Answers to Selected Problems - West Valley College

Problem Sets Physics 4D
In a Millikan experiment the distance of rise or fall of a droplet is 0.600 cm and the average time of fall
when E = 0 is 21.0 s. The observed successive rise times when the electric field is on are 46.0 s, 15.5 s,
28.1 s, 12.9 s, 45.3 s, and 20.0 s.
Using the derivation done in class:
q
mg v v
T E
E vT
a) Prove that charge is quantized. (Hint: charge is quantized if ratios of charge are ratios of small whole
numbers: e.g. q1/q2 = 2/5, q2/q3 = 5/7, etc.)
b) Calculate the successive charges on each drop, and, from these results, determine the fundamental
electric charge. Assume a plate separation of 1.60 cm, a potential difference of 4550 V for the
parallel-plate capacitor, and the mass for an oil drop of 1.67 x 10 -14 kg.
Hint: In Millikan’s time, e was known to be between 1.50 x 10 -19 C and 2.00 x 10 -19 C. Dividing q1 through
q6 by each of these values gives the range of integral charges on each drop. For example, if q1 = 8.39 x
10 -19 C, then the range for q1 is 8.39 / 1.5 = 5.6 and 8.39 / 2.0 = 4.2 multiples of electric charge e. Since 5
is an integer between 4.2 and 5.6. we can conclude that q1 = 5e, or e = q1/5. In like manner, determining e
for each charge, then averaging them, will generate the value of the electronic charge as determined in
Millikan’s day to be e = 1.69 x 10 -19 C.
Answer(s): (see Excel File, “p4d_charge_quantization_computation.xls” for calculations)
a)
q1/q2 = 5/8; q1/q3 = 5/6; q1/q4 = 5/9; q1/q5 = 1; q1/q6 = 5/7
q2/q3 = 4/3; q2/q4 = 8/9; q2/q5 = 8/5; q2/q6 = 8/7
q3/q4 = 2/3; q3/q5 = 6/5; q3/q6 = 6/7
q4/q5 = 9/5; q4/q6 = 9/7
q5/q6 = 5/7
n1 = 5
n2 = 8
n3 = 6
n4 = 9
n5 = 5
n6 = 7
b)
q1 = n1∙e = 8.38 x 10 -19 C
q2 = n2∙e =1.36 x 10 -18 C
q3 = n3∙e =1.01 x 10 -18 C
q4 = n4∙e =1.51 x 10 -18 C
q5 = n5∙e =8.42 x 10 -19 C
q6 = n6∙e =1.18 x 10 -18 C
eavg = 1.68 x 10 -19 C
Melvin J. Vaughn Page 34 of 61 West Valley College