Buoyant Force Direction in the Millikan Oil Drop Experiment

AI Thread Summary
The discussion centers on the direction of the buoyant force in the context of the Millikan oil drop experiment. Participants agree that the buoyant force acts opposite to the weight of the drop. There is uncertainty about how electrical and drag forces might influence this relationship. The apparent weight equation is mentioned, indicating that buoyancy is affected by the density difference between the oil drop and air. Clarification from more experienced individuals is sought to resolve these uncertainties.
Juan Pablo
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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.
 
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As far as I know the buoyant force acts in the opposite direction than the one of the weight.
 
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})
 
I don't think it changes something, but I prefer to wait for more experienced people to help you.
 

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