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 10

^{4}N C^{-1 }Density of oil, ρ = 1.26 gm/cm

^{3}= 1.26 x 10^{3}kg/m^{3 }Acceleration due to gravity, g = 9.81 m s

^{-2 }Charge on an electron, e = 1.6 x 10

^{-19}CRadius 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 gWhere,

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 ρr= [ 3Ene/4πρg]^{1/3}= [ 3x 2.55 x 10⁴ x 12 x 1.6 x10

^{-19}/4 x 3.14 x 1.26 x 10^{3}x 9.81]^{1/3}=[ 3946.09 x 10

^{–}^{21}]^{1/3}= 9.82 x 10

^{–}^{7}mmTherefore, the radius of the oil drop is 9.82 x 10

^{–}^{4}mm.