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Q2: A parallel plate capacitor of area A and separation d is connected to the battery to charge the capacitor to potential difference Vo, calculate the stored energ...

Question

Q2: A parallel plate capacitor of area A and separation d is connected to the battery to charge the capacitor to potential difference Vo, calculate the stored energy before and after introducing dielectric material?

Q2: A parallel plate capacitor of area A and separation d is connected to the battery to charge the capacitor to potential difference Vo, calculate the stored energy before and after introducing dielectric material?



Answers

A parallel plate capacitor has plates with area $A$ and separation $d$. A battery charges the plates to a potential difference $V_{o} .$ The battery is then disconnected and a dielectric slab of thickness $d$ is introduced. The ratio of energy stored in the capacitor before and after the slab is introduced, is (a) $K$ (b) $\frac{1}{K}$ (c) $\frac{A}{d^{2} K}$ (d) $\frac{d^{2} K}{A}$

Okay question is uh C. V. And you and the Q. R. Capacitance potential difference, energy store and the change of and the charge of baron and played capacities respectively. So the quantities that increase when electric slap is introduced between the plate without disconnecting the betrayal. So we have to find out the quantities which are increases due to the traditional the slip. So we have given the circuit which is we can be drawn as this is a good persistency. And we use the applied voltage when the electric is inserted. So when the electric inserted then okay piston Cds is a delicacy. So therefore capacitance increases. No we discuss about the voltage. So therefore the potential difference which is we his his remain constant because we do not disconnect the battery and the energy store use it will do. Uh huh. See we scared. So therefore you take is equal to as the capacitance increases. And it becomes C. D. We scare and the C. D. H. Is equal to K. See therefore the energy store you day she is equal to Went out of two Casey. We scare and you dish is equal to okay. So the energy also increases. And when we talk about charge, Church is equal to All right, see is came out of C. Is equal to supply voltage is equal to count of C. Is equal to count it out of cd. So therefore some days is able to see the fish out of sea. Thank you. Therefore county is equal to. Okay. Thank you. So charge increases by a factor increases. Bye. Factor. KK

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.

According to this user. Just initial charge. Yeah. On the mhm capacity. Yeah. Yeah. Yes. You're not. Egypt was to we know that killing is equal to see we so see not we and she is a silent or so. Uh Absolutely not. We by the and as the battery is disconnected. Mhm. So judge. Mhm. You mean unchanged. Mhm. Yeah. So music was too, wow. Do you not? And you know the A. F. Cell and not we buy the and after instead. Mhm. Directive capacitance C days is equal to case you know where your directory constant and potential readers as it goes to if you buy cds because I will be the same and electric field. Mm hmm. So for the release of three days we know that do not by case. You know it was too we bike and filled. He calls to three days by day. It was to be by candy and initial energy. Yeah. Initial energy. You know that he goes to half she know we square is used to put the value. C. Note then a I'm sellin or to be Squire by two days and final energy. Mhm. You refugee was to have see days he goes to put the values then get we get half. She says he was the case. You know. Yes, he not into videos. It was to rewrite K square after simply by we get a absolute north re squired by two K. D. And work done scenes well done. Okay. It was cool, decreases decreases in potential energy. And so We can clear w. z equals two minus delta. You which goes to minus final energy minus initial energy. So after the book values you? No, my nephew. Everything was to have A. S. L. In north. We squired by day into 1 -1 by, okay, this is our final onset, and this answer, given in Austin would be hands often be each correct? Okay.

Hi. In the given problem plate easier Of the parallel plate, capacitor is given as a is equal to 0.24 Notary square. The angel of the battery, which is applied across the plates is we is equal to 12.0 Ball and made separation here Is given as B is equal to 5.24 millimeters. Now, in the first part of the problem we have to find the charges all over the place which has given us the product of capacities of the capacity with the potential explosive splits here. The expression for the capacitance is absolutely not by the the expression for the capacitance of a parallel paid capacity. So the expression for the charge here will become absolutely not we by D. So plugging in all known values for excellent northeast was 8.854 into 10 days, par -12, Multiplied by area 0.224. Multiplied by potential. Well world Divided by separation between the plates 5.24 Millimeter or 5.24 into 10 days per -3 meter. So finally, this charge comes out to be 4.54 into 10 dish part minus nine column. Or we can say this charge is is equal to 4.54 nano hola. Answer for the first part of this problem? No. In the second part of the problem we have to find energy stored in this capacity that will be given by the expression of you into V. We know the value of charge. We have just found in the first part of the problem and that was 4.54 into tended to power minus nine, Multiplied by potential apply, which is 12 old. Finally, this energy comes out to be 2.72 Into tennis par -8 jules. Is the answer for the second part a problem. Finally, in the third part of the problem, we have to find electric field between the plates of this capacity. That is given as the potential gradient is this is wealth world invited by 5.24 mm Or disease, 12 old invited by 5.24 into tended bar -3 m. Finally, this electric field and so to be 2000 290 volt four Metre, which is the answer for the third and the last part, it's probably thank you.


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