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.

We start with Newton's third law,

`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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