(©2002, François G. Amar, All rights reserved)

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Statement of  Problem-- Determine the charge on a droplet given the following data:

R= 1.64x10^-4 cm ; E=1.92x10⁵ Nt/coulomb ; r = 0.851 g/cm³ (density of oil).
 

Solution

1. Write down the equation we have to use:

     q= 4pR³rg/3E

2. Convert all quantities to metric system (MKS or S.I. units)

        R =1.64x10^-4 cm x (10^-2 m/1 cm)  = 1.64x10^-6 m

r= 0.851 g/cm³ x (10^-3 kg/1 g) x (10^-6 cm³/m³) =  8.51x10² kg/m³

        g = 9.8 m/s²   (acceleration due to gravity at the surface of the earth)

        E =1.92x10⁵ Nt/C = 1.92x105 kg^.m/s^2.C

3. Insert quantities and do a units check and then do the arithmetic.
 

         (R)3           (r)              (g)                 1/3(E)

 q = 4p(1.64x10^-6 m)³ (8.51x10² kg/m³) (9.8 m/s²)/3(1.92x10⁵ kg^.m/s^2.C)
    = 4p(1.64x10^-6)³ (m)³ (8.51x10²)(kg/m³) (9.8) (m/s²)/3(1.92x10⁵ )(kg^.m/s^2.C)
    = [(4p/3)(4.41x10^-18) (8.51x10²)(kg/m³) (9.8)/(1.92x10⁵)][(m)³(kg/m³)(m/s²)/(kg^.m/(s^2.C))]
    = [8.03x10^-19] [(m)³(kg/m³)(m/s²)/(kg^.m/(s^2.C))]    < -- 1/(1/C) = C is all that doesn't cancel!

we see that the units cancel to yield charge q in coulombs (C) as they must and the final answer is

        q=8.03x10^-19 C

4. Note: if we divide this value of the charge by the now known value for the elementary charge
on the electron,

        e=1.602x10^-19 C     (this value is reported as an absolute value no minus sign)

we get,

    # of electrons on drop = q/e = 8.03x10^-19 C/1.602x10^-19 C = 5.01

which is equal to 5 within experimental uncertainties.

5. Finally, note that step 4 was backwards from a historical point of view. Millikan and his coworkers
observed many oil drops and calculated the charges on them and found that they were all multiples
of the same charge which turned out to be 1.6x10^-19 C. This is how they made the first determination of
the value of the elementary charge.