1) In Millikan's experiment, an oil drop of radius 1.95 μm and density 0.857 g/cm3 is suspended in chamber C (see the figure) when a downward electric field of 1.36 × 105 N/C is applied. Find the charge on the drop, in terms of e.
2)In the figure a “semi-infinite” nonconducting rod (that is, infinite in one direction only) has uniform linear charge density λ = 1.38 μC/m. Find (including sign) (a) the component of electric field parallel to the rod and (b) the component perpendicular to the rod at point P (R = 20.5 m).
1. Volume = 4 pi R^3 / 3
and R = 1.95 x 10^-6 m
Volume = 3.106 x 10^-17 m^3
mass = density x volume
= (857 kg / m^3)(3.106 x 10^-17)
m = 2.662 x 10^-14 kg
so equlibrium. Fnet = Fe - fg = 0
q (1.36 x 10^5) = (2.662 x 10^-14) (9.81)
q = 1.92 x 10^-18 C
n = (1.92 x 10^-18) /(1.6 x 10^-19)
n = 12 ( no units)
2. Ex = Ey = - 605.854 N/C
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