Question

Raindrops acquire an electric charge as they fall. Suppose a 2.4-mm-diameter drop has a charge of +13 pC. 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 of the electric force compare to those of the weight force and what is the dircetion of the electric force?

Answer 1

Answer:

Explanation:

13pC = 13 X 10⁻¹² C

Electric field E = 15000 N/C

Electric force on the charge

= charge x electric field

= 13 x 10⁻¹² x 15000

= 195 x 10⁻⁹ N.

It will act in upward direction as force on positive charge acts in the direction in which electric field exists.

Volume of droplet = 4/3 π R³

R = 2.4 X 10⁻³ m

Volume V = 4/3 x 3.14 x ( 2.4 x 10⁻³)³

= 57.87 x 10⁻⁹ m³

density of water = 1000 kg / m³

mass of water droplet

density x volume

1000 x 57. 87 x 10⁻⁹ kg

= 57.87 x 10⁻⁶ kg .

Weight = mass x g

= 57.87 x 10⁻⁶ x 9.8

= 567.126 x 10⁻⁶ N.

Weight is more than the electric force.

Answer 2

Answer:

567.126 x 10⁻⁶ N

$5.6 x 10^{-6} N$

Explanation:

Thinking process:

$13pC = 13 x 10^{-12} C$

The electric field is given by E = 15000 N/C

Electric force on the charge

= charge x electric field

= 13 x 10⁻¹² x 15000

= 195 x 10⁻⁹ N.

The force acts in upward direction as force on positive charge acts in the direction in which electric field exists.

Volume of droplet = 4/3 π R³

R = 2.4 X 10⁻³ m

Volume V = 4/3 x 3.14 x ( 2.4 x 10⁻³)³

= 57.87 x 10⁻⁹ m³

density of water = 1000 kg / m³

mass of water droplet = density x volume

                                     = 1000 x 57. 87 x 10⁻⁹ kg

                                     = 57.87 x 10⁻⁶ kg .

                        Weight = mass x g

                                    = 57.87 x 10⁻⁶ x 9.8

                                     = 567.126 x 10⁻⁶ N.

Therefore, the weight is more than the electric force.