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# Physics help

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?

### 3 Answers by Expert Tutors

Kirill Z. | Physics, math tutor with great knowledge and teaching skillsPhysics, math tutor with great knowledge...
4.9 4.9 (174 lesson ratings) (174)
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If a droplet remains stationary, net force on it must be zero. There is a force of gravity acting on the droplet,
Fg=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:

Fg=(4/3)ρπR3g

The force, which opposes it, is the electrical force, Fe. It can be found as Fe=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:

Fe=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.
Jim S. | Physics (and math) are fun, reallyPhysics (and math) are fun, really
4.7 4.7 (197 lesson ratings) (197)
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Hey David,

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

weight=g4ρr/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
Amarjeet K. | Professional Engineer for Math and Science TuroringProfessional Engineer for Math and Scien...
4.6 4.6 (8 lesson ratings) (8)
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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
:)