Milikan Oil Drop Expt A Numerical Approach
An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 ×10 4 N/C in Millikan's oil drop experiment. The density of the oil is 1.26 g/cm 3 Estimate the radius of the drop At equilibrium the gravitational force=electrostatic force mg=q E ie volume of drop×density×g=neE (since q=ne) 4/3πr 3 ×ρg=neE r 3 =(3×12×1.6×10 -19 ×2.55 ×10 4 ) / (4× 3.14× 1.26×10 3 × 9.81 )=0.94×10 -18 . . . r= (0.94×10 -18 ) 1/3 =9.81×10 -7 So the radius of the drop is 9.81×10 -7 m