Sunday

August 31, 2014

August 31, 2014

Posted by **Sandra** on Saturday, February 6, 2010 at 1:24pm.

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:

2

3

4

5

6

- Physics -
**drwls**, Saturday, February 6, 2010 at 2:28pmVolume = (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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