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 +21 pC. The centers of the droplets are at the same height and 0.44 cm apart.a. What is the approximate electric force between the droplets?b. What horizontal acceleration does this force produce on the droplets?

Answer 1

Step-by-step explanation:

mass of each drop, m = 1.8 mg = 1.8 x 10^-6 kg

Charge on each drop, q = 21 pC = 21 x 10^-12 C

distance, d = 0.44 cm = 0.0044 m

(a) The formula for the electrostatic force between them is given by

`F =(Kq^(2))/(d^(2))`

`F =(9* 10^(9)* 21* 10^(-12)* 21* 10^(-12))/(0.0044^(2))`

F = 2.05 x 10^-7 N

Thus, the force is 2.05 x 10^-7 N .

(b) Let a be the acceleration.

a = F / m

`a = (2.05* 10^(-7))/(1.8* 10^(-6))`

a = 0.114 m/s^2

Answer 2

a)
`F=2.048* 10^(-7)\ N`

b)
`a=0.1138\ m.s^(-2)`

Step-by-step explanation:

Given:

• mass of raindrops,
  `m=1.8* 10^(-6)\ kg`

• charge on the raindrops,
  `q=+21* 10^(-12)\ C`

• horizontal distance between the raindrops,
  `r=0.0044\ m`

A)

From the Coulomb's Law the force between the charges is given as:

`F=(1)/(4\pi.\epsilon_0) .(q_1.q_2)/(r^2)`

we have:

`\epsilon_0=8.854* 10^(-12)\ C^2.N^(-1).m^(-2)`

Now force:

`F=(1)/(4\pi* 8.854* 10^(-12)) .(21* 10^(-12)* 21* 10^(-12))/(0.0044^2)`

`F=2.048* 10^(-7)\ N`

B)

Now the acceleration on the raindrops due to the electrostatic force:

`a=(F)/(m)`

`a=(2.048* 10^(-7))/(1.8* 10^(-6))`

`a=0.1138\ m.s^(-2)`