Find the number N of excess electrons, which must be on each of two identical water droplets with masses

Find the number N of excess electrons, which must be on each of two identical water droplets with masses m1 = m2 = 12g, so that the force of their electrostatic repulsion would balance the force of their gravitational attraction.

The force of gravitational attraction between drops of the same mass m:
F₁ = Gm² / R².
Coulomb force between drops with the same charge q:
F₂ = kq² / R².
Each charge q is formed by the charge of N electrons, which is to be determined:
q = Ne, where e is the electron charge.
F₂ = kN²e² / R².
By condition F₁ = F₂:
Gm² / R² = kN²e² / R².
After multiplying by R² we get:
Gm² = kN²e²
N = root (Gm² / ke²) = (m / e) * root (G / k) =
= 0.012 kg / 1.6 * 10⁻¹⁹ K * root (6.67 * 10⁻¹¹ m³ * kg⁻¹ * s⁻² / 8.98 * 10⁹ N * m² * K⁻²) ≈ 6.5 * 10⁶.

Answer: N ≈ 6.4 * 10⁶.