Question

A spherical water drop 1.4 um in diameter is suspended In calm air due to a downward- directed atmospheric electic leld of magni-587 NC. (a) What i the magnitude of the gravitational force on the drop? (b) How many excess electrons does it have?

Answer 1

`N=1.5*10^(11)` electrons en excess

Step-by-step explanation:

We use the density of water equal to 1000 Kg/m^3, in order to find the mass and Force gravitational:

`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`

As the water drop is suspended in calm air, the Gravitational force is compensated with the electrical Force:

`F_(e)=mg=1.41*10^(-14)N`

But:

`F_(e)=Q*E=N*q_(e)*E`

We solve in order to find the number of electrons in excess at the water drop:

`N=F_(e)/(q_(e)*E)=1.41*10^(-14)/(1.6*10^(-19)*587*10^(-9))=1.5*10^(11)electrons`