Answer:
[tex]N=1.5*10^{11}[/tex] electrons en excess
Explanation:
We use the density of water equal to 1000 Kg/m^3, in order to find the mass and Force gravitational:
[tex]m=\rho *Vol=\rho*4\pi*d^{3}/24\\ F=mg=g*\rho*4\pi*d^{3}/24=9.81*1000*4\pi(1.4*10^{-6})^{3}/24=1.41*10^{-14}N[/tex]
As the water drop is suspended in calm air, the Gravitational force is compensated with the electrical Force:
[tex]F_{e}=mg=1.41*10^{-14}N[/tex]
But:
[tex]F_{e}=Q*E=N*q_{e}*E[/tex]
We solve in order to find the number of electrons in excess at the water drop:
[tex]N=F_{e}/(q_{e}*E)=1.41*10^{-14}/(1.6*10^{-19}*587*10^{-9})=1.5*10^{11}electrons[/tex]
