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

AI Thread Summary
To reverse the motion of an oil drop with six electronic charges and a mass of 1.6 x 10^-12 g, an electric field must be applied to counteract the gravitational force acting on the drop. The gravitational force is calculated by multiplying the mass by the acceleration due to gravity. The required electric field magnitude can be determined using the formula E = F/Q, where F is the gravitational force and Q is the total charge of the drop. The discussion includes a reference link for further clarification on the topic. Understanding these calculations is essential for determining the necessary electric field to achieve upward motion.
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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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