A 0.80-mu m-diameter oil droplet is observed between two parallel electrodes spa

Question

A 0.80-mu m-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if the upper electrode is 20.0 V more positive than the lower electrode. The density of the oil is 885 kg/m^3. What is the droplet's mass? Express your answer to two significant figures and include the appropriate units. What is the droplet's charge? Express your answer to two significant figures and include the appropriate units. Does the droplet have a surplus or a deficit of electrons? How many? deficit 8 electrons surplus 8 electrons surplus 15 electrons deficit 6 electrons

Explanation / Answer

A)

Density = mass/volume
Mass of oil drop (m)= density of oil x volume of the oil drop
Diameter = 0.8 * 10^-6 m, radius r= 0.4 * 10^-6 m
Volume of the oil drop = (4 x pi x r^3)/3
Volume of oil drop = ( 4 x 3.14 x ((0.4x 10^-6)^3)/3
Volume of oil drop = 2.68x 10^-19 m^3
density =885 kg/m3
Mass of oil drop = 885 x 2.68 x 10^-19 kg
Mass of oil drop = 2.37 x 10^-16 kg

B)
For the drop to be stationary the net force must be zero
Downward force (F = mg) = upward force (F=qE)

d = 11 * 10^-3 m , V= 20 V

E = V/d

E = 20/11 x 10^-3
E = 1818.18 N/C

qE = mg
q = mg/E
q = ( 2.37 x 10^-16 x 9.8)/1818.18
q = 1.277 x 10^-18

C)
Charges come in integer multiples of the electronic charge units of 1.62 x 10^-19 C
n = 1.277 x 10^-18 / 1.62 x 10^-19
n = 7.88
number of electrons is 8 as 8 is the nearest integer and here surplus 8 electrons

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