Electric Field Calculator
Calculate electric field strength from a point charge with E = kQ/r2, or the uniform field between parallel plates with E = V/d.
Electric Field Calculator
Calculator
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Formula: E = kQ/r² (point charge) | E = V/d (parallel plates)
Worked example — 899,000 N/C
Formula
E = kQ/r² (point charge) | E = V/d (parallel plates)
Electric field strength at a point in space due to a charge or between plates.
Worked Examples
Example 1: Point charge
Problem:Q=1μC at r=0.1m
Solution:E=8.99e9×1e-6/0.01=8.99×10⁵ N/C
Result:899,000 N/C
Frequently Asked Questions
What is an electric field?
An electric field is a region in space where a charged particle experiences a force. For a point charge Q, the field at distance r is E = kQ/r², where k = 8.9875×10⁹ N·m²/C². Between parallel plates with voltage V and separation d, E = V/d. The SI unit of electric field is N/C or equivalently V/m.
How do you calculate electric field strength?
For a point charge: E = kQ/r². For Q = 1 μC at r = 0.1 m: E = 8.9875×10⁹ × 10⁻⁶ / 0.01 = 8.9875×10⁵ N/C. For parallel plates with V = 100 V and d = 0.01 m: E = 100 / 0.01 = 10,000 V/m. The force on a test charge q is F = qE.
What units does the electric field calculator use?
Charge Q is in coulombs (C); use 1e-6 for microcoulombs. Distance r is in meters (m). For the parallel-plate mode, voltage V is in volts (V) and plate separation d in meters (m). Output electric field E is in N/C (= V/m), and electric potential is shown in volts (V).
What is a real-world example of electric field?
The Earth's atmosphere has a fair-weather electric field of about 100 V/m pointing downward. Near a thundercloud, this can exceed 10,000 V/m, strong enough to ionize air and trigger lightning. In electronic capacitors, fields of 10⁶–10⁷ V/m are common within the dielectric layer.
What is the difference between electric field and electric potential?
Electric field E is force per unit charge, a vector measured in N/C (equivalently V/m); potential V is potential energy per unit charge, a scalar measured in volts. For a point charge they fall off at different rates: E = kQ/r-squared drops as 1/r-squared while V = kQ/r drops as 1/r. In a uniform field the two are linked by E = V/d, which is why volts per metre is a valid unit for field strength.
How do you find the force on a charge placed in the field?
Use F = qE. If the field is 8.99 x 10^5 N/C and you place a 2 microcoulomb charge there, F = 8.99 x 10^5 x 2 x 10^-6 = 1.80 N. A positive charge is pushed along the field direction and a negative charge against it, which is why field lines are drawn pointing away from positive source charges. The field itself is defined as the force a one-coulomb test charge would feel.
Why is the field between parallel plates uniform when a point charge field is not?
A point charge sends field lines radially outward, so their density - and therefore E - falls as 1/r-squared. On charged parallel plates the contributions from the whole sheet of charge superpose into evenly spaced, parallel lines, so E = V/d has the same value everywhere between them. The uniformity fails near the edges, where fringing curves the lines outward, so E = V/d assumes plate width is large compared with separation d.
What is the electric field inside a conductor in electrostatic equilibrium?
It is zero. Free electrons rearrange until their own field exactly cancels any applied field inside the metal, which takes only nanoseconds. Excess charge therefore sits entirely on the outer surface, and just outside that surface the field is perpendicular to it with magnitude sigma/epsilon-zero, where sigma is the surface charge density. This is the principle behind a Faraday cage shielding its contents from external fields.
When does the point-charge formula E = kQ/r-squared stop being valid?
It holds for a genuine point charge, and by the shell theorem for any spherically symmetric distribution when measured outside it. Other geometries follow different laws: the field near a long charged wire falls as 1/r, and near a large charged plane it is sigma/(2 x epsilon-zero), essentially constant with distance. Inside a charge distribution you must apply Gauss's law to the enclosed charge rather than the total.
How is Coulomb's constant related to the permittivity of free space?
Coulomb's constant is k = 1/(4 x pi x epsilon-zero), where epsilon-zero = 8.854 x 10^-12 F/m, giving k = 8.9875 x 10^9 N m-squared/C-squared - the value Electric Field Calculator uses. Inside a dielectric material the permittivity becomes epsilon-r x epsilon-zero, so the field from a given charge is reduced by the relative permittivity epsilon-r. That reduction is exactly why dielectrics increase a capacitor's capacitance.
Background & Theory
Electric Field Strength in Context
Electric field (E) describes the force per unit charge at a point in space, and it's what determines how a charged particle placed there would accelerate. The two formulas used by this calculator cover the two situations that come up constantly in intro electromagnetism: a single point charge radiating a field outward (E = kQ/r², following an inverse-square law like gravity), and a parallel-plate capacitor where the field is uniform between the plates (E = V/d).
| Environment | Typical field strength |
|---|---|
| Earth's fair-weather atmospheric field | ~100-150 V/m, pointing downward |
| Beneath a thundercloud before a lightning strike | 10,000+ V/m |
| Dielectric breakdown of dry air | ~3×10⁶ V/m (3 MV/m) — beyond this, air ionizes and sparks |
| Inside a ceramic capacitor's dielectric layer | 10⁶-10⁷ V/m, sustained safely because the dielectric's breakdown strength is much higher than air's |
| Field binding an electron to a hydrogen nucleus (Bohr radius) | ~5×10¹¹ V/m |
Because field strength from a point charge falls off as 1/r², moving twice as far away reduces the field (and the force on any test charge) to one quarter — the same inverse-square relationship that governs gravity and light intensity.
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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