Electric Field Calculator

Solve E = k·Q/r², V = k·Q/r, F = qE and Gauss's Law with full working shown.

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The electric field is one of the four fundamental quantities in classical electromagnetism. Wherever a charged object sits, it creates a field in the surrounding space — and any other charge placed in that field experiences a force. This calculator covers every standard scenario you meet from A-level to university physics: point charges, spherical charge distributions via Gauss’s Law, uniform fields between parallel plates, and infinite plane sheets.

Six modes are available, and every mode supports solve-for-variable: you can find any unknown from the remaining knowns, not just compute the “standard” direction of the formula.

How it works

Point-charge electric field

For an isolated point charge Q in vacuum, the field at distance r is given by:

E = k · |Q| / r²

where k = 8.9876×10⁹ N·m²/C² is the Coulomb constant, numerically equal to 1/(4πε₀). The field magnitude depends only on the absolute charge and the distance; the direction is radially outward (positive Q) or inward (negative Q). The calculator solves for E, Q, or r as required.

Electric potential from a point charge

Electric potential V is the work done per unit positive charge in bringing a test charge from infinity to the point of interest:

V = k · Q / r

Unlike the field, V is a scalar — it can be positive or negative depending on the sign of Q. The relationship between them is E = −dV/dr. The calculator solves for V, Q, or r.

Gauss’s Law — spherical symmetry

By Gauss’s Law the flux through any closed surface equals the enclosed charge divided by ε₀. For a spherically symmetric charge distribution (e.g. a uniformly charged sphere or a point charge) the external field is:

E = Q / (4πε₀ r²)

This is mathematically identical to the Coulomb point-charge formula. The Gauss mode frames it differently — you supply the total enclosed charge and the radius of your imaginary Gaussian surface — to make the connection explicit for students working through shell-theorem problems.

Uniform field between parallel plates

For two large conducting plates separated by distance d with potential difference V across them, the field between the plates is uniform:

E = V / d

This is the geometry inside a capacitor, a cathode-ray tube deflector, or a Millikan oil-drop apparatus. The mode solves for E, V, or d.

Infinite plane sheet

An infinite (or large) sheet of surface charge density σ produces a uniform field on each side, independent of distance:

E = σ / (2ε₀)

Two opposing sheets in a capacitor add to give E = σ/ε₀ between the plates and cancel outside. The mode solves for E or σ.

Force on a test charge

Given a known field, the electrostatic force on a charge q is:

F = q · E

Positive q → force in the field direction; negative q → force opposite to field. The mode solves for F, E, or q.

Worked example

A proton (charge +e = 1.602×10⁻¹⁹ C) is located 0.53 Å (the Bohr radius, 5.3×10⁻¹¹ m) from an electron in a hydrogen-like model.

Step 1 — field at the electron’s orbit:

E = k · e / r² = 8.9876×10⁹ × 1.602×10⁻¹⁹ / (5.3×10⁻¹¹)²

E = (1.440×10⁻⁹) / (2.809×10⁻²¹) ≈ 5.13×10¹¹ V/m

Step 2 — force on the electron:

F = q · E = 1.602×10⁻¹⁹ × 5.13×10¹¹ ≈ 8.22×10⁻⁸ N

This matches the Coulomb force calculated directly, confirming the field approach. Enter Q = 1.602e-19 C (or select the “e” unit), r = 5.3×10⁻¹¹ m (53 pm), and the calculator returns the same field.

Formula reference

ScenarioFormulaKey constant
Point charge fieldE = k · |Q| / r²k = 8.9876×10⁹ N·m²/C²
Point charge potentialV = k · Q / rk = 8.9876×10⁹ N·m²/C²
Gauss sphereE = Q / (4πε₀r²)ε₀ = 8.8542×10⁻¹² C²/(N·m²)
Parallel platesE = V / d
Infinite sheet (one side)E = σ / (2ε₀)ε₀ = 8.8542×10⁻¹² C²/(N·m²)
Force on chargeF = q · E

All constants are CODATA 2018 values. Calculations run entirely in your browser — no values are sent to a server.

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