Falling raindrops frequently develop electric charges. does this create noticeable forces between the droplets

The value of developed electric force is

Solution:

As per the question:

Mass of the droplet = 1.8 mg =

Charge on droplet, Q =

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

Now, the Electrostatic force given by Coulomb:

The magnitude of force is too low to be noticed.

The electric force is

Solution:

As we know that the Charge on droplet is,

Distance between the 2 droplets,

r = 0.40 cm = 0.004 m

Now, the Electrostatic force given by Coulomb:

here we will plug in data into above equation

F = 5.83 10⁻¹⁷ N

Explanation:

The electric force is given by

    F = k q₁  q₂ / r²

With Gauss's law electric field flow is equal to the charge inside the Gaussian surface, if we make a spherical surface around each drop, the force independent of small deformations due to air resistance

   q₁ = q₂

   F = 8.99 10⁹ (29 10⁻¹²)² / (0.36 10⁻²)²

   F = 5.83 10⁻¹⁷ N

As the two drops have a charge of the same sign they repel