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Homework Help: Problem involving capacitance and water droplets (answer check)

  1. Jun 24, 2009 #1
    1. The problem statement, all variables and given/known data

    A hollow copper sphere is originally filled with sea water, hung from a nylon
    thread and raised to a potential of 100V with respect to earth by connecting it briefly
    to a voltage source.
    The water then leaks out through a hole in the bottom of the sphere, and runs down a
    long thin vertical copper wire which is soldered to the bottom of the sphere. At the
    end of the wire, the water very slowly forms into a succession of drops, which break
    away and fall to earth when the drop radius reaches 2 mm.
    The external radius of the sphere is 4 cm and you may assume that if the wire and the
    copper sphere were isolated from each other, the capacitance of the wire would be one
    tenth of that of the sphere (the wire is 0.3m long and of 10-2mm diameter).
    How many drops have to fall for the potential of the sphere to reach 1 volt?

    2. Relevant equations



    3. The attempt at a solution


    my attempt

    Capacitance for the shere: Cs=[tex]\epsilon(r)[/tex][tex]\epsilon(0)[/tex]4pi(0.4)

    Capacitance of nylon thread: Cs/10

    Capacitance of Water droplet: Cd=[tex]\epsilon(r)[/tex][tex]\epsilon(0)[/tex]4pi(0.002)


    This is what i think is happening in this problem:
    Initially all the charge and voltage is stored in the sphere. As the water leaks out the bottom it carries a charge with it. This reduces the charge and voltage of the sphere. the droplet will deposit some charge on the copper wire however i dont think this matters to our problem since we only want to know the voltage across the sphere. so long as the water droplet takes the charge away from the sphere it doesnt matter where it deposits it. ( i may be wrong about this.)

    Initial charge on sphere: Qs= V1[tex]\epsilon(r)[/tex][tex]\epsilon(0)[/tex]4pi(0.4)

    Charge on droplet: Qd= V1[tex]\epsilon(r)[/tex][tex]\epsilon(0)[/tex]4pi(0.002)

    Change in voltage:



    So i think this means

    dV/dN = -V(0.002/0.4) where N is the number of droplets

    the solutoin to this is V=100(v)exp(-(0.002/0.4)N)

    solving this for when V=1 gives answer of N = 921 droplets.

    Could someone please just check my answer and let me know if they think its correct or not, and if not, where i might be going wrong?
    Thank you
  2. jcsd
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