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Spherical Raindrop

  1. Feb 25, 2007 #1
    1. The problem statement, all variables and given/known data
    If a spherical raindrop of radius 0.800 mm carries a charge of -1.50 pC uniformly distributed over its volume, what is the potential at its surface?

    Two identical raindrops, each with radius and charge specified in part (A), collide and merge into one larger raindrop. What is the radius of this larger drop, if its charge is uniformly distributed over its volume?


    2. Relevant equations

    [tex] V = \frac {K*q} {r} [/tex]

    3. The attempt at a solution

    I tried:

    [tex] V = \frac {(\frac {1} {4*pi*(8.85*10^{-12}})*-1.50} {0.008} [/tex]

    The online program says I'm off by an additive constant, I'm not sure if I'm using pC value correctly, I've never seen that unit before in our lectures.
     
  2. jcsd
  3. Feb 25, 2007 #2

    Hootenanny

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    pC means picocoulombs which is 10-12 Coulombs

    Note also that [itex]0.800mm \neq 0.008m[/itex]
     
  4. Feb 25, 2007 #3
    Whoops, 0.800mm should be 0.0008 m

    So for pC to C is just [tex] -1.50*10^{-12} [/tex]
     
    Last edited: Feb 25, 2007
  5. Feb 26, 2007 #4
    Two identical raindrops, each with radius and charge specified in part (A), collide and merge into one larger raindrop. What is the radius of this larger drop, if its charge is uniformly distributed over its volume?

    Would it just be 2times the R? I don't think it would be, but I'm not sure how to figure this part out.
     
  6. Feb 26, 2007 #5
    Volume is conserved.

    [tex]
    \frac{8\pi R^3}{3} = \frac{4\pi R_{new}^3}{3}
    [/tex]
     
  7. Feb 26, 2007 #6
    So the radius is still the same?
     
  8. Feb 26, 2007 #7
    No; it's not the same.

    Think about it this way. You have two spheres of clay. You mash them together into a bigger sphere. Is the radius of the new sphere going to be the same as the respective radii of the original spheres?

    The same thing is going on here.
     
  9. Feb 26, 2007 #8
    Okay, that was my original intuition that the radius will grow, and you said the volume is also conserved. Now that we know the radius will grow, how do I go about solving for the new R from knowing that each individual raindrop has a radius of r?
     
  10. Feb 26, 2007 #9
    Once again, volume is conserved.

    [tex]
    \frac{4\pi R^3}{3} + \frac{4\pi R^3}{3} = \frac{8\pi R^3}{3} = \frac{4\pi R_{new}^3}{3}
    [/tex]

    Where [tex]R_{new}[/tex] is the radius of the new raindrop.
     
  11. Feb 26, 2007 #10
    I'm sorry I'm still not fully understanding that. I see from the equation above, you have the sum of the volume of the two raindrops, which gives the 3rd term, you said that the volume is conserved, equating that to 4th term. From what I can understand, [tex] R_{new}^3} [/tex] does not seem to be different from the original [tex] R [/tex].
     
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