Hey, guys. So let's take a look at this problem. We haven't Elsie circuit with some inductions and a capacitance. It begins with a capacity is fully charged. We're supposed to figure out how many seconds it takes for a fully charged plate to transfer all of its charged to the other plate. Now, that's a lot to unpack there. So what I'm gonna do is we're gonna begin with just a simple diagram oven, Elsie circuit. Hopefully I do this, right. So I've got the induct her like this and it goes around and then you have a capacitor like this. Now, remember, from our previous video that we talked about If we have a sort of charged capacity like this with all the charges, then a full cycle of Elsie circuit is when the charge goes all the way around the circuit builds up to a maximum and then piles up on the other side. And then the cycle has to reverse direction and do the same thing over again. But backwards. So the words all the charge has to go from here. It escapes through this plates, and it basically piles it back to where it started. from. And then the whole thing repeats over and over again. So what this is saying is, how long does it take for a charged plate to transfer all of its charge over to the other side of the circuit or the other side of the capacitor? So what that really means is that actually just looking for half of the cycle? So that means in terms of the period, that's the variable t That's gonna be T over, too. So really, what we're looking for in this case is what is the half period in seconds? That's really what we're looking for. But in order to do that in order to figure that out, we just have to figure out what the period is in general. So let's remember, if weaken, let's remember our formulas for the period, the frequency and the angular frequency foreign oscillating system we have that the period and the frequency are in verses of each other. So t equals one over f, and we also have that the angular frequency could be related to the linear frequency by this equation right here. So if I want to figure out what the period is, I have to relate that to the frequency, but I don't know what the frequency is, so I have to related to the angular frequency and that I actually can figure out because remember that the angular frequency Omega can also be related to the induct INTs and the capacitance off this equation of the of the circuit. So let's go ahead and do that. Let's see, uh, the tea is going to be. Well, let's see if I wanted the frequency that's actually going to be omega over. Two pi, if I just moved to the other side is gonna be equal to the frequency. So what I have is that tea is actually equal to one over omega over two pi. So that means that the period is going to be two pi over Omega. So that means finally, that t over to is just going to be pi over omega. So what? One half of this is just gonna be one half of this. So the two just cancels out and we just get pi over omega. So finally, what I can do is that can actually just take this formula right here, which is the square root of one ever Elsie and actually action to plug it back in for the denominator in this equation. And what I get is that t over two is equal to pi times the square root of l times C. So what happens is when you take the inverse of this and this is the denominator, basically, what happens is we're taking pi over one over the square root of L C. And so l c just goes up on the top, right? Got it. So I just wanted to sort of walk us through that because it's been a while since we use these kinds of equations. So really, this is actually just going to be the half period. That's all we have to dio. So that means that T, which is equal to the half period, is just gonna be pi times the square root off the induct INTs, which is zero point zero five times the capacitance, which is fifty military. It's so we have elves equal to zero point zero five. We have the capacitance that's equal to fifty military. It's which is actually zero point zero five as well. So that means we're just gonna have zero point zero five here also, and we just get that the half period is just going to be equal to zero point one five seven seconds. And that's our answer. That's how long it takes in this circuit. For the charge to transfer to the other side. That's half the cycle. It would take another point one five seconds to go back and then basically begin the whole thing over again. Okay, guys, that's it for this one. Let me know if you have any questions.