The key idea
Gravitational fields with the masses swapped for charges — same inverse-square shape, same potential logic, two new twists: charge comes in two signs, and uniform fields between plates behave completely differently from radial ones.
Understand it · 01
Define direction using a positive test charge
An electric field is a region in which a charge experiences an electric force. Field strength is force per unit positive charge, E = F/q, with direction defined by the force on a small positive test charge. A negative charge experiences force opposite to this field direction. Field lines emerge from positive charges and enter negative charges. Closer field-line spacing represents greater field strength; the lines are a model rather than physical tracks that all charged particles follow.
For an isolated point charge, field magnitude is k|Q|/r². Superposition applies to fields from several charges: calculate each field vector and add its components. Equal distances do not guarantee zero resultant because the individual field directions matter. At the midpoint between equal opposite charges the fields point the same way and reinforce; at the midpoint between equal like charges they oppose and cancel.
Understand it · 02
Keep the scalar potential separate from the vector field
Electric potential is potential energy per unit positive charge relative to a chosen zero. For a point charge with zero at infinity, V = kQ/r; the sign follows the source charge. The change in potential energy of a charge q is qΔV. A positive charge naturally accelerated by the field moves towards lower potential, whereas a negative charge accelerates towards higher potential while its potential energy also decreases.
Between large parallel plates well away from the edges, the field is approximately uniform and its magnitude is the potential difference divided by plate separation. Equipotential surfaces are perpendicular to the field, and movement along one requires no work against the electrostatic force. Near plate edges the field bends and is no longer uniform, so the simple E = V/d model is an approximation for the central region rather than a complete description everywhere.
Understand it · 03
Connect electric force to motion and energy
A charged particle entering a uniform field has approximately constant acceleration qE/m if the electric force is the only significant force and the non-relativistic model is valid. If its initial velocity is perpendicular to the field, split the motion into a constant-velocity component and an accelerated component, just as for a projectile. The analogy concerns the mathematics; the direction of electric acceleration can reverse when the charge changes sign.
For a particle accelerated from rest through a potential difference, the gain in kinetic energy is the magnitude of qΔV when the field does positive work. This energy method often avoids calculating transit time or plate spacing. An electronvolt is an energy unit equal to the energy gained by one elementary charge through one volt. Convert electronvolts to joules before combining them with SI mass in ½mv², and state when speeds are low enough for that classical expression.
Key definitions
- Electric field strength
- The force per unit positive charge at a point: E = F ÷ Q. Units: N/C or V/m.
- Electric potential
- The work done per unit positive charge in bringing a small test charge from infinity to the point: V = Q ÷ 4πε₀r.
- Coulomb's law
- The force between two point charges is proportional to the product of the charges and inversely proportional to the square of their separation.
Equations, units & conditions
Coulomb's law
F = Qq ÷ 4πε₀r²
Units: N
Field of a point charge
E = Q ÷ 4πε₀r²
Units: V/m
Uniform field
E = V ÷ d
Units: V/m
Potential of a point charge
V = Q ÷ 4πε₀r
Units: V
Worked examples, step by step
Example 1
An electron enters the field between parallel plates 2.0 cm apart with a p.d. of 400 V. Find the force on the electron. (e = 1.6 × 10⁻¹⁹ C)
Follow the method
- E = V/d = 400 ÷ 0.020 = 2.0 × 10⁴ V/m.
- F = EQ = 2.0 × 10⁴ × 1.6 × 10⁻¹⁹.
F = 3.2 × 10⁻¹⁵ N, towards the positive plate
Practice questions with worked solutions
Try each question on paper before opening the solution. These are original academy questions; the suggested marks are a self-checking guide, not an official exam-board mark scheme. Difficulty labels describe this practice set, not predicted grades.
Build · 3 suggested marks
1. Force between plates
Parallel plates are 0.020 m apart with 400 V between them. Find the field magnitude and the force magnitude on an electron, using e = 1.60 × 10⁻¹⁹ C.
Show solution and self-check: question 1
- E = V/d = 400/0.020 = 2.0 × 10⁴ V m⁻¹.
- F = eE = 1.60 × 10⁻¹⁹ × 2.0 × 10⁴.
