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

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In Millikan's experiment, the charge on a suspended oil drop is determined by balancing the gravitational force and the electric force acting on the drop. The gravitational force can be calculated using the mass derived from the drop's volume and density, while the electric force is given by F=qE, where E is the electric field strength. Since the drop is suspended, the net force is zero, indicating that the upward electric force equals the downward gravitational force. By equating these forces, the charge on the drop can be calculated. Understanding this balance is crucial for determining the charge accurately.
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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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