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

Answer:

W = 2.3 10² $F_{e}$

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 = $\frac{4}{3}$ π r³

         V = $\frac{4}{3}$ 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

          $\frac{W}{F_e}$ = 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