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An electric field is a region of space in which an electric charge experiences a force. Electric fields are created by charged objects and are central to AQA specification 3.7.3. The mathematical treatment of electric fields closely parallels that of gravitational fields, but with crucial differences.
Spec mapping (AQA 7408 §3.7.3 — Electric fields): This lesson covers the definition of electric field strength as force per unit positive charge, Coulomb's law for two point charges, the radial field E = Q/(4πε₀r²), uniform fields between parallel plates E = V/d, the parabolic motion of charged particles in uniform fields (analogous to projectile motion), and the direct comparison with gravitational fields. Refer to the official AQA 7408 specification document for the precise statement of each item and any limitations on the candidate's responsibility (e.g. shielding, dielectric effects are picked up later in §3.7.4).
Synoptic links — where this resurfaces in 7408:
- §3.7.2 Gravitational fields (Lessons 0–1 here) — identical inverse-square mathematics with the key sign and shielding differences; the comparison table is a frequent Paper 2 prompt.
- §3.7.4 Capacitance (Lessons 4–6 here) — uniform parallel-plate fields underpin the derivation of C = ε₀A/d.
- §3.7.5 Magnetic fields — combined E and B fields define velocity selectors (Wien filters); the Lorentz force F = Q(E + v×B) is the foundational synoptic equation.
- §3.2 Quantum physics — Millikan's oil drop experiment (covered in Lesson 3) was the first measurement of e and underpins the quantisation of charge — examined synoptically with the photoelectric effect and atomic physics.
- §3.4.1.4 Newton's laws / §3.4.1.6 Projectile motion — perpendicular-entry deflection of an electron between plates uses the same independent-axis decomposition as projectile motion in gravity.
- §3.2.1.5 Particle accelerators — uniform fields between plates accelerate charged particles in linear accelerators; the relationship between potential difference, charge, and kinetic energy gain is examined in the particles unit.
Electric field lines represent the direction a positive test charge would move if placed in the field (by convention).
Described diagram — Field lines around a positive point charge: Lines radiate outward from the charge in all directions, with arrows pointing away from the charge. The lines are closer together near the charge (stronger field) and further apart at greater distances (weaker field).
Described diagram — Field lines around a negative point charge: Lines point radially inward towards the charge, converging at the centre. A positive test charge would be attracted towards the negative charge.
Described diagram — Field between two parallel plates: Uniform, equally spaced, parallel lines pointing from the positive plate to the negative plate. At the edges, the lines curve outward slightly (edge effects), but in the central region the field is uniform.
Key rules for electric field lines:
Key Definition: The electric field strength (E) at a point is the force per unit positive charge acting on a small positive test charge placed at that point.
E = F/Q
where E is the electric field strength (N C⁻¹ or equivalently V m⁻¹), F is the force (N), and Q is the charge of the test charge (C).
Electric field strength is a vector quantity. Its direction is the direction of the force on a positive charge.
Between two parallel conducting plates separated by a distance d with a potential difference V across them, the electric field is uniform:
E = V/d
where E is the field strength (V m⁻¹), V is the potential difference (V), and d is the plate separation (m).
This is a crucial result: the field between parallel plates depends only on the voltage and separation, not on the plate area or the charge on the plates (as long as edge effects are neglected).
Question: Two parallel plates are separated by 2.0 cm and have a potential difference of 500 V between them. Calculate: (a) the electric field strength between the plates; (b) the force on an electron placed between the plates.
Solution:
(a) E = V/d = 500 / (2.0 × 10⁻²) = 2.5 × 10⁴ V m⁻¹
(b) F = EQ = 2.5 × 10⁴ × 1.60 × 10⁻¹⁹ = 4.0 × 10⁻¹⁵ N
The force is directed from the positive plate towards the negative plate (opposite to the electron's motion, since the electron is negatively charged — it accelerates towards the positive plate).
Key Definition: Coulomb's law states that the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
F = kQ₁Q₂/r² = Q₁Q₂/(4πε₀r²)
where:
Key features:
For a point charge Q, combining E = F/Q_test with Coulomb's law:
E = F/Q_test = (kQQ_test/r²)/Q_test = kQ/r²
E = kQ/r² = Q/(4πε₀r²)
Question: Two point charges of +3.0 μC and −5.0 μC are placed 12 cm apart. Calculate the force between them.
Solution:
F = kQ₁Q₂/r²
F = (8.99 × 10⁹ × 3.0 × 10⁻⁶ × 5.0 × 10⁻⁶) / (0.12)²
F = (8.99 × 10⁹ × 1.5 × 10⁻¹¹) / 0.0144
F = 0.1349 / 0.0144
F = 9.37 N (attractive, since the charges are opposite)
Question: Calculate the electric field strength at a distance of 0.30 m from a charge of +2.0 μC.
Solution:
E = kQ/r² = (8.99 × 10⁹ × 2.0 × 10⁻⁶) / (0.30)²
E = 1.798 × 10⁴ / 0.09
E = 2.0 × 10⁵ N C⁻¹ (or V m⁻¹)
The field points radially outward from the positive charge.
