Question

Two electrons exert a force of repulsion of 1.2 N on each other. How far apart are they? The elementary charge is 1.602 × 10−19 C and Coulomb’s constant is 8.987 × 109 N · m2/C2. Answer in units of m.

Answer 1

The distance between the charges is 13.86 X 10⁴m

Step-by-step explanation:

Given:

Force, F = 1.2N

Charge, q₁ = 1.602 X 10⁻¹⁹ C

k = 8.987 X 10⁹ Nm²/C²

Distance, d = ?

According to Coulomb's law:

`F = k(q_1q_2)/(r^2) \\\\`

Substituting the value in the formula we get:

`1.2 = 8.987X 10^9 X(1.602 X 10^-^1^9 X 1.602 X 10^1^9)/(r^2) \\\\1.2 = (23.06 X 10^9)/(r^2) \\\\r^2 = 19.22 X 10^9\\\\r = 13.86 X 10^4m`

Therefore, the distance between the charges is 13.86 X 10⁴m