First slide
Electric field
Question

An oil drop, carrying six electronic charges and having a mass of 1.6×1012g, falls with some terminal velocity in a medium. What magnitude of vertical electric field is required to make the drop move upward with the same speed as it was formerly moving downward with? Ignore buoyancy.

Moderate
Solution

For the first case, F = mg.
For the second case
F+mg=6eE or 2mg=6eE

 or  E=mg3e=1.6×1015×103×1.6×1019=3.3×104NC1

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