Question

A spherical water drop 1.6 um in diameter is suspended in calm air due to a downward-directed atmospheric electric field of magnitude E = 1000 N/C. 

(a) What is the magnitude of the gravitational force on the drop? (b) How many excess electrons does it have? 

(a) Number  _______  Units _______ 

(b) Number  _______  Units _______

Answer 1

SOLUTION :

a.

Volume of water drop = 4/3 pi r^3 = 4/3 pi (d/2)^3 = pi d^3 / 6 

= pi * (1.6 * 10^(-6)) ^3 / 6

=  1.6^3 * pi / 6 *10^(-18)

= 2.14466 * 10^(-18)  m^3 

Density of water = 1000 kg/m^3 

So, mass of water = 1000 * 2.14466 * 10^(-18) = 2.14466 * 10^(-15) kg 

So, gravitational force on the drop 

= m*g

= 2.14466 *10^(-15) * 9.8  (N)

= 2.1017668 *10^(-14)  N  (ANSWER).

b.

In equilibrium, force of electric field , Fe = Gravitational force, Fg

=> Number of electrons * per electron force * Electric field magnitude = 2.1017668 * 10^(-14)

=> Number of electrons * 1.6 * 10^(-19) * 1000  = 2.1017668 * 10^(-14)

=> Number of electrons = 2.1017668 * 10^(-14) / (1.6 * 10^(-16) 

=> Number of electrons = 131 (ANSWER) 

Answer 2

Answer 3

first part is wrong