In Millikans experiment an oil drop of radius 2346 mu m and

In Millikan\'s experiment, an oil drop of radius 2.346 mu m and density 0.812 g/cm^3 is suspended in chamber C when a downward-pointing electric field of 1.92 times 10^5 N/C is applied. Find the charge on the drop, in terms of e. (Include the sign.) For the balancing, how do the electrostatic force and the gravitational force compare? How is the electrostatic force related to the charge and the electric field?

Solution

E = 1.92 * 10⁵ N/C, g = 9.81 m/s²

R = 2.346 * 10^-6 m , density = 812 kg/m²

F_E = m g

q E = m g, q = mg / E

m = p * 4/3 pi * r³ = 812 * 4/3 pi * (2.346 * 10^-6)

= 4.39 * 10^-14 kg

q = (4.39 * 10^-14 kg * 9.8) / (1.92 * 10⁵)

= 14 e