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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:
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:
Two forces must be equal, thus giving:
From this we obtain:
After some calculations, we obtain:
N≈9975185 electrons.
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.
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