Electric Field Calculator
Calculate the electric force between two charges using Coulomb's law (F=kq₁q₂/r²), the electric field from a point charge (E=kq/r²), the electric field between parallel plates (E=V/d), and fields from extended charge distributions — with full step-by-step working and interactive charge diagrams.
k = 8.988×10⁹ N·m²/C² | Always convert to SI before calculating
Coulomb's Law — Electric Force F=kq₁q₂/r²
Electric Field from Point Charge — E=kq/r²
For uniform electric field between parallel plates. E (V/m) = E (N/C) — these units are identical.
Uniform Electric Field — E = V/d
Calculate the net electric force on a test charge at the origin using the superposition principle. Add up to 4 source charges at specified (x,y) positions.
Net Electric Force — Superposition Principle
Infinite line charge — electric field falls as 1/r (not 1/r² like a point charge)
Infinite Wire Electric Field
Infinite plane — field is uniform, independent of distance from the plane
Infinite Plane Electric Field
Charged ring — on-axis electric field. Maximum at x = R/√2
Charged Ring — Axial Electric Field
Disk of charge — on-axis. As R→∞, approaches infinite plane result.
Disk of Charge — Axial Electric Field
| Source | Formula | Notes |
|---|---|---|
| Point charge | E = kq/r² | Inverse square, radial |
| Coulomb's force | F = kq₁q₂/r² | Attractive or repulsive |
| Uniform field | E = V/d | Parallel plates |
| Infinite wire | E = 2kλ/r | Linear, 1/r dependence |
| Infinite plane | E = σ/(2ε₀) | Uniform, distance-independent |
| Ring on axis | E = kQx/(x²+R²)^(3/2) | Max at x=R/√2 |
| Disk on axis | E = (σ/2ε₀)[1−x/√(x²+R²)] | → plane as R→∞ |
| Unit | Equivalent | Used For |
|---|---|---|
| N/C | = V/m (exactly) | Point charges, Coulomb's law |
| V/m | = N/C (exactly) | Capacitors, voltage/distance |
| kV/m | 1,000 V/m | High-voltage applications |
| MV/m | 10⁶ V/m | Breakdown fields, lightning |
| mV/m | 10⁻³ V/m | Weak fields, sensors |
| Constant | Value | Units |
|---|---|---|
| k = 1/(4πε₀) | 8.9876×10⁹ | N·m²/C² |
| ε₀ | 8.8542×10⁻¹² | C²/(N·m²) |
| Elementary charge e | 1.6022×10⁻¹⁹ | C |
| 1 μC | 10⁻⁶ C | Microcoulomb |
| 1 nC | 10⁻⁹ C | Nanocoulomb |
| 1 pC | 10⁻¹² C | Picocoulomb |
| Object | Charge |
|---|---|
| Electron | −1.6022×10⁻¹⁹ C |
| Proton | +1.6022×10⁻¹⁹ C |
| 1 Coulomb | 6.242×10¹⁸ elementary charges |
| Lightning bolt | ~1–5 C |
| Typical lab charge | 1–100 μC |
| Capacitor charge | nC to mC range |
Coulomb's Law — Electric Force Between Two Charges
This electric field calculator implements Coulomb's law (F=kq₁q₂/r²) for the electrostatic force between charges, the electric field formula E=kq/r² from a point charge, E=V/d for the uniform field between parallel plates, and extended distributions including infinite wire, plane, ring, and disk. Coulomb's law is the foundation of all electrostatics.
Coulomb's constant k = 8.988×10⁹ N·m²/C² = 1/(4πε₀). The electrostatic force obeys the inverse square law: doubling the distance between charges decreases the force by a factor of 4 (F ∝ 1/r²). The sign of the force: same-sign charges (++ or −−) → repulsive; opposite-sign charges (+−) → attractive. The formula F=kq₁q₂/r² gives only the magnitude; direction must be determined by the charge signs.
Why 1 Coulomb is enormous: two 1C charges placed 1 meter apart experience a force F = 8.988×10⁹ N — approximately 9 billion Newtons, roughly the gravitational weight of 900,000 metric tonnes. This is why real electrostatics problems use microcoulombs (μC = 10⁻⁶ C) and nanocoulombs (nC = 10⁻⁹ C). The electrostatic force calculator above handles all unit conversions automatically.
Coulombs law calculator shortcut: Convert all charges to Coulombs first (μC × 10⁻⁶, nC × 10⁻⁹), convert distance to meters, then apply F = 8.988×10⁹ × |q₁| × |q₂| / r². The most common error in Coulombs law calculations is forgetting to convert μC to C, giving an answer 10¹² times too large.
Electric Field Formula — E = kq/r²
The electric field E at a distance r from a point charge q is given by the electric field equation E=kq/r². The electric field E is a vector quantity — its magnitude is given by the electric field strength formula E=kq/r² and its direction is away from positive charges (outward) and toward negative charges (inward). The e field formula applies at every point in space around the charge.
