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Capacitance Homework Word Problem

  1. Feb 17, 2008 #1
    The problem.
    A 2.50uF capacitor is charged to 857V and a 6.80uF capacitor is charged to 652V. These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other. What will be the potential difference across each and the charge on each? (note that charge is conserved)


    2. Relevant equations
    PE= V/Q ; V=Ed; Q=CV; C= Eo (A/D); Vb-Va


    Attempt:
    What I started problem I used the forumla Q=CV to find the charge that is occuring in each case finding Q1 = .0021C and Q2= .0044 C. I really wasnt sure if this was even helpful or where to go from there. It appears to be a uniform feild so I know I can use V=Ed but we don't have a distance or an area in this problem.

    In order to get the potential difference I know you need to work with Vb-Va and possibly the potential energy. Should I use PE= V/Q then subtract the potential energies to find the work and then the poential difference?
     
  2. jcsd
  3. Feb 17, 2008 #2
    I would compute the capacitance of the two capacitors in parallel, then use Q=CV for this capacitor to get the potential difference across both, and then use Q=CV for the 2 origional capacitors to get the charge on each
     
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