A constant potential difference of 12 V is maintained between the terminals of a 0.25-uF, parallel-plate, air capacitor. (a) A sheet of Mylar is inserted between the plates of the capacitor, completely filling the space between the plates. When this is done, how much additional charge flows onto the positive plate of the capacitor (see Table 24.1) ? (b) What is the total induced charge on either face of the Mylar sheet? (c) What effect does the Mylar sheet have on the electric field between the plates? Explain how you can reconcile this with the increase in charge on the plates, which acts to increase the electric field.

Video transcript

Everyone today. We are going to calculate the supplementary charge that will flow onto the plates due to the insertion of the piece of glass and then also the charge generated on each side of the glass piece. And lastly determine the impact of the glass piece on the electric field between the plates. Okay So first what we want to do is to probably start by identifying all the necessary information given. So first what we have is the actual capacities itself which is going to be 75 and f. Which is Nano fraud. And remember that nano, we have to multiply it by 10 to the power of minus nine to become the S. I. Unit for rod. And then the fault or the potential difference is gonna be nine fold which is given here and the glass has a dielectric constant Of K. equals 2 7. And those are all the necessary information given in our problem. Okay so now we can actually start recalling um some of the equations that we're going to use in this particular problem. So to calculate the supplementary charge, what we have to do is to calculate the difference on the charge that inserts inserting the actual glass plate or piece of glass is gonna actually change. So recall that the capacitance of the capacitor. C. Is going to be given by Q over fee with Q being the charge. So in this case we can rearrange this to give us the charge which is going to be C. Multiplied by fee which is the charge stored on the plate and when the space between the plates is completely filled by the dielectric, the ratio of the sea here and see. Not before the insertion is going to be called the dielectric constant of the material which is known as K. Here. So okay it's going to be C over C. Naught and with some rearrangement then C. Will equals to K. Time. See not like so okay so we are given the K two B. Seven and the free to be nine. So with that we can actually start up solving this problem. Okay, so before the filling of the space with the glass then the initial charge is going to be given by this formula here. The initial, I'm just gonna indicate that with Q. Not so Q not will equals to see. Not multiplied by fee. Like so and after the insertion Q. Then will become C. Multiplied by fee. But remember the C. Here, we can substitute to see here with our dielectric formula here. So the sea is essentially just came multiplied by C. Not multiplied by fee. Like so so the supplementary charge that will flow onto the positive plate due to the introduction of the glass piece is going to be represented by delta Q. Which is essentially Q minus Q. Not so after minus before which is going to be equals two, multiplied by C. Not multiplied by v minus C. Not multiplied oops see not multiplied by V. So this will then be C. Not Multiplied by v. K -1. There you go. Now we can actually start plugging all the known values that we have. Right The C. Here is gonna be seen. Not so see not is then going to be 75 9 of a rod or in our case is 75. Multiplied by 10 to the part of minus nine for rod. Like so and then the V. Is gonna be ninefold and delta constant is gonna be seven minus one and delta Q. Is then going to be four point oh five times 10 to the power of minus six. Cool. Um And that will be the answer to the first part of our question which is going to be the supplementary charge that will flow. Okay now that we find that we can move on to the second part which is the charge generate on each side of the glass pieces. Okay so recall that the magnitude of charge per unit area induced on the surfaces of the dielectric is represented by this sigma right here. So sigma i which is the majesty of the surface charge density on the capacitor plates is gonna be represented by the sigma, multiplied by 1/1 over K. With K being the dielectric constant. And recall that there is a formula that correlates sigma and Q. Which Q. I. Is going to equal to sigma. I multiply by A. And Q is going to equals two sigma, multiplied by A. So Q. I. Is the induced charge on the glass and the Q. Is the charge on the capacitor. The sigma is going to be the magnitude of the surface charge density on the capacitor plates. And the A. Is going to be the area of the capacitor and of the piece of glass. So the capacitor and the glass piece will actually have the same area. Because it is known from the question problem statement that the glass is actually going to be totally filling the capacitor. So therefore we actually can um substitute this uh formula into this and cross out the A. Because the A. Is going to be the same. Okay so I'm gonna start with substituting the sigma's with the Q. I. And Q. So sigma I hear is then going to be Q. I over A. And sigma is then going to be Q over A. So sigma is gonna be Q. I over A equals Q. Over A. With multiplied by 1/1 minus one over K. We can cross out the ace and we can get Q. I. Two B equals two. Q. Multiplied by one minus one over K. Just like so okay and then recall that the Q. Here we can we actually have already um calculated previously which is going to equals to K. Multiple basically not multiplied by fee. Which I'm just gonna substitute this formula here. So this is gonna be K. Multiplied by C. Not multiplied by v. Multiplied by 1 -1 over K. Like. So so this is essentially gonna be obtain from here just like so and we actually have all this necessary values that is needed to calculate this. So what we need next is just to pretty much um plug everything in. Okay so the K. Is going to be seven and then see not is gonna be the capacity which is 75 multiplied by 10.10 to the part of minus nine, fraud, multiplied by the potential difference. Nine multiplied by one minus 1/7 with the dielectric constant of seven. And then therefore the Q. I. Is then going to be four point oh five multiplied by 10 to the power of minus six. Coolum. And I'll be the answer to our second part of this problem. Okay so we know that the magnitude of induced charge Q. I. On each phase of the dielectric is going to be four point oh five times 10 to the power of minus six. Cool. Um So now moving on to the third part which is calculating the charge generated on the determining the impact of the class piece on the electric field between the plates. Right? So the third product actually asked us to determine the impact on the electric field and recall that the electric field formula or the electric field can actually be calculated by defining the potential difference over the separation distance. So in this case the fee and the D. Is unchanged, right? Because all we do is pretty much just inserting a glass inserting the glass on to in between the capacitor. So essentially the potential difference remains the same, which is not in fault. And the separation distance is also the same. So because the fee and the D. Are unchanged, the electric field actually between the plates is going to be uniform as well. Its magnitude is the potential difference divided by displayed separation. So the electric field will not change when the dielectric is inserted. So because fi nD is unchanged, then E will then be also unchanged and that will be the answer for the third part. Okay, so we know that first, the supplementary charge that we will flow is four point oh five to the time. Stand to the part of minus six colon. So we will we will be able to cross out option C. And also option D. And then we know that the charge generated on each side of the glass piece is also going to be 4.05 times 10 to the power of -6. Cool. Um well, we then know for the last part that the electric field will not change or doesn't change when the glass is inserted. So the answer is then going to be option a here and that will be all for this particular problem. If you guys are still confused, there are other videos on this similar topics, so make sure to check on that. And that will be all for this particular problem. Thank you.