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Help with electrostatics : Charging and reconnecting a capacitor .

  1. May 17, 2012 #1
    1. The problem statement, all variables and given/known data

    So I came by this question yesterday which was in Solved examples part of my Textbook . I looked at the Solution and was not able to figure out how ?
    The Question says .
    A 5uF capacitor is first charged by applying a Potential Difference of 24 V . a different Capacitor of Capacitance 6uF is charged by applying a Potential Difference of 12 V . These capacitors are now connected such that the Positive plate of the first capacitor is connected to the Negative Plate of Second Capacitor and vice versa(does this mean they are connected in series or is there a different name for such configuration?) . Find the new Potential Difference and the New charge on the capacitors .

    2. Relevant equations

    The only formula used would be Q = CV

    3. The attempt at a solution

    The solution had this C1 = 5uF V1 = 24uF therefore = Q1 = 120uC
    C2 = 6uF V1 = 12uF therefore = Q2 = 72uC

    Then they found out what is called effective potential (thats what confusing me ) . They used this equation

    V{effective}(C1+C2) = V1C1 - V2C2
    V{eff} = 4.36 V
    ################################################
    then they Calculated the V{effective} and the new charges . what I am not able to understand is how did they came by this formula ??
    Also when the configuration is like Positive plate is connected to positive and negative to negative(does this mean they are connected in parallelor is there a different name for such configuration?) the effective Potential difference comes out to be 17.45 V . How ? Please explain I just cant get it .
    Thanks in advance
     
  2. jcsd
  3. May 17, 2012 #2
    The +ve to -ve connection means that two anr now connected in series(That is what basic circuitry says)
    Let us now analyse the question:

    ( E means x10 to power)
    5uF charged at Pd of 24v
    6uF charged at Pd of 12V

    The charge stored in the 5uF capacitor = Q1= CV=5E-6 Fx24=1.2E-4C
    " " " " " 6uF " =Q2 =CV= 6E-6 x 12= 7.2E-5C
    Total charge(Qe)= Q1 +Q2=1.92E-4C ----New Charge
    The total Capacitance (Ce)= (C1xC2)/(C1 + C2)= 2 8/11 uF

    New PD= Qe/Ce = 70.4019V

    Do you now get it friend?
     
  4. May 17, 2012 #3
    But the Pd across them is not 70.4019 its 4.36 V
    http://www.evernote.com/shard/s92/sh/dd550448-5624-437f-a07c-66acce892c66/3eb65a63682c8d778a8b2bd326062c42 [Broken]

    you can see that here
     
    Last edited by a moderator: May 6, 2017
  5. May 17, 2012 #4
    Oh I am just checking the solution you gave up there. The V effective is the resulting PD across the two Capacitors.
    I tend to think that when connected in series, the charge in any two Capacitors tend to oppose one another so that is why they are being subtracted. Oh this is confusing even me now...
     
  6. May 17, 2012 #5
  7. May 17, 2012 #6

    turin

    User Avatar
    Homework Helper

    I don't know about this effective potential, so if you must understand your solution manual, then I cannot help you.

    I would do it this way:

    calculate the initial charges on each capacitor from your basic equation Q=CV (as you did)

    calculate the equilibrium charge after connection Qeq=|Q1-Q2| (they are subtracted because the negative from one plate partially cancels he positive from the other plate, and vice versa)

    calculate the equivalent capacitance of the combination (which would be effectively parallel as far as the resulting voltage is concerned) Ceq=C1+C2

    determine the resulting voltage for the resulting combination from your basic equation Qeq=CeqVeq
     
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