Question

A water droplet of radius 0.027 mm remains stationary in the air.

If the downward-directed electric field of the Earth is 150 N/C, how many excess electron charges must the water droplet have?

Answer 1

Givene:

Radius of the droplet: .000027 m

E = 150 N/C

Since there is no equation that can relate these two, we'll have to first use the radius to find mass of the droplet.

m = d(4/3)pi r^3

d = (density) given = 1000

m = (1000)(4/3)(pi)(.000027)^3

m = 8.24e-11

Now that we have enough variables, we can use an equation to relate E with m and g. Eq is the electric force, mg is the force of gravity.

Eq = mg

(150)(q) = (8.24e-11)(-9.8)

q = -5.4e-12

To find the number of electrons, we just divide by the electron charge, 1.602e-19.

Number of extra electrons =-5.4e-12/1.602e-19 = -3.3e6

I hope help you.....