Question 1.25: An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10 4 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).
Class 12 - Physics - Electric Charges And Fields Electric Charges And Fields Page-48
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in equilibrium , force due to electric field on the drop balance the weight {F = mg} of drop.
e.g., electric force = weight
qE = mg
we know, mass = volume × density
so, m = Vρ
also, volume of spherical drop , V = 4/3 πr³
so, qE = Vρg
qE = 4/3πr³ρg
r³ = 3qE/4πρg
here, q = 12e
q = 12 × 1.6 × 10^-19C = 19.2 × 10^-19C
E = 2.55 × 10⁴ N/C , ρ = 1.26 g/cm³ = 1.26 × 10³kg/m³ and g = 9.8 m/s²
now, r³ =3 × 19.2 × 10^-19 × 2.55 × 10⁴/(4 × 3.14 × 1.26 × 10³ × 9.8)
= 0.947 × 10^-18
taking cube root both sides,
r = 0.981 × 10^-6 m
hence, radius of the drop = 9.81 × 10^-7 m
e.g., electric force = weight
qE = mg
we know, mass = volume × density
so, m = Vρ
also, volume of spherical drop , V = 4/3 πr³
so, qE = Vρg
qE = 4/3πr³ρg
r³ = 3qE/4πρg
here, q = 12e
q = 12 × 1.6 × 10^-19C = 19.2 × 10^-19C
E = 2.55 × 10⁴ N/C , ρ = 1.26 g/cm³ = 1.26 × 10³kg/m³ and g = 9.8 m/s²
now, r³ =3 × 19.2 × 10^-19 × 2.55 × 10⁴/(4 × 3.14 × 1.26 × 10³ × 9.8)
= 0.947 × 10^-18
taking cube root both sides,
r = 0.981 × 10^-6 m
hence, radius of the drop = 9.81 × 10^-7 m
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