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Two identical parallel-plate capacitors, each with capacilance C , are separately charged t0 potential difference AV and then disconnected from the battery. They ar...

Question

Two identical parallel-plate capacitors, each with capacilance C , are separately charged t0 potential difference AV and then disconnected from the battery. They are then connected t0 each other in parallel with plates of Iike sign connected_ (a) Find an expression for the total energy of the system of two capacitors:(b) Next, the plate separation in one of the capacitors is doubled. Find an expression for the potential difference across each capacitor after the plate separation is doubled:(c) F

Two identical parallel-plate capacitors, each with capacilance C , are separately charged t0 potential difference AV and then disconnected from the battery. They are then connected t0 each other in parallel with plates of Iike sign connected_ (a) Find an expression for the total energy of the system of two capacitors: (b) Next, the plate separation in one of the capacitors is doubled. Find an expression for the potential difference across each capacitor after the plate separation is doubled: (c) Find an expression for the total energy of the system (of part| b) after the plate separation is doubled



Answers

Two identical parallel-plate capacitors, each with capacitance $C,$ are charged to potential difference $\Delta V$ and then disconnected from the battery. They are then connected to each other in parallel with plates of like sign connected. Finally, the plate separation in one of the capacitors is doubled. (a) Find the total energy of the system of two capacitors before the plate separation is doubled. (b) Find the potential difference across each capacitor after the plate separation is doubled. (c) Find the total energy of the system after the plate separation is doubled. (d) Reconcile the difference in the answers to parts (a) and (c) with the law of conservation of energy.

So we have to capacitors connected in parallel just like this, and they happen to both Be 10 micro fairy capacitors. We also know that the voltage across them this is gonna be 50 volts. Now, First we want to find the energy of this system. That's gonna be 1/2 times, Theo. Quibbling, capacitance times the square of the potential. Now the equivalent capacitance for this will each capacitor is 10 Micro Fareed's, so we can add the two. It's gonna give us 20 Micro Fareed's Okay, Now we can plug that in 20. Micro Fareed's is also 20 times 10 to the minus six. Farage's so right that here we also know that there's a 50 full potential difference. This comes out to 2.5 times 10 to the minus two jewels. Okay, now we take one of the capacitors. I'll just choose the bottom one, and we're gonna double the plate separation something like this. Now this capacitance is going to change. And since we know that the capacitance is inversely related to the separation distance, if we double the separation distance, that means we're gonna take 1/2 the capacities. So now it's a five micro fared capacitor, and now we want to find the this voltage, which is no longer 50. So we know that we have to have the same amount of charge even when we double the separation distance of the capacitor. And the charge it will has to be equal so we can represent the charge with the equivalent capacitance times the voltage. And I'll say, this is, uh, one that's before the plate separation is doubled. And this this will be to So now the equivalent capacitance. This is the 1st 1 Now the equivalent capacitance is 10 plus five. It's now we have 15 Micro Fareed's so now we can plug some stuff in. Um, the equaling capacitance to start was 20 Micro Fareed's times 50 volts. That has to equal 15 Micro Fareed's times the new voltage. So you two is 20 times 50/15. So now we have, um, 66.7 volts. So the potential difference increased. Now, this is 66.7. Um, the one was 50. Uh, now, to figure out the new energy stored in this system, let's label us again. This is you one. Now this is too. It's gonna be 1/2 times second equivalent capacitance. Uh, this is CQ. Want anyone notices to? So this is 1/2 times The new quibbling capacitance is 15 times 10 to the minus six Farage's and the voltage is 66.7 volts. Ends comes out to 3.3 times 10 to the minus two jewels. So this value we get is greater than the original value, which seems like we're not conserving energy. But in order to double this plate separation, we needed to do some work. And so you take the energy stored in the original system 2.5, and then you do work to increase the plate separation. Now you're storing more energy, and you get this 3.3, uh, which is greater than the original energy stored. So they are equivalent, if you include the work done, which is gonna add potential energy to the system