2.0 × 10⁴ V m⁻¹; 3.2 × 10⁻¹⁵ N opposite the field.
Self-check · one mark per point
- Apply E = V/d.
- Calculate the electron force magnitude.
- State that a negative charge feels force opposite to the field.
Build · 3 suggested marks
2. Electron energy gain
An electron starts from rest and is accelerated through a potential difference of magnitude 250 V. Find its kinetic energy gain in eV and joules, using e = 1.60 × 10⁻¹⁹ C.
Show solution and self-check: question 2
- One elementary charge through 250 V gains 250 eV.
- Energy = eV = 1.60 × 10⁻¹⁹ × 250.
250 eV = 4.0 × 10⁻¹⁷ J.
Self-check · one mark per point
- Identify the energy gain in electronvolts.
- Convert using the elementary charge.
- Give 4.0 × 10⁻¹⁷ J.
Common mistakes & how to avoid them
- Using E = V/d for a point charge, or the inverse-square formula between plates. Radial fields: inverse-square. Parallel plates: uniform, E = V/d. Identify the geometry before choosing the equation.
- Sign carelessness with potential: unlike gravitational potential, electric potential can be positive (near positive charge) or negative (near negative charge). Keep the sign of Q in the formula.
- Field strength falls as 1/r², but potential falls as 1/r. Mixing the powers is the classic error on sketch-graph questions.
- A charged particle with an initial velocity component along the plates follows a parabola in the ideal uniform-field, non-relativistic model. If it starts from rest or moves only along the field, its path is straight instead.
Test yourself
10 questions · instant marking
Question 1 of 10
The electric field strength between parallel plates 5.0 cm apart with a p.d. of 200 V is:
Question 2 of 10
The field between parallel plates (away from the edges) is:
Question 3 of 10
Electric field strength of a point charge falls with distance as:
Question 4 of 10
Doubling the distance from a point charge changes the force on a test charge by a factor of:
Question 5 of 10
A charged particle moving parallel to the plates in a uniform field follows a path that is:
Question 6 of 10
Electric potential near an isolated negative charge is:
Question 7 of 10
The force on an electron (q = 1.6 × 10⁻¹⁹ C) in a field of 5.0 × 10⁴ V/m is:
Question 8 of 10
Unlike gravitational fields, electric fields can:
Question 9 of 10
The work done moving a charge q through a potential difference V is:
Question 10 of 10
Where the equipotential lines around a charge are closest together, the field is:
Understand the why
Questions, explained
Start with the short answer, then follow the reasoning. Each explanation has a permanent link you can share with a classmate or return to when revising.
How are electric and gravitational fields similar and different?
Both point-mass gravity and point-charge electricity have inverse-square field magnitudes and inverse-distance potentials. Gravity is attractive between ordinary positive masses, while electric forces can attract or repel because electric charge has two signs.
The mathematical analogy helps with graphs, superposition and energy changes, but the signs must be reconsidered rather than copied. Electric potential can be positive or negative relative to infinity depending on the source charge, whereas the potential of an isolated positive mass is negative. A negative test charge feels force opposite to electric field direction, while a positive test charge follows it. Field strengths are vectors that add by components; potentials are scalars that add algebraically. A point with zero field need not have zero potential in either type of combined-field problem.
When do you use E = V/d versus the inverse-square formula?
Use E = V/d for an approximately uniform field across a separation d, such as the central region between large parallel plates. Use E = k|Q|/r² for the magnitude of an isolated point charge’s radial field.
The voltage in the plate formula is a potential difference between two positions, whereas V = kQ/r gives potential at one point relative to infinity. Mixing them can accidentally produce the wrong distance dependence. Near plate edges, field lines bend and the field is no longer uniform, so the simple plate expression is an approximation. Outside a spherically symmetric charge distribution the point-charge result can be used with distance from its centre. Inside an extended distribution, the same total-charge formula is not automatically valid; the distribution and conductor conditions matter.
What path does a charged particle take in a uniform electric field?
A charged particle with a nonzero velocity component perpendicular to a uniform electric field follows a parabolic path in the non-relativistic, force-free-in-the-other-direction model. Starting from rest or moving purely along the field produces a straight path.
