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

Answer:

The value of developed electric force is $3.516\times 10^{- 7} N$

Solution:

As per the question:

Mass of the droplet = 1.8 mg = $1.8\times 10^{- 6} kg$

Charge on droplet, Q = $25 pC = 25\times 10^{- 12} C$

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

Now, the Electrostatic force given by Coulomb:

$F_{E} = \frac{1}{4\pi epsilon_{o}}.\frac{Q^{2}}{D^{2}}$

$\frac{1}{4\pi epsilon_{o}} = 9\times 10^{9} m/F$

$F_{E} = (9\times 10^{9}).\frac{(25\times 10^{- 12})^{2}}{0.004^{2}}$

$F_{E} = 3.516\times 10^{- 7} N$

The magnitude of force is too low to be noticed.

Answer 2

Answer:

The electric force is

$F = 3.5 \times 10^{-7} N$

Solution:

As we know that the Charge on droplet is,

$Q =25 \times 10^{-12} C$

Distance between the 2 droplets,

r = 0.40 cm = 0.004 m

Now, the Electrostatic force given by Coulomb:

$F = \frac{kq_1q_2}{r^2}$

here we will plug in data into above equation

$F = \frac{(9\times 10^9)(25 \times 10^{-12})(25 \times 10^{-12})}{(0.004)^2}$

$F = 3.5 \times 10^{-7} N$