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A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 $\mathrm{mm}$ . If the separation is decreased to $1.15 \mathrm{mm},$ what is the energy stored (a) if the capacitor is disconnected from the potential source so the charge on the plates remains constant, and (b) if the capacitor remains connected to the potential source so the potential difference between the plates remains constant?

$a ) U=4.19 J$$b ) U=2 * 8.38=16.76$

Physics 101 Mechanics

Physics 102 Electricity and Magnetism

Chapter 18

Electric Potential and Capacitanc

Kinetic Energy

Potential Energy

Energy Conservation

Electric Charge and Electric Field

Gauss's Law

Electric Potential

Capacitance and Dielectrics

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Hope College

University of Winnipeg

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Hey, so in this question we have a capacitor, and we know how much energy is stored into it. It's eight point three age rules. So you is 8.3 a Jules, and we know the separation between the plates. It's 2.3 millimeters. So it can say D. There's 2.3 times, 10 to the minus three meters, and, um, the separation is decreased to 1.15 millimeters. Okay, so d goes to D over, too. And then we kind of want to know about how does everything else change? So I'm just gonna write out some formulas to make. Everything's dependency is kind of clear, so you is equal to 1/2 c. V. Squared equals, um, 1/2 Q B equals 1/2 Q. Squared, oversee and the other formula. I want to write out ISS. See a sequel to Absalon, Not a over d. Okay, so for part a, let's see. So, um, let's find the original energy story, and then we'll find what's what energies air stored when we make these changes. So another sort of pre emptive background task. So, um, the originally 00 they actually give us the original you and so so Never mind. We don't have to do that. Okay, so now we've have this distance and we want to know what's happened to the energy. And given that we've, um, disconnected it from the source. So if we've disconnected it, then that means Q is the same because no other charge, no charge can get on it or off of it. So Hugh is the same. But what we've done by decreasing Dia's, we've increased capacitance. So see goes up by a factor of two. So, um, let's find the formula with Q and see. So if Q is the same and cute, see goes up by a factor of two, we can see from this one that you goes down by a factor of ah, 1/2. So therefore, you is equal to, um, four point. Let's see one is that so 38 divided by two is 19. So, um, 4.19 Jules and then for part B. We're wondering about what happens. Um, if it's you remain connected to the source and so that these constant so again we want to look to the in C one and so see, goes up by a factor of two V stays the same. So you is gonna double biased on this one. So you goes to to you. So then doubling 38 30 is like the worst number to multiply and divide with. And I had I've decided, um, so it's gonna be 16.76 Jules for anything else wanted from us in the question. Um, nope.

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