The electric field E exists in space regardless of whether a test charge is present. To find the electric force on a test charge q₀ placed in the field: F = q₀E. The formula for electric field intensity E=kq/r² is the fundamental equation — the same point charge equation governs the electric field of a proton (q=1.6×10⁻¹⁹ C) and a charged capacitor plate.
Example 1 — Electric field from q=+5μC at r=0.3m
- E = kq/r² = 8.988×10⁹ × 5×10⁻⁶ / (0.3)²
- = 44,940 / 0.09 = 499,333 N/C
- Direction: away from the positive charge (outward)
Example 2 — Electric field from q=−2μC at r=0.5m
- E = 8.988×10⁹ × 2×10⁻⁶ / (0.5)² = 17,976 / 0.25 = 71,904 N/C
- Direction: toward the negative charge (inward)
Example 3 — Force on test charge q₀=+1μC in field E=499,333 N/C
- F = q₀ × E = 1×10⁻⁶ × 499,333 = 0.499 N
- Direction: same as E field (away from +5μC source)
Electric Field Units — N/C and V/m Are Identical
The electric field units N/C (Newtons per Coulomb) and V/m (Volts per meter) are exactly the same unit. This is not an approximation — it is a mathematical identity. The e field units equivalence follows directly from the definitions:
- From E=F/q: electric field has units N/C
- From E=V/d: electric field has units V/m
- Since 1 V = 1 J/C = 1 N·m/C → 1 V/m = 1 N·m/(C·m) = 1 N/C ✓
- Therefore 1 N/C = 1 V/m exactly
Use N/C when computing electric force on a charge (F=qE). Use V/m when working with voltage and distance (E=V/d). Both are equally correct — your textbook may prefer one over the other. Typical electric field values: inside atoms ~10¹¹ N/C, between capacitor plates 10⁴–10⁶ V/m, lightning discharge ~10⁶ V/m, Earth's surface ~100 N/C downward.
Electric Field from Voltage — E = V/d
For a uniform electric field between parallel plates, the electric field strength equation E=V/d relates the field to the voltage difference V and plate separation d. This is exact for a uniform field. The derivation: work done moving charge q across distance d in field E equals W = qEd. But W = qV (work = charge × voltage). Therefore qEd = qV → E = V/d.
Example 1 — Capacitor: V=1000V, d=1cm
- d = 1 cm = 0.01 m
- E = V/d = 1000/0.01 = 100,000 V/m = 100 kV/m
- Force on +1μC: F = qE = 10⁻⁶ × 100,000 = 0.1 N
Example 2 — Find voltage for E=10⁶ V/m, d=1cm
- V = E × d = 10⁶ × 0.01 = 10,000 V = 10 kV
Example 3 — Car battery: V=12V, d=1cm
- E = 12/0.01 = 1,200 V/m
What Is a Coulomb? — The Unit of Electric Charge
The Coulomb (C) is the SI unit of electric charge. 1 Coulomb = charge of 6.242×10¹⁸ protons, or equivalently, the charge transported by a current of 1 Ampere flowing for 1 second (Q = I×t → 1C = 1A × 1s). The elementary charge of an electron is −1.602×10⁻¹⁹ C, and a proton carries +1.602×10⁻¹⁹ C (exact by definition since 2019 SI revision).
Why 1 Coulomb is enormous: the force between two 1C charges 1m apart is F = kq₁q₂/r² = 8.988×10⁹ N — roughly equal to the gravitational pull on 900,000 metric tonnes. Real electrostatic charges are in μC (10⁻⁶ C) to nC (10⁻⁹ C) range. A 1 Ampere wire carries 6.242×10¹⁸ electrons per second past any cross-section.
Electric Field Lines — Direction and Density
Electric field lines are the visual language of electric fields. They always start on positive charges and end on negative charges — they never cross. The density of field lines represents field strength: closely packed lines indicate a stronger electric field. For a point charge: field lines are radially symmetric, spreading uniformly in all directions.
- Single positive charge: Field lines point radially outward in all directions — the electric field always points away from positive charges
- Single negative charge: Field lines point radially inward — the electric field always points toward negative charges
- Electric dipole (+q and −q): Field lines arch from the positive to the negative charge — strong field between charges, weaker field outside
- Two like charges (+q and +q): Field lines repel — no field lines cross, and there is a zero-field point between the charges
- Parallel plates: Uniform, parallel electric field lines between the plates (except near edges) — this is the geometry where E=V/d applies exactly
Electric Field of Extended Distributions — Wire, Plane, Ring, Disk
Beyond the point charge, four standard extended distributions appear in physics courses. Each has a specific geometry that determines how the electric field falls off with distance.
Infinite wire (E=2kλ/r): Electric field of a wire falls as 1/r — not 1/r² as for a point charge. The linear charge density λ is in C/m. For λ=10μC/m at r=0.2m: E = 2×8.988×10⁹×10⁻⁵/0.2 = 899,755 N/C.
Infinite plane (E=σ/2ε₀): The electric field of an infinite plane is remarkable — it is completely uniform, independent of distance. The surface charge density σ is in C/m². Both sides of the plane have equal field strength directed away (positive σ) or toward (negative σ).
