A charged oil drop is suspended in uniform field of 3 × 10 4 V/m so that it neither falls nor rises. The charge on the drop will be (Take the mass of the drop = 9.9 × 10 –15 kg and g = 10 m/s 2 )
Text Solution
Verified by ExpertsThe correct answer is:
A
In steady state electric force on drop balances the weight of the drop.
In steady state,
electric force on drop = weight of drop
∴ qE = mg ⇒ q = 
= 
= 3.3 × 10 –18 C
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