I have everyone. This is the problem based on parallel combination off tapestries on energy stored in it. Here it is given to Cap Astres. Yeah, each off. Kappa States see, are connected in parallel as soon as I figured. Mhm if cross the potential difference off people to cap it. Stirs are charged to data B Boult. Then they're connected in battle as soon their figure. Yeah, the separation between the parade off, one off the capstan each developed. So new cap pistols becomes see y two because it is direct inversely proposed, not do distance between the plates. We have to fight total energy off the system before the separation between the plate is to be doubled. Second part, we have toe quite in potential difference after w the separation. She part total energy after doubling up the grade separation. Yeah. Oh, yeah. Cool. Indeed. Explain the difference off energy in two cases. Uh huh. I'm first case, you know, See? Wait. Oh, really? Total campus trains will be see policy. That is twice something. Yeah. So energy author system before half see, accumulated into data be swept, so it becomes see into data be square. So this is the answer off part eight? No, in part B way have toe file. Yeah. See this? That is see by two. See, potential difference becomes Do you have the final that we're finding? Yeah. No. In this case, CE equivalent Rickles C plus c my to that is three c vital que Before it's que often. See if the equivalent before that is to see into delta V three C by two in tow. Delta v file it. So from data, we finally you will get four Delta bi by three. Yeah, Sea park. Yeah. Total potential energy. After doubling goddess trees, the separation between the bridge half see equivalent and toe data be finally script. So it will be three c by two and took. So you will get group Oh, for three CEO Web D o Gipper. Total energy, after doubling and before doubling is great. So there is los off energy, which is usedto Yeah. Do the work toe to increase that separation. Yeah, then between the bridge. So yeah. Conservation off energy. Mhm. Yeah, to be followed that sort. Thanks for watching it

We have total energy stored equals to off them times have C v squared equals. See fisk word. Now the altered compositions is see prime equal See over to So we have to see Delta V equals. See Delta v Prime. Thus, see over to that the prime equals three C over to Delta v Prime. So we have Delta v prime it waas four third Delta V No. Now, in this party killer case, we will have you prime equals half see Delta v Prime squared thus half times half see dans delta v prime squared and substitute the value off Delta v Prime. Substituting this value, we will have the chance to energy or the altar energy is foresee over three times Delta v.

Hi. And this given problem late area in a parallel plate capacitor that is given as a separation between the plates. That is given as D. And potential difference applied between the place that is given as delta V. So its capacities will be given by C is equal to epsilon. Not A by D. Hands charges toward access charges toward over this capacity will be he was able to see we mean sea into delta v or we can say absolutely not uh by D into delta V. This is the access charges store over the plates of the capacitor initially. Now on disconnecting the battery, the charge store over the capacity will remain same. And now when the separation is doubled, separation between the plates so it's new capacitance that will be given us C dash is equal to absolute note A by to be So in the first part of the problem we have to find the new potential difference between the players in terms of the original potential difference delta V that delta v dash will be given bear will be given as charge. Delta Q access of charged will take you divided by C dash. New capacities for delta Q. This is absolutely not a delta V by D. And for capacities this is Absoluteal not a by two D. Now we may write it like have saleh not a delta B by D into to leave by. Absolutely not A on making a cross multiplication, cancelling this D by D absolute note by absolute note. Finally we get the new potential difference across the plates of this capacity to be equal to Device of the Original one. And that is the answer for the first part of this given problem. Now we have to find in the second part of the problem. Initial and final electrostatic potential energy is stored in this capacity. So for initial energy you either is given as half cv square or we this is delta V square. So here it will be have into ab silent, not A by D. Into delta. We square or rearranging the terms. We may write it like absolutely not A. There has to be square Divided by two d. initial energy and final energy store. That will be given us. Uh C dash delta v dish having a square of it. So it is have or C. Dash. That is absolutely not a bye judy. And for delta video square, this is two delta V having a square of it. So it will be half into absolute not a by two D. Into four delta v squared. So this is to to solve for canceling taste. Final energy hair comes out to be absolutely not a delta v squared divided by B. Which may also be compared with the initial energy and we may write it as U. F. Is equal to twice off you. I means the energy electricity potential energy store in the plates of the capacitor will be doubled if we make the separation between the players double. Finally in the 3rd part of the problem, we have to find work done to increase the gap between the plates and using work energy to from this work and will be equal to in France in energy. You f minus you. I So here it will be drives off U. I. Or you have. This is twice of you, I minus you. I mean this is you. I work done is equal to or we can say this is absolutely not a delta be square divided by device of B, Which is the answer for the 3rd and the last part of this problem. Thank you.


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