Free Pearson Edexcel International A Level Physics lesson
Electric fields and potential
Connect force maps to energy per charge without confusing vectors and scalars.
What you’ll learn
- Use Coulomb’s law and field strength.
- Relate uniform and radial fields to potential.
- Interpret field lines and equipotentials.
Force per charge describes a field
An electric field is a region in which a charged particle experiences an electric force. Its strength E = F/q is defined using a positive test charge small enough not to disturb the source arrangement. The field direction is the force direction on a positive charge; a negative charge feels a force opposite to E.
For point charges in vacuum, Coulomb’s law gives force magnitude k|Qq|/r², with k = 1/(4πε₀). Like signs repel and unlike signs attract. Dividing the force by the test charge gives a radial source field magnitude k|Q|/r². Distance is measured from the source charge, and the inverse-square relation is not a formula for every extended charge geometry.
Fields from several sources superpose as vectors. Draw directions before adding components, particularly between opposite signs where both fields can point the same way. A graph of E against 1/r² should be linear for an isolated point-source model. Nearby conductors, finite-size sources or charge leakage can spoil a classroom inverse-square comparison.
E = F/q; F = k|Qq|/r²; |E| = k|Q|/r²; k = 8.99 × 10⁹ N m² C⁻²
Symbols and units
- E is electric field magnitude in N C⁻¹
- F is force magnitude in N
- Q and q are charges in C
- r is separation in m
- k is 8.99 × 10⁹ N m² C⁻².
How the equation explains the idea
- For a point source charge, Coulomb’s force is k|Qq|/r². Divide by positive test-charge magnitude to obtain |E| = k|Q|/r².
- The field points away from positive Q and toward negative Q. Force on a negative test charge opposes the field. Doubling separation quarters the magnitude in this point-charge model.
Conditions. Isolated positive point charge Q = 1.0 nC in vacuum, k = 8.99 × 10⁹ N m² C⁻². Field magnitude is plotted from 1 m to 4 m; field direction is radially outward.
Distance from charge / m; Field magnitude / N C⁻¹.
At 1 m the field is 8.99 N C⁻¹; at 2 m it is 2.2475 N C⁻¹, one quarter. The graph omits the singular point r = 0 and does not imply a uniform field.
Open the interactive lesson to view the plotted curves and their coordinate tables.
Worked example
A small positive source
- Q = +2.0 nC and r = 0.30 m.
- E = 8.99 × 10⁹ × 2.0 × 10⁻⁹/0.30² = 200 N C⁻¹ approximately.
- An electron experiences a force opposite the outward field.
The field is radially outward; the electron force is radially inward.
Keep this idea. Use signs to determine direction and inverse-square magnitudes only for the stated source model.
Potential is energy per charge, not force
Electric potential gives electric potential energy per unit charge relative to a reference. A potential difference therefore links work and charge: ΔU = qΔV. A positive charge moving freely along the field loses potential energy; a negative charge can move the opposite way and still lose potential energy. Keep the sign of q when reasoning about energy changes.
With zero potential at infinity, a point source has V = kQ/r. Potential is a scalar and retains the sign of Q. The field is related to the spatial slope of potential: in one dimension Eₓ = −ΔV/Δx in a small interval. A steep potential slope means a strong field; a large potential value alone does not determine the local force.
For several sources, add potentials algebraically rather than as arrows. At a location with zero net field, potential can remain nonzero. In a nonuniform radial field, use potential differences to calculate energy changes instead of multiplying a single field value by an arbitrary distance. If the gain in kinetic energy approaches the rest-energy scale, a classical speed calculation ceases to be valid.
V = kQ/r; ΔU = qΔV; Eₓ = −dV/dx; electric work = −ΔU
Symbols and units
- V is electric potential in V = J C⁻¹
- U is potential energy in J
- Q and q are signed charges in C
- r is distance in m
- Eₓ is a field component in V m⁻¹.
How the equation explains the idea
- With zero potential at infinity for a point source, V = kQ/r keeps the sign of Q. Moving a charge gives ΔU = q(Vfinal − Vinitial).
- Electric work is −ΔU, so a positive charge moving to lower potential gains kinetic energy if no other work occurs. The field points down the potential gradient: Eₓ = −dV/dx.
Conditions. Isolated positive point charge Q = 1.0 nC, k = 8.99 × 10⁹ N m² C⁻², zero potential at infinity. The plotted potential is a signed scalar; field is not plotted on this voltage axis.
Distance from charge / m; Electric potential / V.
At 1 m V = 8.99 V and at 2 m V = 4.495 V, one half. The outward field magnitude comes from the negative radial slope, kQ/r², not from the potential height itself.
Open the interactive lesson to view the plotted curves and their coordinate tables.
Worked example
Two radial positions
- For Q = +4.0 nC, V at 0.20 m is 179.8 V and at 0.40 m is 89.9 V.
- For q = +2.0 nC moving outward, ΔU = 2.0 × 10⁻⁹(89.9 − 179.8).
- ΔU = −1.798 × 10⁻⁷ J; the electric force can supply that magnitude of kinetic energy.
Moving a positive charge outward from a positive source lowers its electric potential energy.
Keep this idea. Add potentials as signed scalars and use qΔV for energy changes.
Read uniform fields and equipotentials
Between broad parallel plates, sufficiently far from the edges, field lines are approximately straight, parallel and equally spaced. They point from the higher potential positive plate towards the lower potential negative plate. The potential falls linearly with distance along the field, giving field magnitude E = |ΔV|/d for separation d.
Equipotentials join positions with equal potential and cross field lines at right angles. Moving a charge along an equipotential changes no electric potential energy. For a radial point field, the equipotentials are concentric spheres, shown as circles in a plane diagram. Equal potential steps become farther apart at greater radius because the field weakens.
A particle entering a uniform electric field with a velocity perpendicular to the field has constant acceleration qE/m in the field direction and unchanged velocity component perpendicular to it, if other forces are negligible. This links field motion to projectile analysis. Near plate edges the uniform approximation breaks down; field-line diagrams should show fringing rather than claim E = V/d everywhere.
|E| = |ΔV|/d in a uniform plate field; a = qE/m; ΔU along an equipotential = 0
Symbols and units
- E is electric field in V m⁻¹ = N C⁻¹
- ΔV is potential difference in V
- d is plate separation in m
- q is signed charge in C
- m is mass in kg.
How the equation explains the idea
- In an approximately uniform plate field away from edges, the potential changes linearly, so field magnitude is |ΔV|/d.
- Use F = qE and a = F/m to obtain signed acceleration a = qE/m. Motion along an equipotential has ΔV = 0 and hence ΔU = 0, even when the field is nonzero.
Worked example
Deflection plates
- Two plates differ by 600 V and are separated by 0.020 m.
- E = 600/0.020 = 3.0 × 10⁴ V m⁻¹.
- A +2.0 μC test charge would experience 0.060 N along the field.
V m⁻¹ and N C⁻¹ are equivalent field units.
Keep this idea. Field arrows point down the potential slope, perpendicular to equipotentials.
Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.
Browse Physics Unit 4 lessons and resources