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Questionparal plate capacitor of capacitance fully charged by battery of potential difference 23 vvhile the capacitor still being connecteo battery, two identical d...

Question

Questionparal plate capacitor of capacitance fully charged by battery of potential difference 23 vvhile the capacitor still being connecteo battery, two identical dielectric slabs of dielectric constants and are inserted on the leit and right sides respectively Till up tne entire gap betieen tne parallel plates the capacitor. Find te change stored energy (in J) pefore and aiter Introducing tne dielectrics

Question paral plate capacitor of capacitance fully charged by battery of potential difference 23 vvhile the capacitor still being connecteo battery, two identical dielectric slabs of dielectric constants and are inserted on the leit and right sides respectively Till up tne entire gap betieen tne parallel plates the capacitor. Find te change stored energy (in J) pefore and aiter Introducing tne dielectrics



Answers

A parallel-plate capacitor with plate area $A=2.0 \mathrm{m}^{2}$ and plate separation $d=3.0 \mathrm{mm}$ is connected to a $45-\mathrm{V}$ battery (Fig. 40 $\mathrm{a} )$ . (a) Determine the charge on the capacitor, the electric field, the capacitance, and the energy stored in the capacitor. (b) With the capacitor still connected to the battery, a slab of plastic with dielectric strength $K=3.2$ is placed between the plates of the capacitor, so that the gap is completely filled with the dielectric. What are the new values of charge, electric field, capacitance, and the energy $U$ stored in the capacitor?

So this question they're asking for a lot the capacitance, the charge, the electric field, the potential energy and the potential energy after and after we're including a di electric. So let's just write down our givens. We have a an area of two point zero meters squared and then we have a separation distance of three point zero millimeters. We can say this is equal to point zero zero three meters. Ah, and we have niece of oh equaling three five volts. So the first thing we should find, the capacitance, the capacitance, Sisa Bo is going to be equal to Absalon, not the electric primitive ity of free space times area divided by the separation. So this would be equal to eight point eight five times ten to the negative twelfth and then times two, divided by the separation distance of point zero zero three. So this is going to give us five point nine times ten to the negative ninth carrots. So this would be one answer for party. Ah, we could probably do que ce of theirs. The charge, Q subzero something any people to see subzero about the times he's observed. So this would be five point nine times ten to the negative nine and then times thirty five VOCs. So this is going to give us two point one times ten to the negative seventh. Um, cool ums. So this would be your second answer for party. We need e subzero. So this would simply be equal to the subzero over the separation distance. So this would be thirty five volts, divided by point zero zero three. And we're getting one point one seven times ten to the fourth volts per meter. So this would be our third answer for party. And then our final answer for party would be the potential energy. So this would be equal to one half a cease abo times Visa Bos, where this would be one half and then this would be five point nine times ten to the negative and ninth. And then this would be thirty five squared. So we have a potential energy of three point six one times ten to the negative six jewels. So this would be our final answer for part A. Now, for part B, they're asking us to include the dialect trick. So include Die electric and the dye electric constant K is going to be equal to three point two. So we can say that the new capacitance is equal to the dye electric, constant times the sea. Not so. This would be a three point two times five point nine times ten to the negative nine and we're getting this is equal to one point eight nine times ten to the negative eighth Claridge's and we know that the voltage doesn't change. So essentially, voltage V rather, cue the charge. Rather is going to be equal to see the um, at this point, we found both of these, so we can have one point eight nine times ten to the negative eighth and then the voltage doesn't change again. So this is only thirty. So this is only thirty five bolts once again. And this again equal six point six one times ten to the negative seventh Cool arms. So this would be one answer for me. This would be your second answer for me again. The battery's still connected. The voltage is current Stant. So the electric field e does not change on the separation distance hasn't changed as well as the voltage. So the electric field doesn't change. So e equals e not equals one point one seven times ten to the fourth of Volts Premier again, This would be your third answer for Part B. Um, And then your final answer for part B would simply be equal to half. And then this would be C V squared. So be won over to and then one point eight nine times ten to the negative eighth. This would be thirty five squared. So our final answer for part B the potential energy would be one point two times ten to the fifth. Rather tends to the negative fifth. Jules, this would be our final answer for part B. That is the end of the solution. Thank you for watching.

