Question

INT Raindrops acquire an electric charge as they fall. Suppose a 2.0-mm-diameter drop has a charge of 12 pC; these are both very common values. In a thunderstorm, the electric field under a cloud can reach 15,000 N/C, directed upward. For a droplet exposed to this field, how do the magnitude and direction of the electric force compare to those of the weight force

Answer 1

The magnitude and direction of the electric force acting on a raindrop exposed to an electric field in a thunderstorm can be compared to the weight force acting on the droplet. The electric force is greater in magnitude and opposite in direction to the weight force.

Step-by-step explanation:

The magnitude and direction of the electric force acting on a raindrop exposed to an electric field in a thunderstorm can be compared to the weight force acting on the droplet. In this case, the electric force is given by Fe = qeE, where qe is the charge of the droplet and E is the electric field strength. The weight force is given by Fw = mg, where m is the mass of the droplet and g is the acceleration due to gravity.

In the given scenario, the magnitude of the electric force can be calculated as Fe = (12 pC)(15,000 N/C) = 180,000 pN. The weight force can be calculated as Fw = (volume of the droplet)(density of water)(g), where the volume of the droplet is (4/3)(π)((diameter/2)^3) and the density of water is 1000 kg/m^3 (1 pN = 10^-12 N).

The direction of the electric force is upward, while the direction of the weight force is downward. Therefore, the magnitude of the electric force is greater than the magnitude of the weight force and the two forces act in opposite directions.

Answer 2

W = 2.3 10²
`F_(e)`

Step-by-step explanation:

The force of the weight is

W = m g

let's use the concept of density

ρ= m / v

the volume of a sphere is

V =
`(4)/(3)` π r³

V =
`(4)/(3)` π (1.0 10⁻³)³

V = 4.1887 10⁻⁹ m³

the density of water ρ = 1000 kg / m³

m = ρ V

m = 1000 4.1887 10⁻⁹

m = 4.1887 10⁻⁶ kg

therefore the out of gravity is

W = 4.1887 10⁻⁶ 9.8

W = 41.05 10⁻⁶ N

now let's look for the electric force

F_e = q E

F_e = 12 10⁻¹² 15000

F_e = 1.8 10⁻⁷ N

the relationship between these two quantities is

`(W)/(F_e)` = 41.05 10⁻⁶ / 1.8 10⁻⁷

\frac{W}{F_e} = 2,281 10²

W = 2.3 10²
`F_(e)`

therefore the weight of the drop is much greater than the electric force