A 3 F capacitor is given a potential difference across its plates of 10 V. What is the charge built up on its plates? If the source of the potential difference across the plates is removed, but the plates maintain their charge, what is the new potential difference across the capacitor if the distance between the plates is doubled?

Alright, guys. Hopefully, you were to take a stab at this one your own. We have a 3 farad capacitor. It's given some potential difference. And the first part here, we're supposed to figure out what is the charge that is built up on its plates. So, in other words, that q is our target variable. We know that we can relate the charge to the capacitance and the voltage for any capacitor using q equals CV. Right? So we have q equals CV. We have the capacitance and we have the voltage, the potential difference. So this is gonna be 3 times 10, and that equals 30 coulombs. And let me rewrite that a little bit clearer. So it's 3 times 10 is 30 coulombs. Now what happens is you're gonna remove that potential difference to the battery, whatever it's hooked up to, but the plates are still gonna maintain their charge. And then we're supposed to figure out what is the new potential difference across the path capacitor if the distance between the plates is doubled. And we're also gonna assume that it's a parallel plate capacitor here. So in other words, what we're really trying to find is now that we have this 30 coulomb charge, which is going to be maintained, what is the new potential difference after you have basically doubled the distance. In other words, d new is gonna equal 2 of what it did before, whatever distance it did. So this is gonna be sort of like a situation where we're kinda gonna be working with numbers and variables. So let's go ahead and check out how that's gonna work out. So we know that the new voltage is gonna be equal to well, let's see. If q equals c v, then that means that v equals q over c. Now what happens is we know that this q remained the same, so we won't have to do anything with it. But we know that this capacitance, which by the way is equal to epsilon naught times A over D, what happens is the distance will change. So in other words, this distance is gonna be d nu. We need to figure what the new capacitance is going to be. So we could just plug in our expression for d nu over here. We're just gonna plug in this 2 d down into this formula, and we have epsilon not a times 2, and then we have d. Now what basically happens is we know that this capacitance is equal to 3 farads. Oh, sorry. We we know that the old capacitance was equal to 3 farads. Now, what happens is when we pull out this 2 that's in the denominator right here, we're basically gonna get 1 half of epsilon naught a over d. So it's gonna be 1 half of the original capacitance. The original capacitance was 3 Farads. So that means that the new capacitance, so C new, which I wanna put over here, C new, is going to be, let's see. It's gonna be 1.5 Farads. So, that means that the new voltage is gonna be the original coulombs, which is 30 coulombs, divided by the new capacitance, which is gonna be 1.5. So that means that the new voltage is just gonna be 20 volts instead of 10. So that's what happens when you change the distance in a capacitor. Alright? Let me know if you guys have any questions.

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