Question

A 0.80-?m-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if the upper electrode is 16.0 V more positive than the lower electrode. The density of the oil is 885kg/m3.?

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?

Answer 1

We know that

Density = mass/volume
or

Mass of oil drop = density of oil x volume of oil drop

volume is given by
Volume of oil drop = (4*pi*r³)/3 , d= 0.80 um , r=d/2 = 0.40 um =0.40 x 10^-6 m
V = ( 4 x 3.14x ((0.40 x 10^-6)³)/3
or
Volume of oil drop = 0.2679 x 10^-18 m³ = 2.679 x 10^-19 m³

thus
Mass of oil drop = 885 x 2.679 x 10^-19 kg
= 2.371 x 10^-16 kg

Now for the oil drop to be stationary

the net force must be zero i.e,
the downward force (F = mg) = upward force (F=qE)

aslo

E = V/d
E = 16.0/(11 x 10^-3)
E = 1454.54 N/C

qE = mg
q = mg/E
q = ( 2.371 x 10^-16 x 9.81)/1454.54
q = 0.01599 x 10^-16 = 16 x 10^-19 C

we know that the charge comes only in the integer multiples of the electronic charge units of 1.6 x 10^-19 C
thus

n = 16 x 10^-19/ 1.6 x 10^-19
n = 10
thus the droplet have a surplus of electrons.