Calculating the radius of a water drop suspended in the earth's electric field:

To calculate the radius of a water drop that would remain suspended in the earth's electric field of 300 V/m when charged with one electron, we need to use the formula for the electric force on a charged particle in an electric field.

The formula is:

F = qE

Where F is the electric force, q is the charge on the particle, and E is the electric field.

Step 1: Calculate the electric force on a single electron in the earth's electric field of 300 V/m.

We know that the charge on a single electron is -1.6 x 10^-19 Coulombs, and the electric field in this case is 300 V/m.

Plugging these values into the formula, we get:

F = (-1.6 x 10^-19 C) x (300 V/m) = -4.8 x 10^-17 N

The negative sign indicates that the force is acting in the opposite direction to the electric field.

Step 2: Calculate the weight of the water droplet.

To determine the radius of the water droplet, we need to know its weight. For simplicity, we will assume that the droplet is a perfect sphere and calculate the weight using the formula:

W = (4/3)πr^3ρg

Where W is the weight, r is the radius, ρ is the density of water, and g is the acceleration due to gravity.

Assuming a density of 1000 kg/m^3 for water, and an acceleration due to gravity of 9.8 m/s^2, we get:

W = (4/3)πr^3(1000)(9.8) = 41.2r^3 N

Step 3: Equate the electric force on the electron with the weight of the water droplet.

Since the droplet is in equilibrium, the electric force on the electron must be equal and opposite to the weight of the droplet. Therefore, we can equate the two forces and solve for the radius of the droplet:

-4.8 x 10^-17 N = 41.2r^3 N

r^3 = (-4.8 x 10^-17 N) / (41.2 N)

r^3 = -1.166 x 10^-18 m^3

Taking the cube root of both sides, we get:

r = 2.8 x 10^-7 m

Therefore, the radius of the water droplet that would remain suspended in the earth's electric field of 300 V/m when charged with one electron is approximately 2.8 x 10^-7 m.

Explanation:

The electric field in the earth's atmosphere is caused by the separation of positive and negative charges in the atmosphere. When a water droplet is charged with one electron, it experiences an electric force in the opposite direction to the electric field. If this force is equal and opposite to the weight of the droplet, the droplet will remain suspended in the air. By equating the electric force on the electron with the weight of the droplet, we can calculate the radius of the droplet that would remain suspended in the electric field.