Electric Fields: Complete A Level H2 Physics Cheatsheet
A Level Physics (H2), Chapter 14 · 9478 Topic 14 · Read time: ~12 minutes
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Key Formulas
Coulomb's law and field strength
Two point charges exert a force F = Q₁Q₂/(4πε₀r²) along the line joining them: attractive for unlike charges, repulsive for like charges.
Electric field strength is the force per unit positive charge. For a point charge, E = Q/(4πε₀r²), pointing away from a positive charge and towards a negative one.
Electric potential and potential energy
Electric potential (V) at a point is the work done per unit positive charge by an external force in bringing a small test charge from infinity to that point. V = Q/(4πε₀r): positive near a positive charge, negative near a negative charge.
The potential energy of two point charges is U = Q₁Q₂/(4πε₀r). Work done moving a charge q = qΔV.
E = −dV/dr: field strength is the negative of the potential gradient.
Trap: E is a vector falling as 1/r²; V is a scalar falling as 1/r. When combining several charges, add potentials as signed numbers, but add field strengths as vectors.
Uniform fields and charged particles
Between parallel plates, the field is uniform: E = V/d, the same everywhere (ignoring edges).
A charge q in the field feels a constant force F = qE, so it has constant acceleration qE/m.
A particle entering perpendicular to the field moves like a projectile: constant velocity along the plates, uniform acceleration across them, giving a parabolic path.
Trick: Treat a beam between plates exactly like projectile motion, with qE/m in place of g. The same two-component method works.
Capacitance
Capacitance is the ratio of charge stored to potential difference, C = Q/V, in farads.
The energy stored is the area under the V-Q graph, a triangle: U = ½QV = ½Q²/C = ½CV².
A battery charging a capacitor does work QV, but only ½QV is stored. The other half is dissipated in the circuit's resistance.
Trap: U = QV is the work done by the source, not the energy stored. The stored energy always has the ½.
Worked Example
Two parallel plates 2.0 cm apart have a p.d. of 400 V across them. An electron (charge 1.60 × 10⁻¹⁹ C, mass 9.11 × 10⁻³¹ kg) enters midway between them, moving parallel to the plates.
(a) Calculate the electric field strength between the plates. [1]
E = V / d = 400 ÷ 0.020 = 2.0 × 10⁴ V m⁻¹
(b) Calculate the force on the electron and its acceleration. [2]
F = qE = 1.60 × 10⁻¹⁹ × 2.0 × 10⁴ = 3.2 × 10⁻¹⁵ N
a = F / m = 3.2 × 10⁻¹⁵ ÷ 9.11 × 10⁻³¹ = 3.5 × 10¹⁵ m s⁻²
(c) Describe the path of the electron between the plates. Explain. [2]
A parabola curving towards the positive plate.
Constant velocity along the plates, and constant acceleration perpendicular to them, as in projectile motion.
(d) A 470 μF capacitor is charged to 12 V. Calculate the charge and energy stored. [2]
Q = CV = 470 × 10⁻⁶ × 12 = 5.6 × 10⁻³ C
U = ½CV² = 0.5 × 470 × 10⁻⁶ × 144 = 3.4 × 10⁻² J
Why this question is a good test of the topic: it reuses the projectile model in a new field, and the capacitor part checks the ½ in the stored energy formula.
Frequently Asked Questions
What is the difference between electric field strength and potential?
Field strength is force per unit positive charge, a vector, falling as 1/r². Potential is work done per unit positive charge from infinity, a scalar, falling as 1/r.
How are E and V related?
E = −dV/dr. The field strength is the negative of the potential gradient.
Is the field between parallel plates uniform?
Yes, apart from near the edges. E = V/d everywhere between the plates.
Why is a capacitor's stored energy ½QV?
The p.d. rises from 0 to V as it charges, so the energy is the area under the V-Q graph, a triangle of area ½QV.
Is capacitance in 9478 Electric Fields?
Yes. 9478 places capacitance and stored energy under Electric Fields, and RC charging and discharging under Circuits.
Still confusing field strength and potential?
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