Ch 1 · 18 Electric Fields · A Level Physics 9702
Topic 1 of 2 in Electric Fields and Capacitance — A Level Physics 9702 — 50 lessons.
Introduction
Electric fields are built exactly like gravitational fields, with one big difference: charges can attract or repel. The same pattern of field strength, inverse-square laws and potential appears again, and it explains everything from lightning to the deflection of electron beams. (Syllabus topic 18: sections 18.1–18.5.)
Key ideas — 18.1 Electric fields and field lines
An electric field is a field of force: a region in which a charge experiences a force.
🧑🏫 Teacher's explanation
Imagine you are standing in a room filled with invisible fans blowing air. If you release a tiny, lightweight ping-pong ball, its movement depends on where you release it and how strong the wind is at that exact spot. An electric field is just like that 'wind' of force. Instead of air molecules, it is a region of space where a charge feels a push or a pull.
📖 Definitions
🔍 Worked example — Easy
A charge of +4.0 × 10^-6 C experiences a force of 0.80 N when placed in an electric field. Calculate the electric field strength E.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — 18.2 Uniform fields
Between two parallel plates with p.d. Δ V and separation Δ d, the field is uniform (parallel, equally spaced lines from + to −, apart from the edges):
🧑🏫 Teacher's explanation
Imagine a uniform electric field as a steady, invisible wind blowing through a room. If you throw a tiny balloon (a charge) into this wind, the force it feels doesn't change whether the balloon is near the left wall or the right wall; the 'wind' is just as strong everywhere. This is what we mean by a 'uniform' field.
📖 Definitions
🔍 Worked example — Easy
Two parallel metal plates are separated by a distance of 8.0 mm. A potential difference of 240 V is applied across them. Calculate the magnitude of the uniform electric field strength between the plates.
🔍 Worked example — Standard
An electron (charge e = 1.60 × 10^-19 C, mass m = 9.11 × 10^-31 kg) enters a uniform electric field of 5.00 × 10^3 V m^-1 perpendicular to the field lines. Calculate the acceleration of the electron.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — 18.3 Coulomb's law
For points outside a spherical conductor, the charge on the sphere can be treated as a point charge at its centre.
🧑🏫 Teacher's explanation
Imagine two magnets being pushed toward each other. You can feel the 'push' or 'pull' even before they touch. In electrostatics, charges do the exact same thing. They create an invisible 'influence' in the space around them. If you place another charge in that space, it feels a force.
📖 Definitions
🔍 Worked example — Exam-style
Two identical point charges are placed 0.10 m apart. The force between them is 0.90 N. Calculate the magnitude of the charge on each sphere.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
(a) State the type of force acting between them. (b) Write an expression for the magnitude of the force F in terms of Q, d, and varepsilon_0. [2 marks]
Key ideas — 18.4 Field of a point charge
E = Q/4πvarepsilon_0 r^2
🧑🏫 Teacher's explanation
Imagine you are standing in a room filled with a scent. The closer you are to the source of the scent, the stronger it smells. As you move away, the intensity drops off very quickly. An electric field works similarly: a charge creates an 'influence' in the space around it. If you place a tiny test charge there, it feels a force.
📖 Definitions
🔍 Worked example — Standard
A point charge Q produces an electric field strength of 2.0 × 10^3 N C^-1 at a distance of 0.20 m. Calculate the magnitude of the charge Q.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) Calculate the electric field strength at a distance of 0.20 m from the charge. [2] (b) Calculate the electric potential at this same point. [2] [4 marks]
Key ideas — 18.5 Electric potential
The electric potential V at a point is the work done per unit positive charge in bringing a small test charge from infinity to that point.
🧑🏫 Teacher's explanation
Imagine you are hiking up a hill. The height of the hill at any point is like the electric potential. If you are standing at the base (infinity) and want to reach a peak, you have to do work against gravity. The higher the peak, the more work you do per kilogram of your mass. In electricity, the 'height' is the electric potential V.
📖 Definitions
🔍 Worked example — Standard
An electric field is created between two parallel plates separated by 0.020 m. The potential difference between the plates is 120 V. Calculate the magnitude of the electric field strength E.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) Calculate the electric potential at a point 0.20 m from the charge. [2] (b) Calculate the electric field strength at the same point. [2] [4 marks]
Worked example
Example 1: parallel plates. Plates 20 mm apart have a p.d. of 500 V. Find the field strength and the force on an electron between them.
🔍 Worked example — Easy
Two parallel metal plates are separated by a distance of 20 mm. A potential difference of 500 V is applied across them. Calculate the electric field strength between the plates and the force acting on an electron placed between them.
🔍 Worked example — Standard
Calculate the magnitude of the force between two point charges of +2.0 muC and -3.0 muC separated by a distance of 0.10 m in free space. (Use varepsilon_0 = 8.85 × 10^-12 C^-2 m N^-1)
🔍 Worked example — Exam-style
A point charge of +4.0 nC is located at the origin. Determine the electric potential at a point 0.20 m directly above the charge.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) Field strength: force per unit positive charge; potential: work done per unit positive charge in bringing a small test charge from infinity to the point. (2) Weight = qE = 3.2 × 10^-19 × 800/0.0050 = 5.1 × 10^-14 N. (3) A parabola: constant velocity perpendicular to the field and constant acceleration towards the positive plate. (4) 8.99 × 10^9 × 1.0 × 10^-6/0.050^2 = 3.6 × 10^6 N C⁻¹. (5) E = -Δ V/Δ x: the field is the negative of the potential gradient.