A parallel-plate air capacitor is to store charge of magnitude 240.0 pC on each plate when the potential difference between the plates is 42.0 V. (a) If the area of each plate is 6.80 cm^2, what is the separation between the plates? (b) If the separation between the two plates is double the value calculated in part (a), what potential difference is required for the capacitor to store charge of magnitude 240.0 pC on each plate?

Video transcript

Hi everyone today we are going to determine the distance d separating the two parallel capacitor plates. And also the new value of the potential difference for the new that will charge each plate to the same charge. So what we want to do first is to probably create a list of what is given in our problem. So first we have the initial potential difference which is 35 fold and then we have the area of the place which is 7.20 centimeters squared which in S. I. Unit we can multiply it by 10 triple of minus four And that will give us 7.20. Um The time stand to the power of -4 m squared and then last. We are also given the charge which is cute, which is going to be 300 PICO column, multiplied by 10 to the power of minus 12 column over PICO column And as a unit, this will then be 300 times then to the point of - column. Just like so okay, so now we can actually start solving for this problem by recalling what kind of formulas we want to use. So using the parallel plate capacitor equation that gives us the separation distance d. We can use the formula of capacity will equals to the permitted city of space epsilon not multiplied by the area over the separation distance D. Now we know that out of all of what is given here, we do not know what C. Is right. The absolute not as a constant value. The area is given while the D. Is the one that we want to look for. So we will have to find what the capacity is recall to calculate capacity. We can use this film right here and we can just find it by multiplying dividing the charge by the potential difference. So combining these two formula should give us the separation distance here. So I will do exactly that. Um so I will just equal this to formula here. So this will be key over female equals to absolutely not A over D. And therefore our D will equals 23 times epsilon not A over Q. Just by rearrangements. We actually know all of this value from here. So we can just pretty much plug in all the values and the absolute not as essentially a constant that we need to remember. So that is going to be 8.854 times 10 to the power of minus 12. How long over newton times meter square and A is 7.20 times 10 to the minus four m squared. Along with the fighting off this with our charge Q which is times 10 to the power of minus 12 column. And this will actually give us a distance of 7.44 times 10 to the power of minus four m. So in this unit here in millimeters distance will then be 7.44 times times four minus four meters times 10 to the bar of three millimeters to meters. That's just the conversion that we have. This will be 0.744 millimeters like. So. Okay so now we know that we can actually eliminate option C. And also option D. And we can start with actually solving and finding the new value of the potential difference we knew that will charge each plate to the same charge. Okay so let's start with figuring out our D. New. So we know that the new is gonna be three times our initial distance and solving this problem. We actually don't really need the exact value. So I'm just gonna leave it as three D. Here. So the new is three times the original de and the way we want to solve this is to actually still use these two formulas right here and see the comparison with our new potential difference for new. So let's start with some plugging in D. New into our first formula here. So um see new will be E. Not times A over the new. And then we know that in all times a over D. New. The new is three times D. Three times D. And recall from our previous formula here that all of these things are essentially see or our old C. So The sea new is essentially just see over three. Okay, now we know what the new capacitance of the new capacity is. And then we can actually pluck see new into our second formula to get what our fee knew or what our new potential difference is. So I'm going to do that here. Uh see new, it's gonna be Q over fee new. We know that the Q. Is gonna be the same. So then fi new it's gonna be Q over seeing you like so so the Q. Is the same and the C. Nu sc over three Q over C over three. And this will then be three Q oversea. And we know from last time that Q oversee, let me change that to a rat que oversee will equals two ft from here, fee equals two Q oversee. Like so so essentially our free new is just going to be three times free, which is going to be three Times 35-fold because that's the original potential difference. And fee new will then be 1054 just like so okay, now we can actually solve this problem because we know that D. Is essentially 0.744 millimeter and is one of five and that will leave us with option B. So option B will be our answer. Let me know if you guys still have any confusion and we definitely have a lot of other videos in this particular topic. So make sure to check it out as well and that will be all for this particular practice problem. Thank you