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

Answer:

a) $F=2.048\times 10^{-7}\ N$

b) $a=0.1138\ m.s^{-2}$

Explanation:

Given:

• mass of raindrops, $m=1.8\times 10^{-6}\ kg$

• charge on the raindrops, $q=+21\times 10^{-12}\ C$

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

A)

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

$F=\frac{1}{4\pi.\epsilon_0} .\frac{q_1.q_2}{r^2}$

we have:

$\epsilon_0=8.854\times 10^{-12}\ C^2.N^{-1}.m^{-2}$

Now force:

$F=\frac{1}{4\pi\times 8.854\times 10^{-12}} .\frac{21\times 10^{-12}\times 21\times 10^{-12}}{0.0044^2}$

$F=2.048\times 10^{-7}\ N$

B)

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

$a=\frac{F}{m}$

$a=\frac{2.048\times 10^{-7}}{1.8\times 10^{-6}}$

$a=0.1138\ m.s^{-2}$

Answer 2

Answer:

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

$F =\frac{Kq^{2}}{d^{2}}$

$F =\frac{9\times 10^{9}\times 21\times 10^{-12}\times 21\times 10^{-12}}{0.0044^{2}}$

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

$a = \frac{2.05\times 10^{-7}}{1.8\times 10^{-6}}$

a = 0.114 m/s^2