# A Millikan apparatus is set up in which oil droplets are observed between two parallel electrodes spaced 12 mm apart. An oil droplet with 15 excess electrons hangs motionless when the upper electrode is 97 V more positive than the lower electrode. The density of the oil is 860 kg/m3. What is the radius of the droplet?

This is a third law problem in disguise. You have two forces: electric force and gravitational force. Using the equations for gravity, electric force, electric potential, density, and volume of a sphere, you can solve for radius.

`SigmaF=0`

and substitute in the equations for gravity and electric force.

`F_(g)-F_(e)=0`

`F_(g)=F_(e)`

`mg=qE`

Because gravity is negative, I am ignoring the minus.

Next, solve for mass.

`m=qE/g`

Then substitute in the equation for electric potential.

`m=(V/y)q/g`

We proceed to substitute in density, with X for volume,

`dX=(V/y)q/g`

and finally substitute for volume of a sphere.

`d[(4/3)pir^3]=(V/y)q/g`

At long last we can isolate the radius and solve with numbers.

`r=root(3)((3/4pi)Vqd/yg)`

`r=root(3)((3/4pi)(97)(2.403*10^-18)(860)/(.012)(9.8))`

`r=6.681*10^-10` meters.

Does this answer make sense? Oil droplets are very small, so a radius that small is logical. Did we need terminal velocity? Not for this solution, but only because the particle is suspended. If it were to travel down past the plate, fall, and then float back up, you would be given the distance it had fallen and the terminal velocity as well. That problem could be solved with some integrals.

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