Question

In Millikan's oil drop experiment an oil drop carrying a charge Q is held stationary by a p.d. 2400v between the plates. To keep a drop of half the radius stationary the potential difference had to be made 600v. What is the charge on the second drop?

(A) [(3Q)/2] 

(B) (Q/4) 

(C) Q 

(D) (Q/2)

Answer 1

The correct option (D) (Q/2)

Explanation:

F = qE also F = mg

in balanced condition, qE = mg  (1)

as E = Electric field = [(Potential)/(distance)] & ρ = [(mass)/volume)]

∴ E = (v/d) & m = ρ ∙ v

from (1)

(qv/d) = ρ ∙ (4/3)πr³g

q = (4/3)πgρd(r³/v)

q ∝(r³/v) ------ other quantities are constant

∴ (q₁/q₂) = (r₁/r₂)³ × (v₂/v₁)

= [r₁/(r₁/2)]³ × [(600)/(2400)]

= 8 × (1/4) = 2

∴ q₂ = (q₁/2)

∴ q₂ = (Q/2)