Electric Field & Potential: Formulas & Worked Examples

The electric field formula is E = F/q — electric field strength equals force divided by charge. It tells you the force a +1 coulomb test charge would experience at that point in space. Electric field is measured in volts per metre (V/m) or equivalently newtons per coulomb (N/C).

The companion formula is V = kQ/r — electric potential equals Coulomb's constant times charge divided by distance. Where E gives you force per unit charge, V gives you potential energy per unit charge. These two quantities are linked: the electric field is the negative gradient of the potential, which is why field lines always point from high to low potential.

What Is Electric Field Strength? E = F/q

The electric field strength E at a point in space is defined as the force per unit positive charge placed at that point:

where F is the force in newtons on a test charge q in coulombs. The unit of E is N/C (newtons per coulomb), which is equivalent to V/m (volts per metre). Electric field is a vector — it has both magnitude and direction. By convention, field lines point in the direction a positive test charge would feel force: away from positive charges, toward negative charges.

Rearranging gives the force on any charge placed in a known field:

A positive charge experiences force in the direction of E; a negative charge experiences force opposite to E.

Electric Field from a Point Charge: E = kQ/r²

A point charge Q creates an electric field that falls off with the square of distance r:

where k = 8.99 × 10⁹ N·m²·C⁻² is Coulomb's constant and ε₀ = 8.854 × 10⁻¹² F/m is the permittivity of free space. This is an inverse-square law — double the distance and the field drops to one-quarter.

For positive Q: field points outward from Q (away from the source charge).
For negative Q: field points inward toward Q (toward the source charge).

The force between two charges q₁ and q₂ separated by r follows from E = kQ/r²: the field of q₁ at the location of q₂ is E = kq₁/r², so the force on q₂ is F = q₂E = kq₁q₂/r² — Coulomb's Law.

Electric Potential: V = kQ/r

The electric potential V at a point is the electric potential energy per unit charge at that point. For a point charge Q at distance r:

Units: volts (V) = joules per coulomb (J/C). Note the key differences from the field equation:

Because V is a scalar, finding the total potential from multiple charges is easier than finding the total field — just add the individual potentials algebraically (with signs for positive/negative charges).

V = kQ/r — The Electric Potential from a Point Charge

This equation is one of the most important in electrostatics. At distance r from a point charge Q:

This is derived by integrating the work done against the electric field to bring a unit positive charge from infinity to distance r:

Potential is defined as zero at infinity. For a positive charge Q: V is positive everywhere and increases as r decreases (you must do work to bring a positive test charge closer). For a negative charge Q: V is negative everywhere (a positive test charge is attracted and gains energy approaching).

Relationship Between E and V

Electric field and potential are related — E is the negative gradient of V:

For the uniform field between parallel plates separated by distance d with potential difference V:

This is why V/m and N/C are equivalent units for electric field. If you know how potential varies with position, you can find the field by differentiation; if you know the field, integrate to find the potential.

Worked Example 1: Electric Field from E = F/q

A charge of 2.0 × 10⁻⁶ C placed at a point in an electric field experiences a force of 0.10 N. Find the electric field strength.

Worked Example 2: Field and Potential from a Point Charge

A point charge Q = +3.0 × 10⁻⁹ C (3 nanocoulombs). Find E and V at r = 0.20 m from the charge.

Note: E falls as 1/r², so at r = 0.40 m: E = 674/4 = 168.5 N/C. V falls as 1/r: V = 134.9/2 = 67.4 V.

Worked Example 3: Work Done Moving a Charge

Find the work done moving a charge q = +2.0 × 10⁻⁶ C from point A (V_A = 200 V) to point B (V_B = 50 V).

Positive work done by the field (charge moves from high to low potential, like rolling downhill). The charge gains kinetic energy of 3.0 × 10⁻⁴ J — confirmed by the work-energy theorem.

Worked Example 4: V = kQ/r — Two-Charge System

Two charges: Q₁ = +5.0 nC at origin, Q₂ = −3.0 nC at x = 0.30 m. Find the potential at x = 0.10 m (between them).

Distance from Q₁: r₁ = 0.10 m. Distance from Q₂: r₂ = 0.20 m.

Because potential is scalar, we simply add the contributions with their signs — no vector components needed. This is a major practical advantage of using V rather than E for multi-charge problems.

Equipotential Surfaces

An equipotential surface is a surface on which V is constant everywhere. No work is done moving a charge along an equipotential (W = qΔV = 0). Electric field lines are always perpendicular to equipotential surfaces — if they weren't, there would be a component of E along the surface, doing work to move a charge along it, which would contradict V being constant.

For an isolated point charge: equipotentials are concentric spheres. For a uniform field between parallel plates: equipotentials are flat surfaces parallel to the plates. In conductors at electrostatic equilibrium: the entire conductor surface is an equipotential (E = 0 inside).

Real-World Applications

Capacitors: store energy in an electric field between two parallel plates. E = V/d tells you the field from the plate separation and voltage. Energy stored: U = ½CV² = ½QV = Q²/(2C). Capacitors in phones, CPUs, and power supplies all rely on E = V/d.

Lightning: a thundercloud builds up charge, creating an electric field between cloud and ground. When E exceeds the dielectric breakdown strength of air (~3 × 10⁶ V/m), the air ionises and a lightning bolt carries charge to ground. Tall buildings and trees lower the distance d in E = V/d, increasing the local field and making them more likely strike points.

Particle accelerators: charged particles are accelerated through potential differences. A proton accelerated through V = 1,000 V gains kinetic energy KE = qV = 1.6 × 10⁻¹⁹ × 1000 = 1.6 × 10⁻¹⁶ J = 1 keV. The LHC accelerates protons to 6.5 TeV — equivalent to 6.5 × 10¹² eV, achieved by repeated passage through accelerating fields.

Biology — nerve impulses: neurons maintain a potential difference of about −70 mV across their membrane at rest (inside negative relative to outside). An action potential involves the membrane potential briefly swinging to +40 mV — a 110 mV change driven by ion channels. The electric field across the 7 nm membrane: E = V/d = 0.11 / (7 × 10⁻⁹) = 1.57 × 10⁷ V/m. That's five times the breakdown field of air — one of the strongest sustained electric fields in nature.

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