Class-12th Physics, CBSE and UP Board
Electric Charges and Fields,
Chapter-1, Exercise – 1, Additional Exercise, Q-1.26
NCERT Solutions for Class 12th Physics
An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10⁴ NC⁻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).
Share
Excess electrons on an oil drop, n = 12
Electric field intensity, E = 2.55 x 104 N C-1
Density of oil, ρ = 1.26 gm/cm3 = 1.26 x 103 kg/m3
Acceleration due to gravity, g = 9.81 m s-2
Charge on an electron, e = 1.6 x 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 = 4/3 x π r³ x ρ x g
Where,
q = Net charge on the oil drop = ne
m = Mass of the oil drop = Volume of the drop x Density of oil =4/3 π r³ x ρ
r= [ 3Ene/4πρg]1/3
= [ 3x 2.55 x 10⁴ x 12 x 1.6 x10-19 /4 x 3.14 x 1.26 x 103 x 9.81]1/3
=[ 3946.09 x 10–21]1/3
= 9.82 x 10–7 mm
Therefore, the radius of the oil drop is 9.82 x 10–4 mm.