Physics
posted by Sandra on .
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 setup. 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

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