Electric Field Calculator

Definition

E is force per unit positive test charge:

In SI units the field has units of N/C, equivalent to V/m (volts per metre) — both are used. The two are dimensionally identical (1\ \textN/\textC = 1\ \textV/\textm).

\vecE(\vecr) = \frac\vecFq_0 = (k q)/(r^2) \hatr

Coulomb’s constant k

k = 1 / (4π ε₀) ≈ 8.9875 × 10⁹ N·m²/C². The vacuum permittivity ε₀ ≈ 8.854 × 10⁻¹² C²/(N·m²) is one of nature’s defining constants and gives rise to k.

k = (1)/(4π \varepsilon_0)

Potential and Work

Because E drops as 1/r², the potential scales as 1/r: V(r) = k q / r. You can view V(r) and E as paired, with E = −dV/dr. The relationship carries into electronics, where voltage and field strength are two views of the same physics.

V(r) = (k q)/(r), \quad E(r) = -(dV)/(dr)

Superposition

The field of multiple point charges is the vector sum of the individual fields at the point of interest. This composes the entire medium of multipole expansions (dipole, quadrupole, …) — useful for modelling antennas, molecules, and far-field radiation patterns.

\vecE_\texttotal = Σ_i (k q_i)/(r_i^2) \hatr_i

Inside Dielectrics

Inside a uniform linear dielectric, replace k with k/ε_r (the relative permittivity of the material). Vacuum and air are essentially identical (ε_r ≈ 1.0006); water has ε_r ≈ 80 at low frequency, glass ~5–10, ceramics ~10–10⁴. The field inside the dielectric is therefore weaker than in vacuum.

k_\textmedium = (k)/(\varepsilon_r)

Reality Check — Air Breakdown

Atmospheric air breaks down (a spark forms) at roughly E ≈ 3 × 10⁶ V/m. Lightning fields near the ground approach this value before the discharge starts. The calculator’s output should almost never be in that regime — a 1 µC charge at 1 m produces only 9000 V/m, well below the breakdown limit.

E_\textbreakdown, air ≈ 3× 10^6\ \textV/m

Superposition and Field Direction

The field from several charges is the vector sum of the individual fields — each source contributes independently, with no interference between them. This principle is what makes otherwise intractable arrangements solvable.

The direction convention matters: the field points away from positive charge and towards negative charge, defined by the force a positive test charge would feel. When summing, resolve each contribution into components before adding; adding magnitudes is valid only when the fields are collinear and pointing the same way. For two equal and opposite charges, the fields reinforce along the axis between them and partially cancel outside — the dipole pattern that describes everything from water molecules to antenna radiation.

\vecE_\texttotal = Σ_i \vecE_i = (1)/(4π\varepsilon_0)Σ_i (q_i)/(r_i^2)\hatr_i

Field, Potential and Uniform Fields

Electric field and electric potential describe the same physics in different language: the field is the negative gradient of the potential. For the common case of a uniform field between parallel plates, this reduces to E = V/d — which is why field strength is quoted in volts per metre.

That relationship is the everyday working form. A 9 V battery across a 1 mm gap produces 9000 V/m. A 400 kV transmission line at 10 m ground clearance gives a field of tens of kV/m near the conductor. And a cell membrane, with about 70 mV across roughly 7 nm, sustains an internal field of around 10⁷ V/m — stronger than the breakdown field of air, and a reminder that field strength depends as much on the smallness of the distance as on the size of the voltage.

\vecE = -\nabla V, \qquad E = (V)/(d) \ \text(uniform field)

Conductors, Shielding and Practical Notes

Inside a conductor in electrostatic equilibrium the field is zero: mobile charges rearrange themselves until they cancel any interior field. All excess charge sits on the surface, and the external field meets that surface at right angles.

This gives the Faraday cage — a conducting enclosure shields its interior from external fields, which is why a car is a reasonably safe place in a lightning storm and why sensitive instruments live in metal boxes. It also concentrates the field at sharp points, where surface charge density is highest, which is the operating principle of the lightning rod. Two habits keep calculations honest: use metres for distance (a centimetre slip changes the answer by a factor of 10,000 through the inverse square), and remember that k = 1/4π\varepsilon_0 ≈ 8.99×10^9\ \mathrmN\,m^2/C^2 applies in vacuum — inside a dielectric, divide by the relative permittivity.

E_\textinside conductor = 0, \qquad k = (1)/(4π\varepsilon_0) ≈ 8.99×10^9\ \mathrmN\,m^2/C^2

Worked Example

Inputs

• charge: 1e-9
• distance: 0.1

Result: |E| = 899 N/C = 899 V/m (potential V = 89.9 V at that distance)

Explanation

A single elementary charge (~1.6 × 10⁻¹⁹ C) at 1 cm gives an absurdly small field; a 1 nC charge is much friendlier to think about. With |q| = 10⁻⁹ C and r = 0.10 m, |E| = (8.99 × 10⁹)(10⁻⁹)/(10⁻²) ≈ 899 V/m. Doubling the distance to 20 cm drops the field to 225 V/m (a factor of 4). Doubling the charge doubles the field at every distance — useful for designing ion sources and ion thrusters.

Second Scenario

Inputs

• charge: 1e-9
• distance: 0.075

Result: |E| = 899 N/C = 899 V/m (potential V = 89.9 V at that distance)

Explanation

This scenario uses different inputs (charge = 1e-9, distance = 0.075) to show how changing one variable affects the electric field result. Run the calculator above with these values to get the exact updated output with step-by-step work.