How Is the Charge on a Suspended Oil Drop Determined in Millikan's Experiment?

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SUMMARY

In Millikan's experiment, the charge on a suspended oil drop is determined using the relationship F=qE, where F is the gravitational force and E is the electric field. The oil drop has a radius of 1.64x10^-6 m and a density of 0.851 g/cm^3, and it is suspended in an electric field of 1.92x10^5 N/C. Since the drop is not accelerating, the net force acting on it is zero, indicating that the electric force balances the gravitational force. This allows for the calculation of the charge on the drop by equating the forces.

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In Milikan's experiment, an oil drop of radius 1.64x10^-6 m and density of 0.851 g/cm^3 is suspended in a chamber where the downward electric field of 1.92x10^5 N/C is applied. Find the charge on the on drop.

I know that F=qE, but the drop is suspended so it can't really a force because it has no acceleration. I also know I can find a mass from the info, but, again that doesn't help my force because the drop is suspended. Should I try to use the idea of an electric field due to a point charge?

Thanks for any help!
 
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If the drop is suspended (not accelerating) that means that the forces are balanced (net force is zero).
 

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