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Coalesce of mercury dropssix at different potential

  1. Aug 29, 2015 #1
    Six mercury drops of equal size given potential of +3v and two other drops are given -3v potential. If they coalesce what is final potential

    This question had been eating my brain i know solution if they are given same potential but what to do when they have different potential
     
  2. jcsd
  3. Aug 29, 2015 #2

    TSny

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    Hello and welcome to PF!

    If you can work the problem for the case where all 8 drops initially have the same potential, then you should find that it's not much harder to deal with the case of different initial potentials.

    Can you show how you get the answer when all 8 drops have the same initial potential?
     
    Last edited: Aug 30, 2015
  4. Aug 29, 2015 #3
    Um i solve it using formula
    V=n^(2/3) vs where n is no of drops and vs is potential of each drop but it works for condition when all are provided same potential
     
  5. Aug 29, 2015 #4

    TSny

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    Do you understand how to derive the formula V = n2/3 vs for equal initial potentials?

    If so, then you can use essentially the same method of derivation to get the result for unequal initial potentials.

    If not, my hint would be to consider how the charge on a spherical drop is related to the potential and radius of the drop.
     
  6. Aug 29, 2015 #5
    I know but this methode only useful for similar charge only searched whole internet cant find right answer. All links leads to same potential problem only i tried it myself bt couldnt
     
  7. Aug 29, 2015 #6

    TSny

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  8. Aug 29, 2015 #7
    Thank you guys for idea
    Finally solved it first used formula 4/3 pi R^3= n 4/3 pi r ^3
    Which gave R= n^(1/3)r ........(1)
    And potential V= k Q/R ........(2)
    where Q = 6q-2q since they are provided potential of different sign
    And gave
    Combining 1 and 2 and puting value of Q and n =8
    I came up with answer 6 volts
     
  9. Aug 30, 2015 #8

    TSny

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    Looks good!
     
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