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The potentia difference between the plates of leaky (meaning that charge leaks from one plate to the other) 4.5 AF capacitor drops to one-fourth its initial value i...

Question

The potentia difference between the plates of leaky (meaning that charge leaks from one plate to the other) 4.5 AF capacitor drops to one-fourth its initial value in 4.4 What is the equivalent resistance between the capacitor plates? 73e06 Did you use the charging or the discharging equation? Did you substitute for the V? Do you know how to take the inverse function of an exponential?Additional MaterialsSection 27.4

The potentia difference between the plates of leaky (meaning that charge leaks from one plate to the other) 4.5 AF capacitor drops to one-fourth its initial value in 4.4 What is the equivalent resistance between the capacitor plates? 73e06 Did you use the charging or the discharging equation? Did you substitute for the V? Do you know how to take the inverse function of an exponential? Additional Materials Section 27.4



Answers

The potential difference between the plates of a leaky (meaning that charge leaks from one plate to the other) 2.0$\mu \mathrm{F}$
capacitor drops to one-fourth its initial value in 2.0 $\mathrm{s}$ . What is the
equivalent resistance between the capacitor plates?

Hi. And this given problem the passage dance off the leaky capacitor. That has been given us 2.0 Micro Parrot potential across its plates is reduced to 1/4 of the initial potential In a time. T is equal to 2.0 seconds. So we have to find the resistance through which this capacitor is getting leapt. For which we use an equation for the potential over the plates of the capacitor which is getting discharged. And that is given as we is equal to be not into a race to the power dynasty by doubt where this now is the time constant off the series R. C. Circuit such formed and that is given as a product of resistance with the capacitance. Now for we hear this is we not by four it's going to be not paris par minus 2.0 2nd divided by doubt cancelling this we note. And this one x 4 maybe written has 0.25. Now converting this exponential form into look at me when we get log of 0.25 at BS. E means this is the natural algorithm And that is equal to -2.0 by Joao. So this minus 2.0 Byte out. That is equal to natural algorithm of 0.25 Means the algorithm of 0.25 at best. E. And that comes out to be -1.386. So canceling this negative sign. How comes out to be 2.0 divided by 1.386 means this is 1.4 or 2nd. No, this is the product of resistance with the capacities and that is 1.44 2nd. So resistance here will be given by 1.44 divided by the capacities, which is 2.0 into And each 1 -6. Farid. So finally, this resistance here is zero 0.72 Into 10 days, part six home. Or we can say this is 0.72 mega or which is the answer for the given problem here. Thank you.

That's what we know. Capacitance is sick with the charge to buy, thereby potential difference with his own regiment, your Web que, which is the charge of the school capacitance tested potential difference. We know the capacitance disc, given 3.47 Michael Ferre, which is equal to 0.47 times, tend to pollinate six. Fair because one Michael Vera is he was attention pounding. Six. Fair and the potential difference final question scheme is 1 20 boats, therefore charge. Q. Should we go to 0.47 times tender part next six there, UH, 10 times 120 votes and this will give us the charges. About five point 64 times. Tend to apartment 86 cooler.

Hi there. So for this problem we have uh an ending capacity or and the one that is in here at the left And has a capacitance of 3.2 Micro Farhad's and it is connected to a battery of 12 bowls. Now after that there is a dielectric material With the electric constant 4.5 that is inserted between the plates of this capacitor. So what we need to determine is the magnitude of the service charge on the dielectric that is adjusting to either played off this capacity. So um in that sense, what we want is the church q minus cues minus Q zero where Q zero corresponds to the charge of the anti capacitor and Q corresponds to the charge of the capacitor will get the dielectric. Now we know the charge to zero on the anti capacitor is related to its capacity means that we're gonna call C zero and that is quite the following formula or equation. The capacitance times the electric potential between the place. No. And the charge Q on the capacity that is filled with the dielectric is related also with its capacity. So we will have something similar and we know that the presence of the dielectric increases the capacities such that we know that the capacities of these new capacity or equity dielectric has a value of K. Times where K. Is the dielectric constants times the initial capacities. So and it's the magnitude of the surface charge on the electric is equal to the difference in the charge on the place with and without the electric, we will have the following That q minus the initial charge to use zero is equal to c V. There's definition in here and the other definition for this study is minus zero P. Now we know we use this again in here so we will find that this is equal to the dielectric constant times zero v minus zero B. We can simplify this border by taking out the common terms. So we will find that this is, it is different in the charge is equal to The initial capacities without the dielectric times. They bolted edged times did electric -1. So we have all of these information. So what we need to do is to put these values and we obtain this surface the surface charge on the dielectric. So that will be the capacities which is 3.2 Micro for its micro means tends to the minus seats for its The voltage in here. What is 12? Yeah. Mhm. And the dielectric Constant which is 4.5 -1. So we will obtain that the surface area of the surface charge on the dielectric is going to be equal to 1.3 times 10 to the -4 columns. And this is the result for this problem

