Question

A 0.80-μm-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if the upper electrode is 26.6 V more positive than the lower electrode. The density of the oil is 885kg/m3.

Part A

What is the droplet's mass? Express your answer to two significant figures and include the appropriate units.

Part B

What is the droplet's charge?

Express your answer to two significant figures and include the appropriate units.

Part C

Does the droplet have a surplus or a deficit of electrons? How many?

deficit 6 electrons

surplus 6 electrons

surplus 13 electrons

deficit 4 electrons

Answer 1

Part A.

We know that mass of the droplet is given by:

Mass = Volume*density

density of the oil, = 885 kg/m^3

Volume of spherical droplet = (4/3)*pi*r^3

r = radius of droplet = Diameter/2 = 0.80 m/2 = 0.40*10^-6 m

So,

Mass = (4/3)*pi*r^3*

Mass = (4/3)*pi*(0.40*10^-6)^3*885 = 2.3725*10^-16 kg

In two significant figures

Mass = 2.4*10^-16 kg

Part B.

Using force balance on the droplet in electric field between electrodes, given that droplet hangs motionless between the electrodes, So

F_net = Fe - W = 0

Fe = W

q*E = m*g

q = m*g/E

E = electric field between electrodes = V/d

V = Potential difference between plates = V2 - V1 = 26.6 V

d = distance between electrodes = 11 mm = 11*10^-3 m

So,

q = m*g*d/V

q = 2.3725*10^-16*9.81*11*10^-3/26.6 = 9.625*10^-19 C

Since upper electrode is at higher potential, So direction of electric field is always in from higher to lower potential, which means in this case electric field is downward, Now since electric force should be in upward direction, so that it can balance the downward gravity force, So the charged particle should be negative, because on the negative charged particle, electric force will be in opposite direction of electric field.

In two significant figures

q = -9.6*10^-19 C

Part C

So droplets will have a surplus of electrons, since it is negatively charged.

Now we know that

q = n*e

n = number of electrons on particle = ?

e = charge on single electron = 1.6*10^-19 C

So,

n = q/e

n = 9.6*10^-19/(1.6*10^-19)

n = 6 surplus electrons

Let me know if you've any query.