Charged ring on axis (E=kQx/(x²+R²)^(3/2)): The on-axis electric field of a ring of charge Q with radius R. The field is zero at the center (x=0) and at infinity, with a maximum at x = R/√2. At x=0 the ring symmetry cancels all field components.
Disk on axis (E=(σ/2ε₀)[1−x/√(x²+R²)]): The electric field of a disk of charge transitions from the point-charge formula (when R is small) to the infinite-plane result (as R→∞). The electric field of a disk is always less than σ/(2ε₀).
Common Mistakes in Electric Field Calculations
Mistake 1 — Confusing E (field) and F (force)
- ❌ Wrong: "The electric field on q₀ is F=kq₁q₂/r²"
- ✅ Correct: E=kq/r² is the field (force per unit charge). Force on q₀: F=q₀×E. The electric field exists in space without a test charge; the force requires a charge in the field.
Mistake 2 — Not converting μC to C before using Coulomb's law
- ❌ Wrong: F = 8.988×10⁹ × 2 × 3 / (0.05)² = 2.157×10¹⁰ N (using μC as C)
- ✅ Correct: Convert first: 2μC = 2×10⁻⁶ C, 3μC = 3×10⁻⁶ C → F = 8.988×10⁹×2×10⁻⁶×3×10⁻⁶/0.0025 = 21.57 N
Mistake 3 — Wrong inverse power (1/r instead of 1/r²)
- ❌ Wrong: E = kq/r (linear, not inverse square)
- ✅ Correct: E = kq/r² for a point charge. Note: E∝1/r for an infinite wire, E∝1/r² for a point charge — know which geometry you're using.
Mistake 4 — Wrong direction for electric field
- ❌ Wrong: "Electric field points toward positive charge"
- ✅ Correct: Electric field points AWAY from positive charges and TOWARD negative charges. The electric field is defined as the force on a positive test charge — a positive test charge near a positive source is repelled (pushed away), so E points away.
Mistake 5 — Adding force magnitudes instead of vectors
- ❌ Wrong: F_net = F₁ + F₂ (scalar addition)
- ✅ Correct: Resolve each force into x and y components, sum components separately: Fx_net = F₁x + F₂x, Fy_net = F₁y + F₂y, then |F_net| = √(Fx²+Fy²). Forces are vectors — only collinear forces can be added as scalars.
Worked Examples — 8 Complete Problems
1. Coulomb's law: q₁=+1μC, q₂=+1μC, r=0.3m
- F = k|q₁||q₂|/r² = 8.988×10⁹ × 10⁻⁶ × 10⁻⁶ / (0.3)²
- = 8.988×10⁻³ / 0.09 = 0.0999 N ≈ 0.100 N
- Same sign (++): REPULSIVE
2. Coulomb's law: q₁=+2μC, q₂=−3μC, r=5cm
- r = 5 cm = 0.05 m, r² = 0.0025 m²
- F = 8.988×10⁹ × 2×10⁻⁶ × 3×10⁻⁶ / 0.0025 = 8.988×10⁹ × 6×10⁻¹² / 0.0025
- = 0.053928 / 0.0025 = 21.57 N — ATTRACTIVE
3. Electric field: q=+5μC, r=0.3m
- E = kq/r² = 8.988×10⁹ × 5×10⁻⁶ / (0.3)² = 44,940 / 0.09 = 499,333 N/C outward
4. E from voltage: V=500V, d=2mm
- d = 2 mm = 0.002 m
- E = V/d = 500/0.002 = 250,000 V/m = 250 kV/m
5. Net force on q₀=+1μC at origin, q₁=+2μC at (0.3,0), q₂=−2μC at (0,0.4)
- F₁ = 8.988×10⁹×10⁻⁶×2×10⁻⁶/0.09 = 0.200 N, direction: q₁ repels q₀ → −x direction, so F₁x=−0.200N
- F₂ = 8.988×10⁹×10⁻⁶×2×10⁻⁶/0.16 = 0.112 N, direction: q₂ attracts q₀ → +y, so F₂y=+0.112N
- |F_net| = √(0.04+0.01254) = √0.05254 = 0.229 N at 150.7°
6. Infinite wire: λ=10μC/m, r=0.2m
- E = 2kλ/r = 2×8.988×10⁹×10⁻⁵/0.2 = 179,751/0.2 = 898,755 N/C
7. Ring: Q=1μC, R=5cm, x=12cm
- E = kQx/(x²+R²)^(3/2) = 8.988×10⁹×10⁻⁶×0.12/(0.0144+0.0025)^(3/2)
- = 1078.56/(0.0169)^(3/2) = 1078.56/0.002196 = 491,152 N/C
- Max E at x = R/√2 = 5/√2 ≈ 3.54 cm
8. Parallel plates: find V for E=10⁶ V/m, d=1cm
- V = E × d = 10⁶ × 0.01 = 10,000 V = 10 kV