A charged oil drop is suspended in uniform field of 3 x 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 x 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
A.
3.3 × 10-18 C
In steady state,
electric force on drop = weight of drop
q E = mg