Electric Field – Point Charge Calculator

Calculate the electric field strength and force at any distance from a point charge.

1 μC = 1×10⁻⁶ C
For force calculation
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What a Point Charge Electric Field Represents

An electric field tells how strongly a source charge would push or pull a positive test charge at a location. For an isolated point charge, spherical symmetry spreads the field over larger and larger spherical areas as distance increases. Because sphere area grows as r2, field magnitude falls as 1/r2.

The field is a property of the source charge and position; it does not depend on the value of a test charge placed there. Once a test charge q is introduced, the electric force is F = qE. The sign of q determines whether the force points with or opposite the field. A positive source charge produces an outward field, while a negative source charge produces an inward field.

E = k|Q|/r2     F = qE
SymbolMeaningWhy it appears / units
QSource chargeCoulombs; its magnitude sets field strength and its sign sets direction
rDistance from source chargeMeters, measured from the point charge; squared in the denominator
kCoulomb constant≈ 8.99 × 109 N·m2/C2 in vacuum
EElectric-field magnitudeN/C, equivalent to V/m

If distance doubles, field magnitude becomes one quarter; if distance triples, it becomes one ninth. Superposition is required when multiple source charges are present: calculate each field as a vector and add the vectors. A single point-charge expression cannot capture cancellations or directional combinations from several charges.

Worked Examples

Example 1: 1 μC charge, r=0.1 m
E = 8.99e9×1e-6/0.01
Result: 899,000 N/C
Strong field close to charge
Example 2: Proton at r=1 Å (Bohr radius)
Q=1.6e-19 C, r=5.29e-11 m
Result: 5.14×10¹¹ N/C
Field experienced by electron in hydrogen
Example 3: Nanocoulomb source charge
Q = +3.0 nC = 3.0 × 10−9 C, r = 0.20 m → E = kQ/r2
Result: E ≈ 674 N/C outward
The nanocoulomb-to-coulomb conversion is essential; missing 10−9 changes the answer by a billion.
Example 4: Force on a negative test charge
E ≈ 674 N/C, q = −2.0 μC → |F| = |q|E
Result: |F| ≈ 1.35 × 10−3 N
Because q is negative, the force direction is opposite the electric-field direction even though the magnitude uses |q|.

Common Mistakes

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Forgetting to square the distance

Point-charge field follows 1/r2, not 1/r. Doubling distance reduces magnitude by a factor of four.

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Mixing microcoulombs or nanocoulombs with coulombs

Convert prefixes first: 1 μC = 10−6 C and 1 nC = 10−9 C.

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Using the test charge inside the field formula

E is determined by the source charge Q. The test charge q is used afterward in F = qE.

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Ignoring direction

The scalar expression gives magnitude. State whether the field points away from a positive source or toward a negative source.

Frequently Asked Questions

What are units of electric field?
N/C (Newtons per Coulomb) = V/m (Volts per meter). Both are equivalent. V/m is more common in practical electrical engineering.
How does distance affect field strength?
Electric field follows an inverse square law (E ∝ 1/r²). Doubling the distance reduces the field to 1/4 its original value.
Does changing the test charge change the electric field?
No, in the ideal test-charge concept. The field at a point is determined by the source charges and geometry. Changing q changes the force F = qE on the test charge, while E itself remains the same provided the test charge does not disturb the source configuration.
Why is electric field proportional to 1/r2?
For an isolated point charge, the field has spherical symmetry. The area of a sphere grows as 4πr2, so the same electric flux is distributed over an area proportional to r2. That geometric spreading produces the inverse-square dependence.
How do I determine electric-field direction?
Imagine placing a small positive test charge at the point. The field points in the direction that positive charge would accelerate. Therefore field lines point away from positive source charges and toward negative source charges.
Can I use the point-charge formula for a charged sphere?
Outside a spherically symmetric charge distribution, the field is the same as if the total charge were concentrated at the center. Inside the sphere, the result depends on how charge is distributed, so E = kQ/r2 using the total charge is generally not valid there.

Formula Explorer connections

Interpretation: This formula links charge, voltage, current, resistance, capacitance, power or circuit time response. Assumption: Confirm DC versus AC conditions, RMS versus peak values, component topology and steady-state versus transient behavior. Ideal components may be assumed.

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