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?

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/s^{2}--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/m

^{3}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=4/3πR^{3}; Final formula for the force of gravity is as follows:F

_{g}=(4/3)ρπR^{3}g_{ }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}=EeNTwo forces must be equal, thus giving:

(4/3)ρπR

^{3}g=eENFrom this we obtain:

N=(4/3)ρπR

^{3}g/(eE)After some calculations, we obtain:

N≈9975185 electrons.