NEET-XII-Physics
01: Electric Charges And Fields
- #25An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 104 N C-1 in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm-3. Estimate the radius of the drop. (g = 9.81 m s-2; e = 1.60 × 10-19 C).
Ans : Excess electrons on an oil drop, n = 12
Electric field intensity, E = 2.55 × 104 N C-1
Density of oil, ``\rho`` = 1.26 gm/cm3 = 1.26 × 103 kg/m3
Acceleration due to gravity, 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 oil drop (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.