energy stored in the capacitors when they are connected to the applied potential in series and in parallel. (b) Compare the maximum amount of charge stored in each case. (c) Energy storage in a capacitor can be limited by the maximum electric field between the plates. What is the ratio of the electric field for the series and parallel combinations?
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Textbook Question
A parallel-plate capacitor has capacitance C0 = 8.00 pF when there is air between the plates. The separation between the plates is 1.50 mm. (a) What is the maximum magnitude of charge Q that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00x10^4 V/m? (b) A dielectric with K = 2.70 is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed 3.00x10^4 V/m?
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Textbook Question
A constant potential difference of 12 V is maintained between the terminals of a 0.25-uF, parallel-plate, air capacitor. (a) A sheet of Mylar is inserted between the plates of the capacitor, completely filling the space between the plates. When this is done, how much additional charge flows onto the positive plate of the capacitor (see Table 24.1) ? (b) What is the total induced charge on either face of the Mylar sheet? (c) What effect does the Mylar sheet have on the electric field between the plates? Explain how you can reconcile this with the increase in charge on the plates, which acts to increase the electric field.
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