A $7.7$-$\mu \textrm{F}$ capacitor is charged by a 165-V battery (Fig. 17–43a) and then is disconnected from the battery. When this capacitor $(C_1)$ is then connected (Fig. 17–43b) to a second (initially uncharged) capacitor, $C_2$, the final voltage on each capacitor is 15 V. What is the value of $C_2$? [Hint: Charge is conserved.]

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. Initially this capacitor one is charged up to some voltage V i and that will accumulate some charge Q one initial equals to its capacitance C one times its initial voltage V i and then sometime later the battery is cut out of the circuit and it's the second capacitor C two is put in its place. And when that happens the voltage across capacitor one is going to equal the voltage across capacitor two because these two plates are connected by a conductor and so they'll be at the same potential because charge will distribute between these two plates here. Such that there's becomes no voltage between them. Because if there is a voltage, that would mean that there's an electric field between these two plates which the electric field produces a force on charges and then that does happen at the small instant when they're initially connected but then eventually the charges actually move and distribute such that there is no electric field and therefore no voltage. These plates here are also connected by conductor and so they have no... they're at the same potential as well. And so the potential difference between these two sides here is equal to the potential difference between these two sides here. And so that is to say that V one final equals V two final and so let's just call it V final there's no need to distinguish between one and two since both the same. Now the final amount of charge on capacitor one plus the charge that a capacitor two accumulates is going to equal the total that capacitor one had initially because there's no charge being destroyed here, there's nothing's being neutralized it's just that all the charge that this thing initially accumulated is now going to get distributed between these two capacitors and the total will remain the same. And so we know that Q one final is capacitance one times final voltage and the charging capacitor two is going to be capacitor two times the final voltage. And so we can substitute into this line here by saying Q one f is C one times V f and Q two is C two times V f nd that equals C one times V initial for Q one initial. We're going to solve this for C two. We'll take this term to the right hand side which makes it a minus C one V f and then factor out the C one and you get C two V f equals C one times V initial minus V final and then divide this by V final and divide this by V final and you get capacitance two is capacitance one times the difference between the initial and final voltage divided by the final. So that's 7.7 times ten to the minus six farads times a 165 volts minus 15 volts divided by 15 which is 7.7 times ten to the minus five farads for the capacitance on C two.