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Finding the Electric Potential

  1. Nov 6, 2014 #1
    1. The problem statement, all variables and given/known data
    8 Charged drops of water each of radius 1 mm and having a charge of 10^(-10) Coloumbs are combined to form a bigger drop.Determine the potential of the bigger drop.

    2. Relevant equations
    Does the radius becomes 4*10^(-3)m,when it combines?
    Does the charge sum up to 9*10^(-10)?


    3. The attempt at a solution
    Used the formula -V=9*10^(9)*Q/r
    Got the ans as 1800 V.
     
    Last edited by a moderator: Nov 6, 2014
  2. jcsd
  3. Nov 6, 2014 #2

    Mark44

    Staff: Mentor

    I don't think so. It's probably easier to work with the mm units that were given.
    What's the volume of one rain drop?
    If you combine 8 rain drops into one bigger drop, what is the volume of this larger drop? What's its radius?
     
  4. Nov 7, 2014 #3
    I assume the volume to be (4/3)*(pi)*r^3,where r is 1 mm.
    But what potential has to do with Volume.i am quit weak in this subject
     
  5. Nov 7, 2014 #4

    Mark44

    Staff: Mentor

    Or more simply, 4/3 ##\pi##, the volume of one rain drop,
    From the formula you wrote in post #1, the potential V is a function of the radius r.
    If 8 rain drops combine into one larger drop, what's the volume of the larger rain drop? What's the radius of the larger rain drop?
     
  6. Nov 7, 2014 #5
    i hope it adds up linearliy,that means ,radius of larger frop =8* 1mm (radius of each smaller drop)
     
  7. Nov 7, 2014 #6

    Mark44

    Staff: Mentor

    No, it doesn't.
    The volumes add like you think they would, but the radius of the combined raindrop is NOT 8mm.
     
  8. Nov 8, 2014 #7
    the bigger radius becomes twice the smaller radius ie 2 mm.But what about the charge ?
     
  9. Nov 8, 2014 #8

    Mark44

    Staff: Mentor

    2 mm is more like it. I'm pretty sure the charge is additive.
     
  10. Nov 13, 2014 #9
    THank you Mark44,for helping me to reach to the answer.You made me think,which i already knew
     
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