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gizmo_the_mogwai

3 months ago

6

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?

2 answers:

Sav [3.1K]3 months ago

7 0

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)

According to Coulomb's Law, the force acting between the charges is expressed as:

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

we obtain:

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

Now for the 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)

Next, we find the acceleration on the raindrops due to the electrostatic force:

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

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

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

Sav [3.1K]3 months ago

5 0

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 equation 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 amounts to 2.05 x 10^-7 N.

(b) Let a signify the acceleration.

a = F / m

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

a = 0.114 m/s^2

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