CalculateY
Quick Answer
Two 1 μC charges 1 m apart experience a force of approximately 0.009 N. A proton and electron at 0.053 nm (Bohr radius) experience approximately 8.2 × 10⁻⁸ N of attractive force.
Common Examples
| Input | Result |
|---|---|
| q₁ = 1×10⁻⁶ C, q₂ = -1×10⁻⁶ C, r = 1 m | F = 8.99×10⁻³ N (attractive) |
| q₁ = 1 C, q₂ = 1 C, r = 1 m | F = 8.99×10⁹ N (repulsive) |
| q₁ = 1.602×10⁻¹⁹ C, q₂ = -1.602×10⁻¹⁹ C, r = 5.3×10⁻¹¹ m | F ≈ 8.2×10⁻⁸ N (attractive) |
| F = 0.009 N, q₂ = 1×10⁻⁶ C, r = 1 m | q₁ ≈ 1×10⁻⁶ C |
How It Works
Coulomb’s law
\[F = k \frac{|q_1 q_2|}{r^2}\]Where:
- F = electrostatic force in Newtons (N)
- k = Coulomb’s constant = 8.9875 × 10⁹ N m²/C²
- q₁ = first charge in Coulombs (C), can be positive or negative
- q₂ = second charge in Coulombs (C), can be positive or negative
- r = distance between the charges in meters (m)
The absolute value signs mean force magnitude is always positive. The direction (attractive or repulsive) is determined separately by the signs of the charges.
Attractive vs. repulsive forces
Opposite charges (one positive, one negative) attract each other. Same-sign charges (both positive or both negative) repel each other. This follows from the sign of the product q₁q₂: a negative product indicates attraction, a positive product indicates repulsion.
Coulomb’s constant
k = 8.9875 × 10⁹ N m²/C² is related to the permittivity of free space (ε₀) by:
\[k = \frac{1}{4\pi\varepsilon_0}\]where ε₀ = 8.854 × 10⁻¹² C²/(N m²).
Inverse square law
Force decreases with the square of distance. Doubling the distance reduces the force to one-quarter of its original value. This is the same relationship as Newton’s law of universal gravitation, though gravitational forces are always attractive and are far weaker than electrostatic forces.
Rearranged forms
-
Force: $$F = k\frac{ q_1 q_2 }{r^2}$$ -
Distance: $$r = \sqrt{\frac{k q_1 q_2 }{F}}$$ -
Charge (q₁ or q₂): $$ q_1 = \frac{F r^2}{k q_2 }$$
Common charge values
The elementary charge (proton or electron) is 1.602 × 10⁻¹⁹ C. Practical charges in lab settings are often measured in microcoulombs (μC, 10⁻⁶ C) or nanocoulombs (nC, 10⁻⁹ C). A charge of 1 C is extremely large; a lightning bolt transfers roughly 1 to 5 C total.
Worked example
Two point charges, +3 μC and -5 μC, are separated by 0.2 m. Find the electrostatic force.
\[F = 8.9875 \times 10^9 \times \frac{|3 \times 10^{-6} \times (-5 \times 10^{-6})|}{(0.2)^2}\] \[F = 8.9875 \times 10^9 \times \frac{15 \times 10^{-12}}{0.04} = 8.9875 \times 10^9 \times 3.75 \times 10^{-10}\] \[F \approx 3.37 \text{ N (attractive)}\]Because the charges have opposite signs, the force is attractive.