What Electric Field Magnitude Is Needed to Reverse an Oil Drop's Motion?

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SUMMARY

The discussion focuses on calculating the electric field magnitude required to reverse the motion of an oil drop carrying six electronic charges, with a mass of 1.6 x 10^-12 g. The relevant equation used is E = F/Q, where F is the gravitational force acting on the drop and Q is the total charge. The gravitational force is calculated by multiplying the mass by the acceleration due to gravity (g). The solution involves determining the electric force needed to counteract the gravitational force to achieve upward motion at terminal velocity.

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Homework Statement



an oil drop carries six electronic charges, has a mass of 1.6 x 10^-12 g, has a mass of 1.6 X10^-12g, and falls with a terminal velocity in air. what magnitude of vertical electric field is required to make the drop move upward with the same speed as it was formerly moving downward?

Homework Equations



E=F/Q

The Attempt at a Solution



what i did is multiply mass by g, gives the electric force, and then divided by the charges.am i doing it right?
 
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See,

http://www.antonine-education.co.uk/physics_a2/options/Module_8/Topic_2/topic_2.htm

and,
 

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