Physics help

Question

13. A water droplet of radius 0.018mm remains stationary in air. If the downward electric field in 150N/C, how many excess electrons must the droplet have?

Answer 1

If a droplet remains stationary, net force on it must be zero. There is a force of gravity acting on the droplet,

F_g=mg, where g=9.81 m/s²--acceleration due to gravity, m is the mass of the droplet. It can be found by the formula:

m=ρV; here V is the volume of the droplet, ρ=1000 kg/m³ is the density of water. Volume of a droplet is the volume of a sphere of radius R=0.018mm=1.8*10^-5 m. V=4/3πR³; Final formula for the force of gravity is as follows:

 

F_g=(4/3)ρπR³g

 

The force, which opposes it, is the electrical force, F_e. It can be found as F_e=Eq, where E is the electric field, q is the charge. Overall charge equals the charge of one electron, e, (e=1.6*10^-19 C) times the number or excess electrons, N. So, electric force is:

 

F_e=EeN

 

Two forces must be equal, thus giving:

 

(4/3)ρπR³g=eEN

 

From this we obtain:

 

N=(4/3)ρπR³g/(eE)

 

After some calculations, we obtain:

 

N≈9975185 electrons.

Answer 2

Hey David,

 

weight of droplet = force of the electric field i.e. the sum of the forces are zero.

 

weight=g4ρr³ /3 where ρ=density of water r=radius of the drop g=acceleration due to gravity

 

electric force =qE where q is the charge and e is the magnitude of the electric field. Now if e is the charge on an electron then q=ne where n is the number of electrons on the drop

 

Equating the two terms gives

 

g4ρr3 /3 = neE you can now solve for n

 

In case you don't know it a physicist named Millikan won a Nobel prize for doing an experiment like this.

 

Regards

Jim

Answer 3

Volume of droplet = 4/3 x pi x r^3 = 4/3 x 3.14 x (0.018 x 10^-3)³ = 2.44 x 10^-14 m³

 

Density of water = 1000 kg/m³

 

Mass of water droplet = density x volume = 1000 x 2.44 x 10^-14 kg = 2.44 x 10^-11 kg

 

Wt of water drop let - mass x g = 2.44 x 10^-11 x 10 = 2.44 x 10^-10 Newton

 

This wt is being balanced by Electric force = charge x electric field intensity = n x e x electric field intensity

 

n = is number of excess electrons , e is charge of each electron

 

2.44 x 10^-10 = n x 1.6 x 10^-19 x 150

 

n = 0.01 x 10⁹ = 10,000,000

:)