So we have to passengers connected in parallel. And this is just the general version of the last question without any specific values for the capacitance or voltage. So with the capacitors are just see And this potential is originally built The V before we changed the plate separation. So I'll say the energy stored before the plate separation you one is gonna be 1/2 times the equivalent capacitance before the plate separation times Delta V squared And the equivalent capacitance is just the some of the truth capacitors because they're in parallel. That's to see we have 1/2 times to see time still to be squared, which comes out to sea till two b squared and now we double thesis ation distance of the plates on one of the capacitors. Now, there's gonna be a new potential across the two capacitors, but we're gonna maintain the same charge. So if we know that the charge has to be the same, then the equivalent capacitance before the separation times at this still TV, it's gonna be the same. And this is this represents the total charge. So that has to be the same as the equivalent capacitance. After the plate separation times the potential difference. Which I'll just say's I don't know. V two Doesn't really matter what we call it now. OK, so this is the first equivalent capacitance. Now we want to find what happens after be moved the plates, so this is no longer a capacitance C. We know that the capacitance is related to separation distance inversely So if we double the separation distance, that's cutting the capacitance in half. So now we have a capacitor off 1/2 see, So the equivalent capacitance now is C plus 1/2 see, which is three have C. So now we can plug in down here. You have to see times Delta V equals three have seen times B two. Now it's isolated. Be to so divide by three have C So v two it's gonna be 4/3. That's two times 2/3 The seas cancel left with 4/3 Delta V. So the potential difference increased once we doubled the plate separation. Now let's find the energy of this new configuration. So collect use up to is 1/2 times this equivalent capacitance times The new potential difference, which is 1/2 times three halves. Uh, see, Time's or thirds Delta V squared. This should be squared. Okay, so now 1/2 times three halves, that's 3/4 c. And the 4/3 gonna be squared. And so it's gonna end up canceling with 3/4 and we're left with just 1 4/3 So we get 4/3 see times Delta V squared, and this is greater than CV square we got from the first part. That's because we had to do work to increase the separation distance attended plate that takes work. And if we add that to the original energy stored, um, before we changed the capacitance, you add the energy and the work, then you get the energy stored after you separate the plates.

Okay. This question has three parts. In the first part of the question, we have to find out the a dielectric. Okay and in the second part of the question we have to find out the energy stored in the get busy dad. In the third part of the question we have to find out the energy changes in the capacitor. So we are given the initial capacities are the key upstairs. Yes, it would be 1.5 Michael fettered in asia. Both bridges given us twin world and if I never voltages system too. Fight what? So the directorate constant is defined as we know told tough avi So K is equal to 12. We divide by five. Okay, He's given us two x 4. This is the dielectric constant. The second part of the question is to find out the energy stored in the vista. We know that when a dielectric is inserted between the plates of the capacitor then final energy. I need energy degrees is by keita get that scenes. K. Is greater than burn. Therefore the final energy is also less than the initial energy. We know that the energy is given is half. See not either. Are scared. If we put the value over here then we we get one out of two. I want to play by 1.5. multiplied by 10, raised to the power minus six months and multiply you by 24 scared. So we get the finding the energy. Yes one by zero eight. Multiply by 10, raised to the power minus for jews. Yes. Fortune. Yeah. And the dielectric and the capacitance increases by Okay dad so see he is equal to to find four multiply by 1.5. Multiply by 10 raised to the power minus six. So the capacitance Is equal to 3.6. Multiply by 10. Raised to the power minus six figured. So the changing energy is given this, you c minus you not is equal to 4.5 multiplied by 10, raised to the power minus five minus one x 0, eight. Multiplied by 10 raised to the power power minus name. So we get change in energy. He is equal to six fine, three multiplied by 10 raised to the power eight Junior.

Have a parallel plate capacitor, so basically and obviously initially they are. They are expressed. Within them are vacuum and a potential difference off Delta. Be over the ridge applied on the Let's Say, the capacitance you see. Basically, let's see the capacitance. There's a potential difference Delta over there. So we need to find that person for the Chargers in our the story in the capital. So basically a Espace between them is vacuum. So the charge on them that will be nothing but the Q equals to see into Delta B C, the capacitance and on the energy you sort, it will be half into see into Delta V A square. Okay, under there is other formulas. Well, but we're going dealing with this one because you already know this. Maybe the next part when the direct question Cajun start between them like this one. The constant care. So basically now new capacitance C that will be but the few times off, See, obviously, you know, get multiple advocate and see so new energy, you will be half times off. By this logic, we can say that new energy you that will be half times off C and C H multiple cases. Okay, times off initial see into Delta V s were so silly. Newer genital collapsing


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