An oil drop, carrying six electronic charges and having a mass of 1.6×10−12g, 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
105NC−1
b
104NC−1
c
3.3×104NC−1
d
3.3×105⋅NC−1
answer is C.
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Detailed Solution
For the first case, F = mg.For the second caseF+mg=6eE or 2mg=6eE or E=mg3e=1.6×10−15×103×1.6×10−19=3.3×104NC−1