Electric Field of a Point Charge Calculator

Electric Field from a Point Charge

Point Charge Electric Field Formula

Formula

\[ E = \frac{k \, |Q|}{\varepsilon_r \, r^2} \quad \text{where} \quad k = \frac{1}{4\pi\varepsilon_0} \approx 8.988 \times 10^9 \ \tfrac{\text{N}\cdot\text{m}^2}{\text{C}^2} \\ \varepsilon_0 \approx 8.854 \times 10^{-12} \ \tfrac{\text{F}}{\text{m}} \]

Where:

\(E\) — electric field magnitude (N/C, or equivalently V/m)
\(k\) — Coulomb's constant
\(|Q|\) — magnitude of the source charge (C)
\(\varepsilon_r\) — relative permittivity of the medium (1 for vacuum, ≈1.0006 for air)
\(r\) — distance from the charge to the field point (m)

This formula, derived from Coulomb's law, gives the strength of the electric field that a point charge creates in the surrounding space. The field points away from a positive charge and toward a negative charge; the formula returns only the magnitude.

Point Charge Electric Field – Calculation Example

Example: a charge of 1 µC sits in air. What is the electric field 10 cm away from it?

\[ E = \frac{k\,|Q|}{\varepsilon_r\, r^2} = \frac{8.988\times 10^9 \times 1\times 10^{-6}}{1 \times (0.10)^2} \]

\[ E = \frac{8.988\times 10^{3}}{0.01} \approx 8.988\times 10^7 \ \text{N/C} \]

Result: the electric field magnitude is approximately \(8.99 \times 10^7\) N/C, or \(8.99 \times 10^7\) V/m — the two units are identical. Because the charge is positive, the field points radially away from it.

How to Use the Calculator

Enter the Source Charge (Q) — the magnitude of the charge creating the field. Use only the absolute value; the sign affects the field's direction, not its strength.
Choose the charge unit from the dropdown: nC, µC, mC, or C.
Enter the Distance (r) from the charge to the point where you want to know the field, and pick the unit: mm, cm, m, in, or ft.
Optionally set the Relative Permittivity (εr) of the medium. Leave it at 1 for vacuum or air; use ≈2.2 for oil, ≈4.7 for glass, or ≈80 for water.
Read the Electric Field Magnitude in N/C and the equivalent value in V/m, along with a step-by-step breakdown in the Calculation Details block.

What Is the Electric Field of a Point Charge?

An electric field is the region of space around a charged object where another charge would feel a force. You don't need a second charge to define the field — it exists whether or not anything is there to detect it. For a point charge, the field spreads out uniformly in all directions, so its strength depends only on how much charge is present and how far away you measure it.

The formula \( E = k|Q|/(\varepsilon_r r^2) \) shows two key features. First, the field grows in direct proportion to the charge: double the charge and the field doubles. Second, it obeys an inverse-square law: double the distance and the field falls to a quarter of its value. This is the same mathematical behavior as gravity — you can explore that analogy with our Gravitational Force Calculator, which uses an analogous law with mass instead of charge.

The medium matters too. Inside a dielectric such as water or glass, the charges in the material partially cancel the field of the source charge, weakening it by a factor of \(\varepsilon_r\). That is why the same charge produces a field about 80 times weaker in water than in air. If you want the full force between two charges rather than the field of one, see the Coulomb's Law Calculator — the field is simply that force divided by the test charge.

The results are given in both N/C and V/m. These are the same unit written two ways: 1 N/C = 1 V/m, since a volt is a joule per coulomb and a newton is a joule per meter. For reference, the Earth's fair-weather field at the surface is roughly 100 V/m, so the example above — nearly \(10^8\) V/m — is an enormous field, comparable to what causes air itself to break down and spark.

