In Millikan’s 1909 oil drop experiment that made the first measurement of the ch

Question

In Millikan’s 1909 oil drop experiment that made the first measurement of the charge of an electron, Millikan applied uniform vertical electric fields to small charged droplets of oil. As the oil moved through the air it experienced a linear (viscous) drag force F(air) on oil = Cv opposite to its motion, a downward gravitational force and an electric force. Each oil drop quickly reached a terminal velocity (where the sum of the forces acting on it add to zero). For this problem, take the strength of the electric field to be E = 2.5 × 10^5 N/C, the drag constant to be C = 5.6×10^10 N/(m/s) , and the mass of the oil droplet to be m(oil) = 1.44×10^14 kg, m/s and there to be 9 excess electrons on the oil droplet. Each electron has a charge q = 1.602 × 10^19 C.

1. (a) What size electric force in Newtons will the oil droplet experience in this electric field?

1. (b) What size gravitational force will the oil droplet experience?

1. (c) If the uniform electric field points upwards, in what direction will the oil droplet accelerate until it reaches its terminal speed, vE up? Please explain your answer thouroughly

1. (d) Draw a free body diagram for the forces acting on the oil droplet once it reaches its terminal speed.

1. (e) Write the Newton’s second law equation corresponding to your free body diagram and solve for the terminal speed of the drop, vE up.

1. (f) If the uniform electric field is now reversed to point downwards, in what direction will the oil droplet accelerate until it reaches its terminal speed? Please explain your answer.

1. (g) Draw a free body diagram for the forces acting on the oil droplet once it reaches its terminal speed with the electric field pointing downwards.

1. (h) Write the Newton’s second law equation corresponding to your free body diagram and solve for the terminal speed of the drop, vE down.

In this way, by measuring vE up and vE down for several oil drops of different charge, Millikan found that the charge on the oil droplets only came in integer multiples of e, the charge of the electro

Explanation / Answer

i would be answering only 4 parts, it's not fair on your part, to ask or post so many parts

a) Electric force = qE =9( 1.602 × 10^19)(  2.5 × 10^5 N/C) = -36.045 x 10^ -14 N

b) Gravatational Force = mg = (1.44×10^14 kg, )(9.8) = 14.112 x 10^-14 N

c) If the electric fiedl is pointing in the downward direction, it implies that positive charged plate is kept lower whilethe negative charged plate is kept higher, so oil drop with a net negative charge will feel force in the downward direction, The oil drop will be accelerted in downward direcvtion , as it experiences electric force and gravational force in the downward direction and drag force in the upward direction,

e) Newton' second law of motion = F = ma

net force on the oil drop at terminal velocity is 0,

qE + mg = cv

+36.045 x 10^ -14 + 14.112 x 10^-14 = 5.6×10^10 (V)

V = 8.95 x 10^ -4 m/s apprx

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