Solve Electrical Potential Homework: 24 kV, Droplet Mass 2.2 x 10^-13 kg

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SUMMARY

The discussion focuses on calculating the charge of an oil droplet with a mass of 2.2 x 10^-13 kg in a 24 kV electric field between parallel plates separated by 1.8 cm. The relevant equations include potential energy (PE) equating to kinetic energy (KE) and the relationship qV = (1/2)mv^2. The user identifies that since the droplet is stationary, its velocity is zero, leading to the conclusion that the charge (q) must also be zero, which is incorrect. The solution requires considering the forces acting on the droplet to find the charge accurately.

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Homework Statement


A potential difference of 24 kV maintains a downward-directed electric field between two horizontal parallel plates separated by 1.8 cm. Find the charge on an oil droplet of mass 2.2 x 10^-13 kg that remains stationary in the field between the plates.

Homework Equations


PE=KE
qV=(1/2)mv^2

V=Ed

E=k(q/r^2)

The Attempt at a Solution


PE=KE
qV=(1/2)mv^2
q=((1/2)mv^2)/V

If velocity is zero, then "q" is zero. Since its stationary. (I think the only way to find velocity is PE=KE)
I'm stucked at here, any solutions?
 
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Consider the forces acting on the droplet.
 

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