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Homework Help: Capacitor efficiency problem

  1. Feb 12, 2007 #1
    1. The problem statement, all variables and given/known data

    A 0.45 µF capacitor is charged by a 1.5 V battery. After being charged, the capacitor is connected to a small electric motor. Assuming 100% efficiency, to what height can the motor lift a 4.7 g mass?

    2. Relevant equations

    I really need some direction to start off this problem. I think that I probably need to use U=1/2QV=1/2CV^2=Q^2/2C and/or Q=CV but I have no idea how to go about doing this.
  2. jcsd
  3. Feb 12, 2007 #2
    The whole idea of this problem is that if the efficiency is 100% no energy will be lost. So all http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html" [Broken] will be used to lift that mass. Now what would be the energy necessary to lift that mass to a height h?
    Last edited by a moderator: May 2, 2017
  4. Feb 12, 2007 #3
    So...would I have to use Q=CV to find out Q and then use U=1/2QV to find out U ?
  5. Feb 12, 2007 #4
    You could use any of the 3 formulas you mention, though since you know [tex]C[/tex] and [tex]V[/tex] the fastest would be to use [tex]U=\frac{CV^2}{2}[/tex]
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