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In the circuit shown in Figure $18.45,$ the potential difference across $a b$ is $+24.0 \mathrm{V}$ . Calculate (a) the charge on each capacitor and (b) the potential difference across each capacitor.

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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

Cornell University

Rutgers, The State University of New Jersey

University of Washington

Hope College

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Okay, So for this problem, we're given a circuit. I'm a shown in the figure they mentioned, so I'm gonna go ahead and redraw list since has a three, 35 and six. Okay, so it goes like this. Think this is the same as the last figure, right? Oh, yeah. Basically is Okay. And then that's supposed to be a capacitor down there. So I'll call this one C one. This one See to this one. C three c one is three. My growth fair on it. We'll check my girl and then see two is five micro fair odds, and C three is six micro fraud. And we want to get the charge on each capacitor, given that the potential difference from A B is 20 bolts and we also want to get the potential difference across each capacitor. So the a B is equal toe 24 0.0 folds. Okay, so let's come up with an equation for the voltage from A to B. So we could kind of pretend like you have ah, battery in here, creating this 24 volts and we could use a cure coughs loop role. So we can't say the voltage across the battery B A B is equal to the bull change across C one. So let's do the outer loop minus the voltage across there. So I plus the voltage across C three because these will be polarized in the same direction. Positive to negative, positive to negative, um, and the problem of squirt gun the charge in each capacitor. Um and then we know that be one is also equal to be, too. He calls me, too, And, um, to those new creation. There's an equation. So now let's come up with some equations of on the charge. So basically two capacitors and Siri's, they're gonna have the same charge on them or excuse me. Two networks in Siri's will have the same charge, so we need to find that So we can say that the total charge in the network of C one and two is equal to see three. So then you can say Q one plus que tu equals Q three. And let's hope that was to be Q one. Let's hope we can can combine these. Also with in general C is equal toe Q over B, um, to to solve for these unknowns. Um, let's see. So there's I mean, I wrote out all these equations, and it's not immediately coming to me what to do without solving, like, a bunch of systems of equations. So I'm gonna kind of put this to the side because we really only know the capacitance is, Well, maybe I could just keep following this approach, but I have a one a backup approach in mind. Um, if this doesn't work, so I'm gonna take I'm going to try to get everything in terms of voltages and capacitance is, um So I'm gonna take this season and plug it in here. So then I got q one. Uh oh, que is ego CV soak. So then this is gonna be, um, see one V one plus C two movie Thio is equal to maybe three. And then V one is equal to be too. So then I got C two plus C one times. V one is equal to be three. Ah, there we go. And then I could take that and I can sub that in for here. So if I sub in that for V three, then I got a B. A B equals the one plus now subbing this in C two plus C one almost applied by the one. Um, no, wait, That that's not right. Oh, that's the three. C three. There we go. Be three C three and then So now if I When I plug in for V three, I actually need to divide by C three here. And so, um yeah, so that we can say V one times one plus C two plus two C one over a C. Three times, all of that. Um, just be one. So then you won is equal to a V a b divided by this whole thing. So one plus c one plus C two over c three. Sorry, this is all just algebra two. So I mean, honestly, if you understood, if you're not bad, are pure. Um, if you'll feel comfortable with algebra, I would just not watch me do the algebra. I would just do it yourself. So it's just I mean, combining all of these equations, I think it's more difficult to watch someone do algebra than to do it yourself sometimes. Okay, so now I'm gonna calculate this on the next page. So then I'm gonna take this 120 volts dividing it by one plus C one plus C two is eight. Micro fair odds divided by, um, six micro parents. Couldn't that movement an integer? Okay, so then this is gonna be our You see this wasthe This is R B one. So I'm gonna go ahead and plug that into a calculator. So 1 20 divided by one plus get over six. So it's 51 point for brawls. Um, great. And so now we have the one, and then we can think it's gonna sort of unravel from here we have. No, they V one and possibly three s add up to 1 20 So then be three is gonna be 100 20 minus the one. So that's gonna be 1 20 minus this. And then I am paying 68.6, and, um, then we want to get all the charges on each capacitor. Oh, yeah? Then this is gonna be equal to be, too. So see is Cuba Rubies soak us CV. And so then, for one, we're just gonna multiply. Um, so, yeah, 35 and six. So, for one, it's just gonna be, um three micro fair odds times the voltage on one, which is 51 point for bulls. So multiplying those two out gives me 1.54 times 10 to the minus for and then for two, we're gonna do the same thing, except for with five micro fair odds. And then they have the same voltage. So that's the calculation we want to D'oh! Um, so that makes it 2.57 times 10 to the minus two. Excuse me. Times 10 to the minus for and then for a three, um, we're gonna do six times 10 to the minus six. So that's the capacitance times the be three. So 68. Oh, goodness. What happened to that, eh? Okay, so that's gonna be 68.6. And so that's actually a six hand. Let's erase this little mark. So 68.6, I'm six times on my six 4.12 times 10 to the minus for oops. And then just kind of checking my work. I'm like, Oh, yes. Cue one. Prosecute. You do equal Q three as required by the equations

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