Q: A charged oil drop is of charge q is falling freely under gravity in the absence of electric field with a
velocity ‘ v ’. It is held stationary in an electric field, as it acquires a charge it moves up with a velocity
3v . Now the charge on the drop is
Sol: mg = 6 π η r v , q E = mg
q’ E = m g + 6 π η r (3v)
q’ E = q E + 3 q E
q’ = 4 q