Question -
Answer -
Excess electrons on an oildrop, n = 12
Electric field intensity, E = 2.55 × 104 NC−1
Density of oil, ρ = 1.26 gm/cm3 =1.26 × 103 kg/m3
Acceleration due togravity, g = 9.81 m s−2
Charge on an electron, e = 1.6 × 10−19 C
Radius of the oil drop = r
Force (F) due to electric field E is equal to the weight of the oildrop (W)
F = W
Eq = mg
Ene 
Where,
q = Net charge on the oil drop = ne
m = Mass of the oil drop
= Volume of the oil drop ×Density of oil
= 9.82 × 10−4 mm
Therefore,the radius of the oil drop is 9.82 × 10−4 mm.