Assumptions and Limitations

Point charge approximation: the source charge is treated as concentrated at a single point. A uniformly charged sphere behaves exactly like a point charge at its center, but only at distances outside the sphere. Inside a charged sphere, this formula does not apply.
Uniform, linear medium: the relative permittivity must be the same everywhere between the charge and the field point. The calculator cannot handle boundaries between different materials (for example, half air, half glass).
No other charges: the result is the field of the source charge alone. With multiple charges present, calculate each field separately and add them as vectors.
Distance must be positive: the field is undefined at the exact location of the charge, where the formula predicts an infinite value.
Static charges: the formula assumes the charge is not moving. Rapidly changing charges radiate electromagnetic waves and require a more complete treatment.

When interpreting results, keep everyday magnitudes in mind: fields below about \(10^6\) V/m are common in electronics and lab equipment, while air breaks down at roughly \(3\times 10^6\) V/m. If your result approaches that range, expect sparking, corona discharge, or other nonlinear effects that this simple formula does not capture.

Frequently Asked Questions

How do I calculate the electric field of a point charge by hand?

First convert the charge to coulombs and the distance to meters. Then multiply Coulomb's constant \(k \approx 8.988 \times 10^9\) by the charge magnitude, and divide by the square of the distance (and by \(\varepsilon_r\) if you're not in vacuum). For example, with \(Q = 2\,\mu\text{C}\) at \(r = 0.5\) m in air: \(E = 8.988\times10^9 \times 2\times10^{-6} / 0.25 \approx 7.2\times10^4\) N/C. Watch the unit prefixes — a µC is \(10^{-6}\) C, not \(10^{-3}\) C.

What is a typical or strong electric field value?

For perspective: household wiring and electronic circuits involve fields of roughly 1–1000 V/m, the Earth's fair-weather field is about 100 V/m, medical imaging and lab equipment reach \(10^3\)–\(10^6\) V/m, and dry air breaks down and sparks near \(3\times10^6\) V/m. A tiny 1 µC charge just 1 cm away already produces about \(9\times10^7\) N/C — strong enough to exceed air's breakdown threshold, which is why static discharges are so common in dry conditions.

How do I convert between N/C, V/m, kV/m, and other electric field units?

N/C and V/m are identical, so no conversion is needed. For larger scales, divide by 1000 to get kV/m and by \(10^6\) for MV/m: \(5\times10^6\) V/m = 5000 kV/m = 5 MV/m. In Gaussian units the field is measured in statV/cm, where 1 statV/cm ≈ \(2.998\times10^4\) V/m; this mostly appears in older physics textbooks, so most everyday work stays in N/C or V/m.

What happens to the field if I double the distance or the charge?

The field obeys an inverse-square law in distance but scales linearly with charge. Doubling the distance quarters the field, because \(r\) appears squared in the denominator: at 2r the field is \(E/4\). Tripling the distance cuts it to \(E/9\). Conversely, doubling the charge doubles the field exactly. This asymmetry is a common source of errors when estimating fields at different distances by hand.

Does the sign of the charge affect the calculated field strength?

No — the sign only sets the direction. A \(-5\) µC charge and a \(+5\) µC charge produce fields of identical magnitude at the same distance; the field of the positive charge points radially outward while that of the negative charge points radially inward. That's why the calculator uses the absolute value \(|Q|\). If you need direction, apply it after computing the magnitude: away from positive charges, toward negative ones.

How do I find the field from several charges at once?

Calculate each charge's field separately with this formula, then add them as vectors, not plain numbers. For charges along the same line, fields pointing the same way add while opposite fields subtract; for charges in 2D or 3D, combine components: \(E_x = \sum E_i\cos\theta_i\), \(E_y = \sum E_i\sin\theta_i\), and the magnitude is \(\sqrt{E_x^2+E_y^2}\). This principle of superposition works because the field equation is linear in charge.

Can I use this formula for a charged sphere, plate, or real object?

A uniformly charged sphere acts exactly like a point charge at its center — but only for points outside the sphere; inside, the field decreases linearly toward the center and this formula fails. Flat plates and long wires follow different laws (constant field and \(1/r\) respectively), so don't apply the point-charge formula to them. Real irregular objects can be approximated as point charges only when the observation distance is much larger than the object's size.