Question
A charged oil drop is suspended in a uniform field of 3 × 104 V/m so that it neither falls nor rises. The charge on the drop will be (take the mass of the charge = 9.9 × 10−15 kg and g = 10 m/s2 )
-
3.3 × 10−18 C
-
3.2 × 10−18 C
-
1.6 × 10−18 C
-
4.8 × 10−18 C
Solution
A.
3.3 × 10−18 C
In steady state, electric force on drop = weight of drop
∴ qE = mg
