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Calculation of radius

  1. Nov 10, 2015 #1
    1. The problem statement, all variables and given/known data
    A spherical drop of water carrying a charge of 3×10^-19 has a potential of 500V at its surface (with V=0 at infinity) (a) What is the radius of the drop? If two such drops of the same charge and radius combined to form a single drop, what is the potential at the surface of the new drop

    2. Relevant equations
    ##V##=##\frac{q}{4πε0r}##

    3. The attempt at a solution
    Here q=3×10^-19 C and V=500 volts.
    ##r##=##\frac{3×10^-19×9×10^9}{500}##
    =##\frac{27×10^-10}{5×10^2}##
    =##\frac{27×10^-10×10^-2}{5}##
    =##\frac{27×10^-12}{5}##
    =5.4×10^-12
    but the answer for r is wrong it should be r=0.54cm.What went wrong .
     
  2. jcsd
  3. Nov 10, 2015 #2

    BvU

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    Typo (or hasty reading :rolleyes:) : 3 x 10 -19 should have been 3 x 10 -9.

    For comparison: 1 single electron has a charge -1.60217662 × 10-19 Coulomb
     
  4. Nov 10, 2015 #3
    Your rolling eyes!
    Neither of them.Actually it is
    :
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    Printing mistake in the textbook.:smile:
     
  5. Nov 10, 2015 #4

    BvU

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    That's what we call a typo :biggrin: . The printer prints what has been typeset
     
  6. Nov 10, 2015 #5
    I just wanted to make sure that's not my mistake :rolleyes:
     
  7. Nov 10, 2015 #6

    BvU

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    Part (b) is going OK ?
     
  8. Nov 10, 2015 #7
    Yes!:smile:
    seriously, loved the way you paid attention to my all the threads today.
     
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