The field produces constant acceleration qE/m along its direction for a positive charge and opposite to it for a negative charge. A perpendicular velocity component stays constant if no other force acts. The shared time links these motions just as in projectile calculations, but the electric acceleration depends on the charge-to-mass ratio and can reverse with charge sign. After the particle leaves the field, the electric force from that ideal region is no longer applied. The subsequent path follows its exit velocity until another force acts; do not keep curving it indefinitely.
Can electric field be zero where electric potential is not zero?
A point can have zero electric field and nonzero electric potential because field contributions add as vectors while potential contributions add as scalars. Cancellation of force directions does not require cancellation of energy per charge.
At the midpoint between two equal positive charges, each field points away from its source. The two fields have equal magnitude and opposite directions, so their resultant is zero. Both potentials are positive, however, and add to a positive total. Between equal opposite charges, the midpoint gives the reverse contrast: potentials cancel, while fields point in the same direction and reinforce. The zero of potential is a chosen reference, while zero field has the physical meaning of no net electric force on a small test charge at that point.
Equal positive source charges
Two +2.0 nC point charges are each 0.10 m from their midpoint. Find the midpoint field and potential, using k = 9.0 × 10⁹ N m² C⁻².
- The equal opposite field vectors cancel.
- Potential from each charge = kQ/r = 9.0 × 10⁹ × 2.0 × 10⁻⁹/0.10 = 180 V.
- Add the two scalar potentials.
Resultant field 0 N C⁻¹; potential +360 V.
Why does an electron accelerate towards the positive plate or higher potential?
An electron accelerated by an electric field moves towards higher electric potential because its charge is negative. Its potential energy qV decreases as V increases, allowing kinetic energy to increase when other energy transfers are negligible.
Field direction is defined using a positive test charge, so the electron’s force is opposite to the field arrow. The energy equation keeps the signs consistent: ΔU = qΔV and the work done by the field is −ΔU. A positive ΔV combined with negative q gives a potential-energy decrease. Do not interpret the phrase higher voltage as universally meaning higher potential energy for every particle. If the electron starts with a velocity against its force, it may slow before reversing, so the direction of instantaneous motion need not immediately match the direction of acceleration.
Electron energy increase
An electron moves through a potential increase of 80 V under the electric force. Find its kinetic-energy gain, using e = 1.60 × 10⁻¹⁹ C.
- ΔU = qΔV = −1.60 × 10⁻¹⁹ × 80 = −1.28 × 10⁻¹⁷ J.
- The field does positive work equal to −ΔU.
Kinetic-energy gain 1.28 × 10⁻¹⁷ J, or 80 eV.
Why do field lines cross equipotential surfaces at right angles?
Electric field lines are perpendicular to equipotential surfaces because displacement along an equipotential gives no potential-energy change for a test charge. Any field component along that surface would imply nonzero work along it.
For a point charge, equipotential surfaces are concentric spheres and the field is radial. Between ideal parallel plates, equipotentials are parallel planes and the field crosses them normally. Closer spacing between equally stepped potential surfaces indicates a stronger field because the potential changes more rapidly with distance. A charged particle is not required to travel along a field line or remain on an equipotential; its trajectory also depends on its initial velocity. Use field arrows to determine acceleration, not as a drawing of every possible path through the region.
Your learning checklist
Use these goals to check your understanding. Requirements vary by specification, tier and exam year; the official references below define your full course.
- Use Coulomb's law: F = Qq ÷ 4πε₀r²
- Define electric field strength E = F/Q; use E = Q ÷ 4πε₀r² for point charges
- Use E = V/d for uniform fields between parallel plates
- Define electric potential; use V = Q ÷ 4πε₀r
- Analyse the motion of charged particles in uniform electric fields
Sources & how to cite these notes
Original explanations, illustrations and practice by Dr Desouky Physics Academy. Official specifications guide the course scope; questions and suggested marks on this page are our own revision material.
Reference this page
Dr Desouky Physics Academy. Edexcel IAL Physics: Electric fields. Updated 2026-09-25. https://www.drdesouky.com/revision/a-level-physics/electric-fields
Use the arrow beside an explanation heading to link straight to that section. Spotted a problem? Contact the academy with the topic and the step you want us to check.