A charged oil drop falls with terminal velocity v 0 in the absence of electric field. An electric field E keeps it stationary. The drop acquires charge 3 q , it starts moving upwards with velocity v 0 . The initial charge on the drop is

Question

A charged oil drop falls with terminal velocity v 0 in the absence of electric field. An electric field E keeps it stationary. The drop acquires charge 3 q , it starts moving upwards with velocity v 0 . The initial charge on the drop is, q 2, q, 3 2 q, 2 q

MARKS App · Accepted Answer

When drop is stationary, then q 1 E = 6 π η r v 0 or q 1 = 6 π η r v 0 / E When drop moves upwards, then 3 q = 6 π η r v 0 + v 0 E = 2 × 6 π η r v 0 E = 2 q 1 ∴ q 1 = 3 2 q

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