Q: A charged oil drop falls with a terminal velocity V in the absence of electric field. An electric field E
keeps the oil drop stationary in it. When the drop acquires a charge ‘q’ it moves up with same velocity.
Find the initial charge on the drop.
Sol: mg = 6 π η r v , q E = mg
E (Q + q) = mg + 6 π η r v ,
E (Q + q) = 2 E Q
Q + q = 2 Q
q = Q