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Coulomb's Law Calculator

Free Coulomb's law calculator for electrostatic force, electric field, and potential energy calculations with step-by-step solutionsfor physics and electrostatics. Perfect for students learning electromagnetic interactions.

Last updated: December 15, 2024

Three calculation types: force, electric field, potential energy
Coulomb's constant and unit conversions
Step-by-step solutions with formulas

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Coulomb's Law Calculator
Calculate electrostatic force, electric field, and potential energy

Coulomb's Law Results

Force:

5.394e+0 N

Coulomb Force

Formula: F = k × |q₁ × q₂| / r²

Force:

5.394e+0 N

Electric Field:

1.8e+6 N/C

Potential Energy:

5.394e-1 J

Step-by-Step Solution:

Step 1: Given parameters
Charge 1 (q₁) = 2.0e-6 C
Charge 2 (q₂) = 3.0e-6 C
Distance (r) = 0.1 m
Step 2: Apply Coulomb's law formula
F = k × |q₁ × q₂| / r²
F = 9.0e+9 × |2.0e-6 × 3.0e-6| / (0.1)²
F = 9.0e+9 × 6.0e-12 / 1.0e-2
F = 5.394e+0 N
Step 3: Calculate electric field
E = F / q₂ = 5.394e+0 / 3.0e-6 = 1.8e+6 N/C

Coulomb's Law Tips:

  • • F = k × |q₁ × q₂| / r² for electrostatic force
  • • E = k × |q| / r² for electric field
  • • U = k × q₁ × q₂ / r for potential energy
  • • k = 8.99 × 10⁹ N⋅m²/C² (Coulomb's constant)
  • • Like charges repel, opposite charges attract

Coulomb's Law Types

Electrostatic Force
F = k × |q₁ × q₂| / r²

Formula

F = k × |q₁ × q₂| / r²

Force between two charged particles

Electric Field
E = k × |q| / r²

Formula

E = k × |q| / r²

Electric field at a point in space

Potential Energy
U = k × q₁ × q₂ / r

Formula

U = k × q₁ × q₂ / r

Electrostatic potential energy

Capacitor Design
Electronics applications

Example

Electric field between plates

E = σ/ε₀ for parallel plate capacitor

Particle Physics
Charged particle interactions

Example

Electron-proton attraction

F = k × e² / r² for hydrogen atom

Electrostatic Precipitators
Air pollution control

Example

Particle collection efficiency

Force on charged particles in electric field

Quick Example Result

Coulomb's law with q₁ = 2.0 × 10⁻⁶ C, q₂ = 3.0 × 10⁻⁶ C, r = 0.1 m:

Force

5.394 mN

Electric Field

2.697 kN/C

Potential Energy

0.539 mJ

How to Calculate Coulomb's Law

Coulomb's law is a fundamental principle in electrostatics that describes the force between two charged particles. Understanding this law is essential for physics, electrical engineering, and electromagnetic applicationswhere charged particle interactions are important.

The Coulomb's Law Process

Step 1: Identify the charges (q₁, q₂) and distance (r)
Step 2: Choose the calculation type (force, field, or energy)
Step 3: Apply the appropriate Coulomb's law formula
Step 4: Use Coulomb's constant k = 8.99 × 10⁹ N⋅m²/C²
Step 5: Calculate and interpret the results

This systematic approach ensures accurate Coulomb's law calculations for any electrostatics problem.

Coulomb's Law Formulas

The three main formulas are: F = k × |q₁ × q₂| / r² for electrostatic force, E = k × |q| / r² for electric field, and U = k × q₁ × q₂ / r for potential energy. Coulomb's constant k = 8.99 × 10⁹ N⋅m²/C² relates the force to the charges and distance. The law follows an inverse square relationship, meaning force decreases as 1/r².

  • Electrostatic Force: F = k × |q₁ × q₂| / r²
  • Electric Field: E = k × |q| / r²
  • Potential Energy: U = k × q₁ × q₂ / r
  • Coulomb's Constant: k = 8.99 × 10⁹ N⋅m²/C²
  • Like charges repel, opposite charges attract

Sources & References

  • Introduction to Electrodynamics - David J. GriffithsComprehensive coverage of electromagnetic theory including Coulomb's law
  • Physics for Scientists and Engineers - Raymond A. Serway, John W. JewettDetailed explanation of electrostatics and Coulomb's law applications
  • Khan Academy - Electrostatics and Coulomb's LawVideo tutorials and practice problems on electromagnetic interactions

Need help with other physics topics? Check out our free fall calculator and kinetic energy calculator.

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Coulomb's Law Example

Step-by-Step Solution
Calculating electrostatic force for q₁ = 2.0 × 10⁻⁶ C, q₂ = 3.0 × 10⁻⁶ C, r = 0.1 m

Given Parameters:

Charge 1 (q₁) = 2.0 × 10⁻⁶ C

Charge 2 (q₂) = 3.0 × 10⁻⁶ C

Distance (r) = 0.1 m

k = 8.99 × 10⁹ N⋅m²/C²

Solution Steps:

  1. Step 1: Given parameters
  2. Charge 1 (q₁) = 2.0e-6 C
  3. Charge 2 (q₂) = 3.0e-6 C
  4. Distance (r) = 0.1 m
  5. Step 2: Apply Coulomb's law formula
  6. F = k × |q₁ × q₂| / r²
  7. F = 9.0e+9 × |2.0e-6 × 3.0e-6| / (0.1)²
  8. F = 9.0e+9 × 6.0e-12 / 1.0e-2
  9. F = 5.394e+0 N
  10. Step 3: Calculate electric field
  11. E = F / q₂ = 5.394e+0 / 3.0e-6 = 1.8e+6 N/C

Final Results:

Electrostatic Force

5.394e+0 N

Electric Field

1.8e+6 N/C

Potential Energy

5.394e-1 J

Coulomb's Constant

8.99 × 10⁹ N⋅m²/C²

Capacitor Design

q₁ = 1.0 × 10⁻⁶ C, q₂ = -1.0 × 10⁻⁶ C, r = 0.01 m

F = 89.9 N (attractive)

Atomic Physics

e = 1.6 × 10⁻¹⁹ C, r = 5.3 × 10⁻¹¹ m (Bohr radius)

F = 8.2 × 10⁻⁸ N

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