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Hi, O tried this question, but I'm getting an answer of 0.99 which rounds to 1, but it was not given as one of the answer choices...can u tell me where I'm going wrong?

What I did was used the following simplified formula:

Volume*density*g =q*E ; where I solved for q.

Then I divided q/e(elementary charge) to find the ratio..

In Robert Millikan's experiment a constant electric field along the vertical axis is obtained with two charged plates, one located above and on below the experimental set-up. The electric field is directed downwards. An oil drop of radius 1.48 μm and density 0.81 g/cm3 is levitated in the chamber when an electric field of 1.7 x 105N/C is applied. Find the charge on the drop as a multiple of the elementary charge e.

the choices were:


  • Physics -

    Volume = (4/3) pi r^3 = 13.58*10^-18 m^3
    Weight = 810 kg/m^3*9.81 m/s^2*13.58*10^-18 m^3
    = 1.079*10^-13 kg

    Q = Weight/E = 6.36*10^-19
    Q/e = 3.97 (call it 4)

    You made a mistake somewhere

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