Buoyant Force Direction in the Millikan Oil Drop Experiment

[Juan Pablo]

What's the direction of the buoyant force? I'm trying to find the equation for the charge on the Millikan oil drop experiment. It requires two equilibrium equations but I'm not sure which sign should the buoyant force have.

My internet searches neither my book have been able to answer this. Thanks.

 

[fluidistic]

As far as I know the buoyant force acts in the opposite direction than the one of the weight.

 

[Juan Pablo]

Yes, but I don't know if the fact that there are electrical and drag forces changes anything. According to Wikipedia the apparent weight equation is:

$$W = \frac{4}{3} \pi r^3 g(\rho - \rho_{air})$$

 

[fluidistic]

I don't think it changes something, but I prefer to wait for more experienced people to help you.