Coulomb's Law: Force Between Two Charges

Coulomb's law describes the force between two stationary point charges. It's the electrostatic analog of Newton's law of gravitation, with the same inverse-square form:

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

where $k = 9 \times 10^{9}\;\text{N}\cdot\text{m}^{2}/\text{C}^{2}$ is Coulomb's constant.

Direction:

• Like charges (both positive or both negative) → repel each other
• Opposite charges → attract each other

The force on each charge has the same magnitude (Newton's third law) but points in opposite directions along the line connecting them.

Step-by-Step Solution

Given: $q_1 = 2\;\text{μC} = 2 \times 10^{-6}\;\text{C}$, $q_2 = 3\;\text{μC} = 3 \times 10^{-6}\;\text{C}$, $r = 0.5\;\text{m}$.

Find: The magnitude and direction of the force between them.

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Step 1 — Plug into Coulomb's law.

$F = \dfrac{k\,q_1\,q_2}{r^{2}} = \dfrac{(9 \times 10^{9})(2 \times 10^{-6})(3 \times 10^{-6})}{(0.5)^{2}}$

Step 2 — Compute the numerator.

$(9 \times 10^{9})(2 \times 10^{-6})(3 \times 10^{-6})$

Combine the powers of 10:

$= 9 \times 2 \times 3 \times 10^{9 - 6 - 6}$

$= 54 \times 10^{-3}$

$= 0.054$

Step 3 — Compute the denominator.

$r^{2} = (0.5)^{2} = 0.25$

Step 4 — Divide.

$F = \dfrac{0.054}{0.25} = 0.216\;\text{N}$

Step 5 — Determine the direction.

Both charges are positive, so they repel each other. The force on $q_1$ points away from $q_2$, and the force on $q_2$ points away from $q_1$. The two forces are equal in magnitude (0.216 N each) and opposite in direction.

Step 6 — Sanity check the magnitude.

0.216 N is about 22 grams of force — comparable to the weight of a small AA battery. Two microcoulombs (millionths of a coulomb) at 0.5 m apart push each other with this force. To put it in context: 1 coulomb is an enormous charge. Two coulombs at 1 m apart would push each other with $F = 9 \times 10^{9}\;\text{N}$ — billions of newtons. That's why we never see free coulombs of charge: the forces are unimaginable.

Step 7 — How does the force scale with distance?

If you doubled the separation to 1 m:

$F = \dfrac{0.054}{1} = 0.054\;\text{N}$

That's $1/4$ of the original — confirming the inverse-square law. Halve the distance to 0.25 m, and the force becomes $4\times$ stronger: 0.864 N.

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Answer:

$\boxed{\;F = \dfrac{kq_1 q_2}{r^{2}} = 0.216\;\text{N}\;}$

The two positive charges repel each other with a force of about 0.216 N, in equal and opposite directions along their connecting line.

Try It

• Adjust the two charges (positive or negative) and the distance between them.
• Watch the force vectors update in real time.
• Try opposite charges — the force becomes attractive.
• The HUD shows both the magnitude and the inverse-square dependence on distance.