Question

Falling raindrops frequently develop electric charges. Does this create noticeable forces between the droplets? Suppose two 1.8 mg drops each have a charge of +25 pC; these are typical values. The centers of the droplets are at the same height and 0.40 cm apart. What is the approximate electric force between them?

Answer 1

The electric force is

`F = 3.5 * 10^(-7) N`

Solution:

As we know that the Charge on droplet is,

`Q =25 * 10^(-12) C`

Distance between the 2 droplets,

r = 0.40 cm = 0.004 m

Now, the Electrostatic force given by Coulomb:

`F = (kq_1q_2)/(r^2)`

here we will plug in data into above equation

`F = ((9* 10^9)(25 * 10^(-12))(25 * 10^(-12)))/((0.004)^2)`

`F = 3.5 * 10^(-7) N`

Answer 2

The value of developed electric force is
`3.516* 10^(- 7) N`

Solution:

As per the question:

Mass of the droplet = 1.8 mg =
`1.8* 10^(- 6) kg`

Charge on droplet, Q =
`25 pC = 25* 10^(- 12) C`

Distance between the 2 droplets, D = 0.40 cm = 0.004 m

Now, the Electrostatic force given by Coulomb:

`F_(E) = (1)/(4\pi epsilon_(o)).(Q^(2))/(D^(2))`

`(1)/(4\pi epsilon_(o)) = 9* 10^(9) m/F`

`F_(E) = (9* 10^(9)).((25* 10^(- 12))^(2))/(0.004^(2))`

`F_(E) = 3.516* 10^(- 7) N`

The magnitude of force is too low to be noticed.