Q1. Two point charges +3 µC and −3 µC are 20 cm apart. Find the force between them.
F = k q₁q₂/r² = 9×10⁹ × 3×10⁻⁶ × 3×10⁻⁶ / (0.2)²
F = 9×10⁹ × 9×10⁻¹² / 0.04
F = 8.1×10⁻² / 0.04 = 2.02 N (attractive)
Everything electrostatic starts here. Use the Coulomb calculator to feel the inverse-square force, then practise field and force numericals.
Set two charges and their separation to read the electrostatic force between them.
Halve the distance and the force becomes four times as strong — that is the inverse-square law.
Coulomb's law and electric-field problems with steps.
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Q1. Two point charges +3 µC and −3 µC are 20 cm apart. Find the force between them.
F = k q₁q₂/r² = 9×10⁹ × 3×10⁻⁶ × 3×10⁻⁶ / (0.2)²
F = 9×10⁹ × 9×10⁻¹² / 0.04
F = 8.1×10⁻² / 0.04 = 2.02 N (attractive)
Q2. How many electrons make up a charge of −1 C?
q = n e ⇒ n = q/e
n = 1 / (1.6×10⁻¹⁹)
n = 6.25×10¹⁸ electrons
| Quantity | Formula | SI unit |
|---|---|---|
| Coulomb's law | F = k q₁ q₂ / r² | N (k = 9×10⁹) |
| Electric field | E = F/q = k Q / r² | N/C |
| Gauss's law | Φ = q / ε₀ | N·m²/C |
| Field of a point charge | E = k Q / r² | N/C |
| Charge quantization | q = n e | C (e = 1.6×10⁻¹⁹) |
The electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them: F = kq₁q₂/r².
The region around a charge where another charge experiences a force; numerically the force per unit positive charge, E = F/q, in N/C.
The net electric flux through any closed surface equals the total charge enclosed divided by ε₀ (Φ = q/ε₀).
Charge exists only in integer multiples of the elementary charge e = 1.6×10⁻¹⁹ C, so q = ne.
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