Question

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).

Chapter 22, Problem 039 x Your answer is incorrect. Try again 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 1 Oil spray Insulating chamber wall Oil C Microscope Numbe Units the tolerance is +/-2% No units SHOW HINT LINK TO TEXT LINK TO SAMPLE PROBLEM VIDEO MINI-LECTURE By accessing this Question Assistance, you will Tearn while you earn points based on the Point Potential Policy set by your instructor. Chapter 22, Problem 033 Your answer is partially correct. Try again In the igure a semi-infinite nonconducting rod that is, infinite in one direction only has uniform linear charge density λ point P (R = 20.5 m) 1.38 μC m. Find including sign a the component o electric field parallel to the rod and b the component perpendicular to the rod at (a) Number 1[605.854 UnitSTN/C or V/m (b) NumberT-605.854 Units TN/C or V/m

Answer 1

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