In Millikan's oil drop experiment,if the electric field between the plates was of just the right magnitude,it would exactly balance the weight of the drop.Suppose a tiny spherical oil droplet of radius 1.3 × 10−⁴ cm carries a charge equivalent to one electron.What electric field is required to balance the weight? (The density of oil is 0.85 g/cm³,e = 1.6 × 10−¹⁹ C. )

Question

Quizplus · Accepted Answer

The weight of the droplet can be calculated as:

W = mg = (4/3)πr3ρg = (4/3)π(1.3 x 10^-14)^3(0.85)(9.81)

W = 4.17 x 10^-14 N

The electric force on the droplet can be calculated as:

Fe = qE

where q is the charge on the droplet (1.6 x 10^-19 C) and E is the electric field.

Since the droplet is in equilibrium, we have:

Fe = W

qE = (4.17 x 10^-14)

E = (4.17 x 10^-14) / (1.6 x 10^-19)

E = 2.61 x 10^5 N/C

Therefore, the best choice is D) 4.8 x 10^15 N/C, which is the closest value to 2.61 x 10^5 N/C.

In Millikan's Oil Drop Experiment,if the Electric Field Between the Plates