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Potential on merging water droplets

  1. Sep 17, 2015 #1
    1. The problem statement, all variables and given/known data
    Suppose you have two identical droplets of water, each carrying charge 2.8 pC spread uniformly through their volume. The potential on the surface of each is 589 Volts.

    Now, you merge the two drops, forming one spherical droplet of water. If no charge is lost, find the potential at the surface of this new large water droplet. in V.

    2. Relevant equations
    V = kQ/r

    3. The attempt at a solution

    I wasn't sure how to approach this problem so I assumed that the total volume will be conserved (since water is not compressible) and therefore if the volume of one initial droplet doubles after merging (V=2XV0) then the radius will increase by 1/2Xr0.

    So by following that reasoning I got rfinal = 6.41 x 10-5 m

    ⇒ Vfinal = k*Qtotal / rfinal

    ⇒ Taking the total conserved charge to be Q1 + Q2 ⇒ (2)*(2.8 X 10-12) = 5.6 x 10-12 C

    Plugging in all the values I got Vfinal = 785 V , and this result turned out to be wrong :oldfrown:
    Last edited: Sep 17, 2015
  2. jcsd
  3. Sep 17, 2015 #2
    I would look at the value of the radius of the coalesced drop that you calculated.

    The volume of a sphere is 4/3 π R3
  4. Sep 17, 2015 #3
    Thanks, now I see it.

    So from Vdroplet = k (2.8x10-12)/r0,
    I got r0 = 4.27 x 10-5

    Plugged that in V0 = 4/3*π*(r0)3 and got V0 = 3.27 x 10-13

    Now Volfinal = 2V0 = 6.54 x 10-13

    ⇒ 6.54 x 10-13 = 4/3*π*R3

    ⇒ R = (1.56 x 10-13)1/3

    ⇒ Vbig droplet = k Qtotal/R

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