Question

A spherical water droplet of radius 35 um carries an excess 236 electrons. What vertical electric field in N/C) is needed to balance the gravitational force on the droplet at the surface of the earth? (Assume the density of a water droplet is 1,000 kg/m". Enter the magnitude.)

Answer 1

Answer:

4.66 x 10^7 N/C

Explanation:

radius of drop, r = 35 micro meter = 35 x 10^-6 m

number of electrons, n = 26

density of water, d = 1000 kg / m^3

The gravitational force is equal to the product of mass of drop and the acceleration due to gravity.

Mass of drop, m = Volume of drop x density of water

$m = \frac{4}{3}\times 3.14 \times r^{3}\times d$

$m = \frac{4}{3}\times 3.14 \times \left ( 35\times10^{-6} \right )^{3}\times 1000$

m = 1.795 x 10^-10 kg

Gravitational force, Fg = m x g = 1.795 x 10^-10 x 9.8 = 1.759 x 10^-9 N

Charge,q = number of electrons x charge of one electron

q = 236 x 1.6 x 10^-19 = 3.776 x 10^-17 C

Electrostatic force, Fe = q E

Where, E is the strength of electric field.

here electrostatic force is balanced by the gravitational force

Fe = Fg

3.776 x 106-17 x E = 1.759 x 10^-9

E = 4.66 x 10^7 N/C