Formula & Calculator
Electric Field of a Point Charge
Calculates the electric field strength at a distance from a point charge using Coulomb's constant.
Interpretation
Electric field of a point charge: E = k·q/r², where k is Coulomb's constant, q is charge, r is distance. It describes the field strength at a point. Example: q=1e-6 C, r=0.1 m → E = 8.99e9×1e-6/0.01 = 8.99e5 N/C.
Variables
| Symbol | Quantity | Unit |
|---|---|---|
| E | Electric field strength | N/C |
| k | Coulomb's constant | 8.99e9 N.m2/C2 |
| q | Source point charge | C |
| r | Distance from the charge | m |
What it means
The electric field E at a point due to a point charge q is the force per unit positive charge placed at that point. The magnitude is given by Coulomb’s law: E = k|q|/r², direction radially outward for positive charges and inward for negative. This field is the basis for electrostatics, used to compute forces on other charges. The concept is extended to continuous charge distributions by integration. Electric fields are essential in understanding capacitance, electric potential, and in designing electronic components. In practice, fields are used in accelerators, particle detectors, and in everyday devices like capacitors and transmission lines. Understanding electric fields is fundamental for electromagnetism.
Worked example
Electric Field of a Point Charge – Two Examples
Real‑World| Parameter | Value |
|---|---|
| q | 1×10⁻⁶ C |
| r | 1 m |
| Parameter | Value |
|---|---|
| q | 5×10⁻⁶ C |
| r | 0.5 m |
Common mistakes
- Coulomb constant k: k = 8.988×10⁹ N·m²/C² – use the correct value.
- Charge q: In coulombs (C) – not microcoulombs without conversion.
- Distance r: From the charge to the point – in metres.
- Field direction: For positive charge, field points away; for negative, toward the charge.
- Superposition: For multiple charges, the total field is the vector sum.
Applications
The electric field of a point charge, E = k·q/r², describes the force per unit charge at a distance r from the charge. This law is the basis for electrostatics and is used to design capacitors, sensors, and particle accelerators. Engineers apply it to determine the forces on charged particles in semiconductor devices, to design electrostatic precipitators for pollution control, and to model lightning and atmospheric electricity. In materials science, it helps understand dielectric breakdown. By calculating electric fields, professionals can ensure insulation adequacy, design efficient energy‑storage devices, and predict the behaviour of charged particles in various applications.
- Design of capacitors and energy storage devices
- Electrostatic precipitators and air cleaners
- Particle accelerator and beamline design
- Semiconductor device modelling
- Lightning protection and electromagnetic compatibility
Frequently Asked Questions
The electric field (E) at a distance r from a point charge q is E = k·q / r², where k is Coulomb's constant (8.99×10⁹ N·m²/C²). The field is radial: directed away from a positive charge and toward a negative charge.
In SI, the unit is N/C (newton per coulomb) or V/m (volt per metre), which are equivalent.
Treating the electric field as a scalar and ignoring its direction. The field is a vector; for multiple charges, you must vectorially add the individual fields.
The field strength follows the inverse‑square law: E ∝ 1/r². Doubling the distance reduces the field to one‑quarter.
The total electric field at a point is the vector sum of the fields produced by each individual charge. This allows calculation of fields from any charge distribution.
For a dipole, the field along the axis is E = (1/4πε₀)·2p/r³, where p is the dipole moment. This falls off as 1/r³, faster than a point charge.
The electric field is the negative gradient of the electric potential: E = –∇V. For a point charge, V = kq/r, and E = –dV/dr = kq/r².
- Design of capacitors.
- Particle accelerator beam optics.
- Electrostatic precipitators.
- Atomic and molecular structure calculations.