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## Main Question or Discussion Point

Can some one help me with this Milliken oil drop problem.

Data from a Milliken oil drop experiment are given below. Calculate the value of the charge on an electron from this data:

Voltage needed to suspend drop (V) Drop diameter (microns)

103------ 1.1

44.7----- 1.2

127------ 1.6

175------ 1.5

11.3----- 0.9

Given:

Plate separation: 0.5cm

Density of oil: 0.95 g/mL

Gravitational constant: 9.80 m/s^2

Ok... this is what I did: q=mg/E

1. I calculated for electric field first since: E= kV/m

2. One microns is equal to 10^-6 m

3. I changed volts into kV.

4. I took the average of all 5 of the electric fields and got 67896.71718 kV/m

5. Now I solved for the mass by cubing the plate separation and cancelled it with the mL... to get grams then I changed grams into kilograms to get 1.1875 x 10^-4 kg.

6. And finally I plug everything back into the equation: q=mg/E and get the answer of 2.706 x 10^-8 coulombs.

Please tell me if I did it right.

Data from a Milliken oil drop experiment are given below. Calculate the value of the charge on an electron from this data:

Voltage needed to suspend drop (V) Drop diameter (microns)

103------ 1.1

44.7----- 1.2

127------ 1.6

175------ 1.5

11.3----- 0.9

Given:

Plate separation: 0.5cm

Density of oil: 0.95 g/mL

Gravitational constant: 9.80 m/s^2

Ok... this is what I did: q=mg/E

1. I calculated for electric field first since: E= kV/m

2. One microns is equal to 10^-6 m

3. I changed volts into kV.

4. I took the average of all 5 of the electric fields and got 67896.71718 kV/m

5. Now I solved for the mass by cubing the plate separation and cancelled it with the mL... to get grams then I changed grams into kilograms to get 1.1875 x 10^-4 kg.

6. And finally I plug everything back into the equation: q=mg/E and get the answer of 2.706 x 10^-8 coulombs.

Please tell me if I did it right.