Question

A spherical water drop 1.20µm in diameter is suspended in calm air owing to a downward–directed atmospheric electric field `E = 462 N/C`. (a) What is the weight of the drop? (b) How many excess electrons does it have?

Answer 1

Weight of the water droplet: `w=8.8668\times10^{-15} N`,
** Number of excess electrons** `N_e=120`.

Explanation:

`m` - mass of the water droplet ( in kg ),
`q` - the total charge on the water droplet ( in coulombs ),
`E=462 N/C` - electric field strength,

`e=1.602176\times10^{-19} C` - fundamental unit of charge,
`g = 9.8 m/s^2` - acceleration due to gravity,
`\rho_w = 1000.0 (kg)/m^3` - density of water,
`d=1.20\times10^{-6} m` - diameter of the water droplet,

(a) Weight of the water droplet :

Volume of the water droplet: `V=4/3\pi(d/2)^3=9.0478\times10^{-19} m^3`
Mass of the water droplet: `m=V\rho_w = 9.0478\times10^{-16} kg`
Weight of the water droplet: `w=mg= 8.8668\times10^{-15}N`

(b) How many excess electrons: To calculate the number of excess electrons first calculate the total charge on the water droplet using the fact that its weight (`mg`) balances its electric force (`F_E=E.q`).

Total charge: `F_E=m.g \qquad \rightarrow q = (mg)/E=1.91922\times10^{-19}C`

If `N_e` is the number of excess electrons, `q=N_e.e`

`N_e=q/e = 120`