# An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 104 N C−1 (Millikan’s oil drop experiment). The density of the oil is 1.26 g cm−3. - Physics

Numerical

An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 104 N C−1 (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).

#### Solution

Excess electrons on an oil drop, n = 12

Electric field intensity, E = 2.55 × 104 N C−1

Density of oil, ρ = 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 = 4/3pi"r"^3 xx rho xx "g"

Where,

q = Net charge on the oil drop = ne

m = Mass of the oil drop

= Volume of the oil drop × Density of oil

= 4/3 pi"r"^3 xx rho

∴ "r" = root(3)((3"Ene")/(4pirho"g"))

= root(3)((3 xx 2.55 xx 10^4 xx 12 xx 1.6 xx 10^-19)/(4 xx 3.14 xx 1.26 xx 10^3 xx 9.81))

= root(3)(946.09 xx 10^-21)

= 9.82 xx 10^-7 m

= 9.82 × 10−4 mm

Therefore, the radius of the oil drop is 9.82 × 10−4 mm.

Concept: Electric Field - Physical Significance of Electric Field
Is there an error in this question or solution?
Chapter 1: Electric Charges and Fields - Exercise [Page 48]

#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 1 Electric Charges and Fields
Exercise | Q 1.25 | Page 48
NCERT Class 12 Physics Textbook
Chapter 1 Electric Charge and Fields
Exercise | Q 25 | Page 48
Share