# The electron charge is e = 1.6 × 10−19 C, and its mass is me = 9.11 × 10−31 kg.

**The electron charge is e = 1.6 × 10−19 C, and its mass is me = 9.11 × 10−31 kg. How many times is the force of the Coulomb repulsion of two electrons greater than the force of their gravitational attraction?**

The force of electrostatic repulsion of two electrons having a charge e located at a distance R is found according to the Coulomb law: F₁ = k ∙ | q₁ ∙ q₂ | / R ^ 2, where q₁ and q₂ are the charges of interacting bodies, the coefficient k = 9 ∙ 10 ^ 9 (H ∙ m ^ 2) / Cl ^ 2. Since q₁ = q₂ = e, then:

F₁ = k ∙ e ^ 2 / R ^ 2.

The force of gravitational attraction is determined using the law of universal gravitation: F₂ = G ∙ m₁ ∙ m₂ / R ^ 2, where m₁ and m₂ are the masses of interacting bodies, the coefficient G = 6.67 ∙ 10 ^ (- 11) N ∙ m ^ 2 / kg ^ 2 – gravitational constant. Since m₁ = m₂ = m, where m is the electron mass, then:

F₂ = G ∙ m ^ 2 / R ^ 2.

To compare how many times the force of the Coulomb repulsion of two electrons is greater than the force of their gravitational attraction, we find the ratio: F₁: F₂ = (k ∙ e ^ 2 / R ^ 2): (G ∙ m ^ 2 / R ^ 2); F₁: F₂ = k ∙ e ^ 2 / (G ∙ m ^ 2). Since the electron charge is e = 1.6 ∙ 10 ^ (- 19) C, and its mass is m = 9.11 ∙ 10 ^ (- 31) kg, then:

F₁: F₂ = (9 ∙ 10 ^ 9 (H ∙ m ^ 2) / Kl ^ 2 ∙ (1.6 ∙ 10 ^ (- 19) Kl) ^ 2) / ((6.67 ∙ 10 ^ (- 11 ) N ∙ m ^ 2 / kg ^ 2 ∙ (9.11 ∙ 10 ^ (- 31) kg) ^ 2));

F₁: F₂ = 4.2 ∙ 10 ^ 42.

Answer: the force of the Coulomb repulsion of two electrons is 4.2 ∙ 10 ^ 42 times greater than the force of their gravitational attraction.