Electric field calculator

What the electric field is

The electric field at a point is the force a charge would feel there, per unit of charge: E = F ÷ Q, measured in newtons per coulomb, which is the same as volts per meter (OpenStax University Physics, Volume 2, section 5.4). A point charge q produces the field E = k·q ÷ r², with k = 8.988 × 10⁹ N·m²/C² and ε₀ = 8.854 × 10⁻¹² C²/(N·m²) in the form k = 1/(4πε₀); it points away from a positive charge and toward a negative one. When several charges are present each makes its own field, and the net field is their vector sum. The electric potential V = k·q ÷ r is a scalar companion that measures the energy per unit charge, and potentials add as ordinary numbers.

The formulas

For each charge i at distance rᵢ from the field point, Eᵢ = k·qᵢ ÷ rᵢ², directed along the line from the charge to the point (away from it when qᵢ is positive). The calculator turns each into x and y components, adds them, and reports E = √(Eₓ² + E_y²) with the angle measured from the +x axis. The potential is V = Σ k·qᵢ ÷ rᵢ, and a test charge Q feels F = Q·E. Between two large parallel plates carrying equal and opposite charge densities the field is uniform, E = σ ÷ ε₀ = V ÷ d, where σ is the charge per unit area on the inner surface, V the potential difference and d the gap (sections 5.5 and 7.2).

Worked example: two equal charges and a point above them

This is problem 61 of OpenStax chapter 5, whose answer key gives 5.12 N; the small difference comes from rounding k to 8.99 × 10⁹. Put these two charges and the field point into the calculator (they are the default values) and the table shows each charge’s part.

Common mistakes

The most frequent one is adding field magnitudes as if they were numbers. Fields are vectors, and two equal fields at 90° give 1.41 times one of them, not twice. OpenStax advises working with the absolute values of the charges to get each magnitude and deciding the direction from the signs afterwards; this calculator does the signs for you, so enter a negative charge with a minus sign. The second mistake is mixing units: the formula wants meters and coulombs, but charges are usually given in µC or nC and distances in cm, so check the unit menus. Third, the distance is from each charge to the field point, not between the charges. Finally, σ ÷ ε₀ is the field between two plates, while one isolated sheet gives σ ÷ 2ε₀.

Limits of this method

The tools here are for point charges at rest. Real charged objects are only point-like if they are small compared with the distances involved: a charged sphere acts like a point charge outside of itself, but a rod, a ring or a flat plate needs an integral over its charge (section 5.5). The charges lie in one plane, so three-dimensional arrangements need a z component. The field between plates is uniform only well inside the gap, where the plates are large compared with the separation; near the edges it bends. Finally, dry air holds only about 3.0 × 10⁶ V/m before it sparks (OpenStax Example 7.7), so a large field strength computed for plates in air is a number that cannot be reached in practice.

What is the formula for the electric field of a point charge?

E = k·|q| ÷ r², where k = 1/(4πε₀) = 8.988 × 10⁹ N·m²/C² and ε₀ = 8.854 × 10⁻¹² C²/(N·m²) is the permittivity of vacuum (OpenStax University Physics, Volume 2, sections 5.3 and 5.4; the book rounds k to 8.99 × 10⁹). The field points away from a positive charge and toward a negative one. In OpenStax Example 5.3 the nucleus of an ionized helium atom (charge 2e) at 26.5 pm gives 4.1 × 10¹² N/C at the electron. A test charge Q in a field E feels the force F = Q·E.

How do I add the fields of several charges?

Add them as vectors: find each charge’s field at the point, split it into x and y components, sum the components, and then take the magnitude and angle of the sum (the superposition principle, section 5.4). Magnitudes alone cannot be added unless the fields are parallel. In OpenStax problem 61 two charges of 5.0 µC sit 24 cm apart and a third point lies 13 cm from both. The x components cancel, the y components add, and the net field is 2.05 × 10⁶ N/C along +y. On the −2.5 µC charge that gives 5.11 N, which the answer key rounds to 5.12 N because it uses k = 8.99 × 10⁹.

What if the charges have opposite signs?

Use the signed charges: a negative charge simply gives a field that points toward it. OpenStax problem 77 puts +50 µC and −25 µC 1.0 m apart and asks for the field halfway between: both fields point toward the negative charge, so they add, E = 2.7 × 10⁶ N/C (the key gives 2.70 × 10⁶ N/C). A 20 µC charge there feels 53.9 N. Between two charges of the same sign the fields point in opposite directions and partly cancel, and on the line outside them they add.

How do I find the electric field between two parallel plates?

For two large plates with equal and opposite charge the field between them is uniform: E = V ÷ d from the voltage and the gap, or E = σ ÷ ε₀ from the charge per unit area (OpenStax Examples 5.9 and 7.8). A 25.0 kV potential difference across 4.00 cm gives 625,000 V/m, and a 0.500 µC charge between the plates feels 0.313 N. If instead 8.0 µC sits on the inner face of a plate 2.0 cm on a side, σ = Q ÷ A = 0.02 C/m² and E = σ ÷ ε₀ = 2.26 × 10⁹ N/C. A single infinite sheet gives half of that, σ ÷ 2ε₀.

How do I find the electric potential of a point charge?

V = k·q ÷ r, measured from zero at infinity, with the sign of the charge; for several charges the potentials just add as numbers, with no vectors (section 7.3). OpenStax Example 7.10: a metal sphere carrying −3.00 nC, observed 5.00 cm from its center, is at −539 V (the book gets −539 V with k = 8.99 × 10⁹). The page shows the potential V of the whole group at your field point as well.

How do I get the charge or the distance from a measured field?

Rearrange E = k·|q| ÷ r²: |q| = E·r² ÷ k, or r = √(k·|q| ÷ E). OpenStax problem 73 says the field is 100 N/C at 50 cm from a point charge: q = 100 × 0.50² ÷ k = 2.78 nC. The "One charge" tab solves for any one of the three from the other two. The sign of the charge cannot come from the size of the field, only from its direction.

Sources: OpenStax University Physics, Volume 2, sections 5.3 “Coulomb’s Law”, 5.4 “Electric Field”, 5.5 “Calculating Electric Fields of Charge Distributions” (Example 5.9), 7.2 “Electric Potential and Potential Difference” (Examples 7.7 and 7.8) and 7.3 “Calculations of Electric Potential”, with the chapter 5 problems and answer key; ε₀ = 8.8541878188 × 10⁻¹² C²/(N·m²) (CODATA 2022) and k = 1/(4πε₀). Related: Coulomb’s law calculator, capacitor calculator, magnetic field calculator and Ohm’s law calculator.