Q.

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.

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a

105NC1

b

104NC1

c

3.3×104NC1

d

3.3×105NC1

answer is C.

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Detailed 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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