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A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 2.55$\mu \mathrm{C}$ . (a) What is the potential difference between the plates? (b) If the charge is kept constant, what will be the potential difference between the plates if the separation is doubled? (c) How much work is required to double the separation?

a) $V=2770 \mathrm{V}$ .b) The capacitance halves and the voltage doubles to 5540 $\mathrm{V}$ .c) $U=3.53 \times 10^{-3} \mathrm{J}$

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

Simon Fraser University

Hope College

University of Winnipeg

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Okay, So this problem, we start with the parallel play capacitor, and it has a 920 PICO fair Add, uh, capacitance. And the charge is 2.55 My group Coolum on each player. And first we want to get the potential difference. So this is for a vehicle's question, mark. So see, is Cuba via Savi Askew oversee. So it's Q over. See, um, so I'm gonna go ahead and fund goes into a calculator. So let's do it by 5 10 minus six. Provided by nine. When e I'm sorry for the minus 12 when I got to thousands, uh, 770 vaults on. And next. We are concerned with what happens if the potential difference if we double the separation. So we want to got V equals question. Mark. He was the same. Um, And then the distance goes to double. So for to figure out what happens, um, mean again, we want evaluate v Askew, oversee the problem, tells us that Q stays the same since he's gonna end up changing when you increase the distance because of C is equal toe Absalon, not a over D. If you double the distance. Then you're gonna have the capacitance. And, um, if you have the capacitance than you'll double the voltage, um, so he goes to to be, um and so be Then it's just gonna be double 2000 770. And so I'll find out by two. And I got 5540 once. Um, another way. You could reason this is that the, um, electric field stays the same if you have the same amount of charge and then, um, volt ages Thean girl of electric field over distance. If you've increased the distance, you have increased the voltage. And so the next we want to know how much work is required to do this. So box this off. So here's a B and then lets you see so we just run, they get the work is equal to a change in energy. So we want to get you final minus you initial. Um, so let's just evaluate this. So you final is gonna be 1/2. Let's call the new capacitance, which is double the or, um, sorry. Half the old one. Um, see, prime. Um, ok, actually, let me. You were wine just a little bit I think I'm gonna use I think I'm just gonna first start by evaluating the potential energy initially and then we'll figure out how it scales. So initially, you is equal. Thio, um, 1/2 q squared overseas. So all used that formula. So when I evaluate that, I'll go ahead and type it in. So 1/2 times my frying pan a minute six all squared, provided by Snee. I obtained, um, 3.53 Millie jewels. So go ahead and write that I say 53 c, Yeah, I did 3.53 times 10 to the minus three jewels. And then Now let's consider what happens when you change the capacitance, so we're gonna keep you the same cities. And then we said we're gonna take C and half it, So if you take half of see it happen, this is in the denominator. That means you're gonna double you. So, um, so if you're doubling a number, then the change in you is gonna be just the same in the original cause you're just gonna do, you know, two times this number minus one times this number, and that's just gonna give you one times that number. So 3.53 times 10 to the minus three jewels. Just double check something. Yeah, sounds great.

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