A 5.00-uF parallel-plate capacitor is connected to a 12.0@V battery. After the capacitor is fully charged, the battery is disconnected without loss of any of the charge on the plates. (a) A voltmeter is connected across the two plates without discharging them. What does it read? (b) What would the voltmeter read if (i) the plate separation were doubled; (ii) the radius of each plate were doubled but their separation was unchanged?

Video transcript

Hi everyone in this practice problem. We have a problem of parallel plate capacitor where the student coupled 10 micro for a capacitor made of two parallel circular plates, circular is a key word here. Each of radios are and separated by a distance of the, connected it to a 15th voltage source and let it charge completely before disconnecting. And this problem for particular asked us to first determine the reading on the fault meter connected to the charge capacitor of 15 fold and then determine what the fault meter will read when the student increases the distance between the two plates 25 D. And lastly what the fall meter reading is going to be when the plate radius is stretched to three are but the distance will remain to D. Okay so to first do this example we're gonna just um start with tackling part one. So for part one we know that when the capacitor is connected to the charge uh source, then enough charge flows onto the plate to make the voltage across the plate will be exactly the same as the voltage source. So I'm just gonna write down same as voltage source Which is going to be 15 fold just like so so we can immediately eliminate option A. And option B. And then next for part two. Um we are asked to determine their fault meter reading when the distance between the two plates is increased to five D. And in this case recall that full capacitance and charge is correlated with one another with this formula. So uh fault age will equals to the charge divided by the capacitance. And the capacitance can be calculated using this formula where it is going to be absolutely not times A over D. Which is the separation distance and this is the area. So the capacitance of the capacitor before modifying the separation distance is going to be represented by this. Now that the student increases the distance between the two place to five D. We want to find what Seenu is. So seeing you will equals two epsilon non absolute just a constant, multiplied by A. Which is going to stay the same in part two and the distance will be increased to five D. So this is going to be the new C. Knew where a quarter of the not absolute absalon, not A and D. And if you guys can observe clearly here this is going to be pretty much see on this site. So see new is essentially just going to be 1/5 0 point to see just like that. Okay, so now that we find what Seenu is, we can find what the new voltage is going to be. So using this formula right here, free new is going to be Q over C. Knew where in this case the Q. Is going to be the same. So if we knew it's going to be Q. Over we're just gonna substitute to see new value here, 0.2 C. And this will give us a value of Ashley five Q oversee And if you guys can see the key oversee is pretty much the old fee here. So we can conclude that fee new or the new foal meter reading is going to be 53 with the fee equals to 15 fold, so this is going to be five multiplied by 15 fold, Which going to equals to 75 volt just like so okay, so we can kind of conclude that part, the option that is going to be our answer, but we are going to first try to tackle part the third part of this problem to make sure. Okay, so the third part of the problem, we want to tackle it pretty much using the same Method as Part two, but in this case we want to note that when the radius as three R. Then the area of the circular which is going to equal to P r pi r squared is going to increase by pi. This is uh I'm gonna do by multiplied by three R squared, this is the new one, this is new and therefore the new a or the new area is going to be nine pi R squared. So if the radius of the place is tripled, then the area will increase by a factor of nine. Okay, so similarly to just how we tackle part two, we just want to start with finding what the C. New is and then substituting it to this formula here, so see new is then going to be epsilon not multiplied by a new, multiplied by same distance initially. And this is going to be absolutely not multiplied by nine pi R squared Divided by D. or essentially nine absolute not A over D. And we know that this part is equals to normalcy now we want to substitute this to the fault, charge and capacities formula to find what the new is. The new equals Q. Oversee new and that's really close to Q over nine C. Just like so and the key overseas is essentially the old fee so it is going to be 1/9 feet And this is going to then be the fee is going to be 15 fold Which is known divided by nine and this will come out to be 1.7 fold just like so Okay, so we know that part three will equals to 1.7 fold the four m reading and then part two, the four m reading really close to 75 the first one is going to equal to so that will correlates to part our option D. Of our answer just like so, so that will be all for this particular problem. If you guys have any questions or any sort of confusion, please feel free to check out our other lesson plan on this or other lesson videos and that will be all. Thank you