This problem are given a capacitor first without an electric and we are dielectric in and the end goal is to find what the surface charge will be on the dielectric. And before we get into the calculation of that, let's talk about how to arises. So here is the situation where there's no doubt that electric and is charged up to a charge Q one. I've used four electric field lines to represent the electric field between the plates. Remember dr field is tangent at any point following the general sense of the arab, now we put it to electric in between. What happens if I remember dialectics are insulators, you don't have free movement of charge, but they can move slightly in their general location. What you get is a polarization where the pluses and minuses would be on top of each other. Their at least their distributions would be. Now they will be slightly off. I'm making this larger than then. In reality they'd be slightly off. So the miners would be closer to the plus plate and the plus I was close to the honest plate. Now, let me just put in a couple, you know, I won't put in the big thing I just put in the end results decide Factory, this surface will seem to have a negative charge, and likewise, this surface will seem to have a positive charge. We're not charging the dielectric. It still it still has no access charge. We're just causing this polarization effect. And it gives us this this charging of the, you know, this appearance of charge surface. Yeah. And in that respect, operationally, now, let's talk about the electric field aspect of this. We know there's electric field, original electric field. Now, with these charges been in induce the factory on the surfaces. Think about what goes on. This would make it look like. So we would have the original field this way that's from here and that looks like this would give me a field back to the left. And so I'm going to get this is my now new field between the place have reduced electric field and that is certainly the case. And how do we represent that? We left a few lines well, remember they begin on positives and and on negative. So there's one line, here's another line you can think of these guys as knocking each other out. You know, if you think of them as separate. A little bit, there's a little, there's a little bit of a arrow from the plus, the minus, but it's all very, very, very small. Uh, in terms of that does not really impact anything. So, notice two lines is before that visually tells me lesser feel. That's what the field lines are. Therefore in addition to giving me direction. So we have a lesser field. Now, let's talk about that. That's the key ingredient here. We have a lesser electric field. Now, this is hooked up to a battery in this problem. 12 votes. Everything's hooked up to a battery. The same situation just you know that we're just going from one step to the other hooked to a battery. Um, so that says between these places be 12V. But I just told you the electric field drops, electric field drops. That means the potential difference between the plates drops, But that can't be. We're hooked up to something that says you're gonna be 12V between those planes. So what has to happen? Well, more charge must flow to bring up the electric field back to the way it used to be. So that we have the same potential difference of 12 votes across the plates. So let me put six. I still got my, Still got my two. So there is our situation charge grew but in combining these together because remember it's all the surface Combining together. It's got to give me the same as Q1 which gave me the original field because we're hooked up to a battery. If you're not hooked up to the battery, that's a different problem. That's not what we're dealing worried about here. Now, what we're looking for in reality is this QUE que Q cap, I should say. People say Q cap capital for the electric counselor, looking for the for this come out and obviously this is just minus Q. K. Q. Capa we should say. Um so getting using that symbol. Let's look at this play here. That tells me what that que tu minus Q. K. I'm thinking cute kappa Q coppa as as a positive number. We always we can pass is always thinking about the talk about the positives. Que tu minus Q. K. Is equal to Q. One. That's how we get the same electric field that we had before. That gives me the same potential difference. So now we can solve for Q Kepa. You kappa you too ask you what now? This is given by the compassionate when we have the dielectric. I see mine is the casualties capacities when we don't have no cause C zero transfer fee. Now what is the capacities when you have a dielectric? Well that's very simply kappa times. See not if you remember see not just see not was epson zero for parallel plate San zero A over D. So this is just going to be kappa times that. And so we can factor out, see not see kappa minus one. And we can put in our numbers now. Mhm 32 times 10 to -6 ferrets 12 votes 4.5 -1. billion. The dielectric constant. And this works out to be 1.3 times 10 to manage for cool apps so that that is a positive charge. That is amount of charge on the electric surface here or minus status, The amount of charge that is seen on the dielectric surface on the left, whichever, wherever you're looking. So that is that is our answer and that is the whole problem.


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