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


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


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.

The electric force is

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


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

F = 5.83 10⁻¹⁷ N


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

Do you know the answer?

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