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Coulomb's law calculator

k = 9 000 000 000

Formula

$$F = k\,\frac{|q_1|\,|q_2|}{r^2}$$

Magnitude of the force with which two point charges in vacuum act on each other

ENT-level problem

Two point charges $q_1 = 2$ uC and $q_2 = 3$ uC are placed $r = 0.1$ m apart. Find the electrostatic force between the charges. Coulomb's constant is $k = 9 \cdot 10^9$ N·m²/C².

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Coulomb's law:

$$F = k \frac{q_1 q_2}{r^2} = 9 \cdot 10^9 \cdot \frac{2 \cdot 10^{-6} \cdot 3 \cdot 10^{-6}}{(0.1)^2}$$

$$F = 9 \cdot 10^9 \cdot \frac{6 \cdot 10^{-12}}{10^{-2}} = \frac{54 \cdot 10^{-3}}{10^{-2}} = 5.4 \text{ N}$$

Explanation

The force with which one point charge acts on another is proportional to the product of the magnitudes |q₁|·|q₂| and inversely proportional to the square of the distance r between them. The signs do not enter the formula — they only set the direction: like signs repel, unlike signs attract (on the scene the arrows F₁ and F₂ either point away from each other or toward each other). The forces on the two charges are equal in magnitude and opposite in direction — Newton's third law. The inverse square is the key part of the formula: half the distance — four times the force, a third of the distance — nine times. Example: q₁ = q₂ = 5 nC = 5·10⁻⁹ C, r = 0.4 m → F = 9·10⁹ · (5·10⁻⁹)² / 0.4² ≈ 1.4·10⁻⁶ N. The law holds for point charges (or balls much smaller than r) in vacuum.

Frequently asked questions

What is the value of Coulomb's constant?

In vacuum k = 9 · 10⁹ N·m²/C²; in air the value is nearly the same.

How does the force depend on distance?

The force falls off with the square of the distance: twice as far — four times weaker (inverse square, like gravity).

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