- #1

Daniel Luo

- 23

- 0

## Homework Statement

When an oil drop falls freely, the velocity first increases, but afterwards it quickly reaches a constant speed, since the air resistance becomes equal to the weight of the oil drop. The air resistance is given by:

F

_{air}=6[itex]\pi[/itex]r[itex]\eta[/itex]v,

where [itex]\eta[/itex] is the viscosity of air with a value of 1.759*10

^{-5}N s/m

^{2}, Oil's density is 918.7 kg/m

^{3}. In Millikan's experiment, the speed of the oil drop is 5.449*10

^{-4}m/s.

- Find the radius and mass of the oil drop.

Afterwards, electricity was sent through the plates, and the speed of the oil drop increases by 5.746*10

^{-4}m/s. The upward electric force acting on the oil drop is given by:

F

_{E}=qE,

where E is the electric field strength: E = 3.178*10

^{5}N/C.

Find the electric charge of the oil drop, q.

## Homework Equations

For no. 1:

Weight = Air resistance

mg = F

_{air}=6[itex]\pi[/itex]r[itex]\eta[/itex]v,

[itex]\rho[/itex]Vg=F

_{air}=6[itex]\pi[/itex]r[itex]\eta[/itex]v,

No. 2:

F

_{E}=qE

mg = F

_{E}+ F

_{air}

## The Attempt at a Solution

All right. What I did in no. 1 was to say that the weight was equal to the air resistance. Assuming the drop to be spherical, I said:

[itex]\rho[/itex](4/3[itex]\pi[/itex]r

^{3})g = 6[itex]\pi[/itex]r[itex]\eta[/itex]v. Then I solved for r:

r = 2.19*10

^{-6}m

And I found m:

m = [itex]\rho[/itex](4/3[itex]\pi[/itex]r

^{3}) = 4.04*10

^{-14}kg (This seems to be too light even for a raindrop?)

For no. 2: I said that the weight was equal to the sum of air resistance and the electric force:

mg = 6[itex]\pi[/itex]r[itex]\eta[/itex]v + qE

Solving for q gave: q = -1.31*10

^{-18}C.

However, this answer seems wrong, as Millikan tried to find the charge of the electron e from this experiment, which is 1.602*10

^{-19}C.

Have I done something wrong? In case I did, what was my mistake?

Thanks a lot!

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