A charged particle in a uniform electric field experiences a constant force, producing constant acceleration. This is analogous to projectile motion in a uniform gravitational field.
Parallel motion: If a charged particle enters a uniform field parallel to the field lines, it accelerates uniformly in a straight line (like a ball thrown vertically in a gravitational field).
Perpendicular motion: If a charged particle enters a uniform field perpendicular to the field lines, it follows a parabolic path (like a ball thrown horizontally in a gravitational field). The component of velocity parallel to the plates remains constant, while the component perpendicular to the plates increases uniformly.
Question: An electron enters a uniform electric field of strength 4.0 × 10³ V m⁻¹ at right angles to the field with a horizontal speed of 2.0 × 10⁷ m s⁻¹. The plates are 5.0 cm long. Calculate the vertical deflection of the electron as it passes through the plates. (Electron mass = 9.11 × 10⁻³¹ kg, electron charge = 1.60 × 10⁻¹⁹ C.)
Solution:
Step 1: Calculate the force on the electron. F = EQ = 4.0 × 10³ × 1.60 × 10⁻¹⁹ = 6.40 × 10⁻¹⁶ N
Step 2: Calculate the vertical acceleration. a = F/m = 6.40 × 10⁻¹⁶ / 9.11 × 10⁻³¹ = 7.03 × 10¹⁴ m s⁻²
Step 3: Calculate the time spent between the plates. t = L/v_horizontal = 0.050 / 2.0 × 10⁷ = 2.5 × 10⁻⁹ s
Step 4: Calculate the vertical deflection. y = ½at² = ½ × 7.03 × 10¹⁴ × (2.5 × 10⁻⁹)² = ½ × 7.03 × 10¹⁴ × 6.25 × 10⁻¹⁸ y = ½ × 4.39 × 10⁻³ = 2.2 × 10⁻³ m ≈ 2.2 mm
| Property | Gravitational Field | Electric Field |
|---|---|---|
| Source | Mass | Charge |
| Force law | F = GMm/r² | F = kQ₁Q₂/r² |
| Field strength definition | g = F/m (N kg⁻¹) | E = F/Q (N C⁻¹ or V m⁻¹) |
| Radial field strength | g = GM/r² | E = kQ/r² |
| Nature of force | Always attractive | Attractive or repulsive |
| Relative strength | Very weak (G ≈ 6.67 × 10⁻¹¹) | Very strong (k ≈ 8.99 × 10⁹) |
| Shielding | Cannot be shielded | Can be shielded (Faraday cage) |
| Uniform field | g near Earth's surface | E = V/d between parallel plates |
Exam Tip: The comparison table above is a favourite exam question. Make sure you can reproduce it from memory. The mathematical structures are almost identical — the key physics difference is that gravitational forces are always attractive, while electric forces can be attractive or repulsive.
Question: Two point charges, Q₁ = +5.0 × 10⁻⁹ C and Q₂ = −5.0 × 10⁻⁹ C, are placed 0.20 m apart in air. Calculate the magnitude of the electric field at a point P that lies 0.10 m directly above the midpoint of the line joining them.
Solution:
Distance from each charge to P: r = √(0.10² + 0.10²) = √0.02 = 0.1414 m.
Magnitude of field due to each charge: E_each = kQ/r² = (8.99 × 10⁹ × 5.0 × 10⁻⁹) / (0.1414)² = 44.95 / 0.02 = 2.25 × 10³ N C⁻¹.
Each field is directed along the line from the source charge to P (away from Q₁, towards Q₂). At point P, the vertical components of the two fields cancel (one up, one down — actually no, both fields lie in the plane through P and the two charges; let me re-orient).
Resolving the two fields: each makes an angle θ with the horizontal where cos θ = (0.10)/0.1414 = 0.707 (i.e. θ = 45°).
The horizontal components of the two fields (both directed from + towards −, i.e. along the −x direction) add: E_net,horizontal = 2 × 2.25 × 10³ × cos 45° = 2 × 2.25 × 10³ × 0.707 = 3.18 × 10³ N C⁻¹.
The vertical components (one upward, one downward) cancel by symmetry.
So |E_net at P| = 3.18 × 10³ N C⁻¹, directed parallel to the line from +Q to −Q.
This is the field of an electric dipole observed on its perpendicular bisector — the field falls off as 1/r³ at large distances (faster than the 1/r² of a single charge) and points anti-parallel to the dipole moment on the bisector. The 45° geometry here is illustrative; for a dipole with charge separation d observed at distance r ≫ d on the perpendicular bisector, the field magnitude is approximately E ≈ kp/r³ where p = Qd is the dipole moment, a vector pointing from the negative to the positive charge. This 1/r³ behaviour explains why polar molecules (water, HCl) exert weaker long-range electrostatic influence on each other than free ions do — though over short distances dipole-dipole interactions are the dominant force in liquids and biological systems.
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