SCALE 1:1 STD NEC · IEC · AS/NZS · ACI · AISC
B
CodePass.PRO
Engineering Calculators
SHEET 10 / 79
Calculator · Electrical · Coulomb · Maxwell · Ohm

Electricity

Coulomb\'s law calculator with the constant k = 8.99 × 10⁹ N·m²/C², plus a reference covering impedance, DC vs AC, two-phase and three-phase systems, and the formulas that connect them all. The physics anchor for every other calculator on this site. Reviewed by a licensed PE.

Use the Coulomb\'s law calculator

Coulomb\'s law is the foundational equation for the electric force between two point charges, with the constant k = 8.99 × 10⁹ N·m²/C² appearing as the proportionality. Enter two charges, the distance between them, and pick a medium — the calculator returns force, electric field, electric potential, and potential energy.

CALC.006 Coulomb's Law · F · E · V · U · k = 8.99 × 10⁹ N·m²/C²

Coulomb's constant in vacuum: k = 8.99 × 10⁹ N·m²/C². In a dielectric medium, the effective k is reduced by factor εr.

Force F between charges
— N
Force direction depends on charge signs: like signs repel, opposite signs attract.
Force direction
Effective k (vacuum: 8.99e9)
Electric field at q₂ position
Electric potential at q₂ position
Potential energy U
FORMULA · F = k · q₁ · q₂ / r² SOURCE · COULOMB 1785 · SI 2019
COULOMB'S LAW · F = k · q₁ · q₂ / r² + q₁ +1.0 µC + q₂ +1.0 µC F F distance r = 1.0 m F = k · q₁ · q₂ / r² k = 8.99 × 10⁹ N·m²/C² repulsive if same sign attractive if opposite TORSION BALANCE · Coulomb 1785 θ ∝ F
Figure 1 — Coulomb's Law: electrostatic force between two point charges, with the historic torsion balance inset

The electricity formulas

Three families of formulas cover most practical electrical engineering: the electrostatic relations (Coulomb), the resistive-circuit relations (Ohm), and the AC impedance relations (Z, X). Each underlies a different class of calculator on this site.

Eq. 01 — Coulomb's law (electric force between two point charges) SI · Coulomb 1785
F=kq1q2r2k=14πε08.99×109Nm2C2F = \frac{k \cdot q_{1} \cdot q_{2}}{r^{2}} \qquad k = \frac{1}{4 \pi \varepsilon_{0}} \approx 8.99 \times 10^{9} \, \frac{N \cdot m^{2}}{C^{2}}
F
magnitude of electric force, N
q_1, q_2
point charges (signed), C
r
distance between the charges, m
k
Coulomb's constant, N·m²/C²
ε_0
permittivity of free space ≈ 8.854 × 10⁻¹² F/m, F/m
Eq. 02 — Ohm's law (resistive circuit) SI · Ohm 1827
V=IRP=VI=I2R=V2RV = I \cdot R \qquad P = V \cdot I = I^{2} \cdot R = \frac{V^{2}}{R}
V
voltage across the element, V
I
current through it, A
R
resistance, Ω
P
electrical power dissipated, W
Eq. 03 — Impedance in an AC circuit SI · Steinmetz 1893
Z=R2+X2XL=2πfLXC=12πfCZ = \sqrt{R^{2} + X^{2}} \qquad X_{L} = 2 \pi f L \qquad X_{C} = \frac{1}{2 \pi f C}
Z
impedance magnitude, Ω
R
resistance (in-phase), Ω
X
net reactance (90° component), Ω
X_L
inductive reactance, Ω
X_C
capacitive reactance, Ω
f
frequency, Hz

Worked example: force between two 1 µC charges

Two point charges of +1 µC each, separated by 50 cm in air. What force acts on each charge?

StepCalculationResult
Convert charges to SI1 µC = 1 × 10⁻⁶ C10⁻⁶ C each
Convert distance to SI50 cm = 0.5 m0.5 m
Apply Coulomb\'s lawF = (8.99 × 10⁹) × (10⁻⁶) × (10⁻⁶) / 0.5²3.6 × 10⁻² N
Air correction (εᵣ ≈ 1.0006)negligible — divide by 1.0006~3.6 × 10⁻² N
Resultboth charges positive → repulsive~36 mN, repulsive
Same charges in water (εᵣ ≈ 80)F / 80~0.45 mN

The 80× reduction in water explains why dielectric materials are used to insulate capacitor plates — they let you store much more charge at the same voltage by reducing the effective Coulomb force trying to push the charges apart.