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Homework Help: Finding Capacitance of a capacitor

  1. Jun 16, 2010 #1
    1. The problem statement, all variables and given/known data
    A 100 μμF capacitor, C1, is charged to a potential difference of 50 Volts. The charging battery is disconnected and the capacitor is connected to a second uncharged capacitor, C2. If the potential across C1 drops to 35 volts, what is the capacitance of C2?


    2. Relevant equations
    Q = CeqΔV


    3. The attempt at a solution
    Q = CeqΔV
    Q = (C1 + C2)ΔV
    Q = (100x10-12F + C2)(-15V)

    I am confused on what to do from this point on, what do I do with the Q?
     
  2. jcsd
  3. Jun 16, 2010 #2
    Re: Capacitance

    I think you're confused about how the capacitors are working here. First, C1 gains charge Q on its plate. Next, C1, still with charge Q on its plate, is hooked up to C2, a capacitor with zero charge. You need to realize that no matter the distribution of charge on the two capacitors' plates, the sum cannot exceed or undershoot how much charge was initially put on C1. Otherwise, you have created or destroyed charge.

    Since the capacitors are in parallel, you know they share the same voltage, so you can calculate how much charge is on C1 after the connection by using 35V and prior to the connection by using 50V. Using total charge in the system (initial charge on C1) and charge on C1 after the connection, you can subtract the two to find how much charge must be on C2. Finally, using the charge-voltage relationship of a capacitor (q = cv), you can calculate the capacitance of C2.

    Using -15V seems really off. The delta V is usually written V and stands for voltage across the capacitor. It's not "change in voltage" as in "I went from 50 to 35 volts. Therefore, -15"
     
    Last edited: Jun 16, 2010
  4. Jun 16, 2010 #3
    Re: Capacitance

    How do you know their in parallel?
     
  5. Jun 16, 2010 #4
    Re: Capacitance

    It's the only way to connect a two capacitors together.

    From the problem:
    If component A has connections A1 and A2 and component B has connections B1 and B2, you can connect them only two ways: A1 to B1 and A2 to B2 or A1 to B2 and A2 to B1. Either way, they share two nodes with each other, the definition of parallel.
     
  6. Jun 16, 2010 #5

    collinsmark

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    Re: Capacitance

    I'm a little uncertain of your notation, "100 μμF." Is that a typo with the extra 'μ' (making the original capacitor 100 μF = 100 x 10-6 F) or do you really mean 100 x 10-12 F as stated in your attempted solution? If you really mean 10-12, that factor is commonly denoted by 'pico', making the capacitance 100 pF (Units of 'μμF' is not common notation [as a matter of fact, this is the first time I've ever seen such notation]. Picofarad, pF, is very common. So I suspect if you were given units of 'μμF' it might be a typo.)

    Anyway, the key to solving this problem is realizing that charge is conserved before and after the capacitors are connected. The charge on the first capacitor, before the capacitors are connected, is the the same as the sum of the charge of both capacitors after the capacitors are connected. With that in mind, how much charge was transferred to the second capacitor? :wink:
     
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