📚 A-Level Physics: Formula Derivation from PH04 June 2022 Examiner Report | A-Level 物理：2022年6月PH04考试报告中的公式推导
The June 2022 AQA PH04 examiner report highlighted that many students lost marks not because they could not recall formulas, but because they were unable to derive them from first principles or explain the steps logically. This article revisits the key derivations that appeared or were implied in that paper, linking each to common errors noted by examiners. Mastering these derivations will strengthen your conceptual understanding and prepare you for the high‑mark questions that demand ‘show that’ or ‘prove’ responses.
2022年6月的AQA PH04考官报告指出，许多学生丢分不是因为他们记不住公式，而是因为他们无法从基本原理推导公式，或者不能条理清晰地解释推导步骤。本文回顾了该试卷中出现或隐含的关键推导，并将每个推导与考官指出的常见错误联系起来。掌握这些推导将加深你对概念的理解，并为那些要求“证明”或“推导”的高分题目做好准备。
1. Centripetal Acceleration Derivation | 向心加速度公式推导
The exam report noted that candidates often stated a = v²/r without showing how the direction change leads to a magnitude expression. For a particle moving with constant speed v in a circle of radius r, consider a small time interval Δt. The velocity vector changes direction from v₁ to v₂, both of magnitude v. The change in velocity Δv points towards the centre. The two velocity vectors form an isosceles triangle with angle Δθ between them. The magnitude of Δv is approximately vΔθ for small Δθ. Since Δθ = vΔt / r, we have Δv ≈ v × (vΔt / r) = v²Δt / r. Acceleration magnitude a = Δv / Δt = v² / r. As Δt → 0, this becomes exact. Using v = ωr, we also obtain a = ω²r.
考试报告指出，考生经常直接写出 a = v²/r，却没有说明方向变化如何导出加速度的大小。对于一个以恒定速率 v 在半径为 r 的圆周上运动的质点，取一小段时间 Δt。速度矢量从 v₁ 变为 v₂，两者大小均为 v，但方向不同。速度变化量 Δv 指向圆心。两个速度矢量构成一个夹角为 Δθ 的等腰三角形。当 Δθ 很小时，Δv 的大小约为 vΔθ。由于 Δθ = vΔt / r，我们得到 Δv ≈ v × (vΔt / r) = v²Δt / r。加速度大小 a = Δv / Δt = v² / r。取 Δt → 0 时的极限即可得到精确结果。利用 v = ωr，还可得到 a = ω²r。
2. Deriving a ∝ −x in Simple Harmonic Motion | 简谐运动中 a ∝ −x 的推导
A common mistake in the PH04 paper was confusing the defining equation of SHM with its consequences. SHM occurs when the resultant force F is directly proportional to the displacement x from equilibrium and always directed towards that equilibrium. Newton’s second law gives F = ma, so ma ∝ −x, which leads to a ∝ −x. For a mass‑spring system, F = −kx, hence a = −(k/m)x. The examiner’s report emphasised that students must state the condition a ∝ −x and then relate the constant of proportionality to ω², so a = −ω²x. This step is crucial for linking period T = 2π/ω.
在PH04试卷中，一个常见错误是将简谐运动的定义式与其推论混淆。当合外力 F 与相对平衡位置的位移 x 成正比且始终指向平衡位置时，物体做简谐运动。由牛顿第二定律 F = ma，可得 ma ∝ −x，即 a ∝ −x。对于弹簧振子系统，F = −kx，因此 a = −(k/m)x。考官报告强调，学生必须明确指出 a ∝ −x 这一条件，并将比例系数与 ω² 对应，即 a = −ω²x。这一步对于建立周期 T = 2π/ω 的关联至关重要。
3. Period of a Simple Pendulum | 单摆周期的推导
Examiners observed that many candidates could not justify why T = 2π√(l/g) is valid only for small amplitudes. Start with the restoring force on a pendulum bob displaced by a small angle θ: the tangential component of weight is mg sinθ. For small angles, sinθ ≈ θ in radians, so the restoring force is approximately mgθ. Displacement along the arc is x = lθ, giving F ≈ −(mg/l)x. This is a restoring force proportional to x, matching the SHM condition. Hence k_eff = mg/l. In mass‑spring SHM, T = 2π√(m/k), so T = 2π√(m/(mg/l)) = 2π√(l/g). Credit was lost when students omitted the small‑angle approximation.
考官发现，许多考生无法解释为什么 T = 2π√(l/g) 仅在小振幅下成立。从单摆摆球偏离一个小角度 θ 开始分析：重力的切向分量为 mg sinθ。当角度很小时，sinθ ≈ θ（弧度制），因此回复力约为 mgθ。沿弧的位移 x = lθ，由此得 F ≈ −(mg/l)x。这是一个与 x 成正比的回复力，符合简谐运动条件。因此有效劲度系数 k_eff = mg/l。在弹簧振子的简谐运动中，T = 2π√(m/k)，代入得 T = 2π√(m/(mg/l)) = 2π√(l/g)。如果忽略了小角度近似，就会被扣分。
4. Gravitational Field Strength and Potential | 引力场强度与引力势的推导
The PH04 report indicated a lack of rigour when deriving g = GM/r² and the potential V = −GM/r. For two point masses M and m at distance r, Newton’s law gives F = GMm/r². Gravitational field strength g is defined as force per unit mass, so g = F/m = GM/r². To derive potential, remember that potential V is the work done per unit mass in bringing a test mass from infinity to that point. Work done against the field is W = ∫∞ʳ −F dr = ∫∞ʳ −(GMm/r²) dr. Evaluating the integral: W/m = [GM/r]∞ʳ = GM/r − 0, but because the force is attractive, the work done by the field is negative, leading to V = −GM/r. Many students mis‑handled the sign.
PH04 报告显示，学生在推导 g = GM/r² 和势 V = −GM/r 时不够严谨。对于相距 r 的两个质元 M 和 m，牛顿万有引力定律给出 F = GMm/r²。引力场强度 g 定义为单位质量所受的力，因此 g = F/m = GM/r²。推导势时，要记住势 V 是将单位检验质量从无穷远处移至该点外力所做的功。克服引力所做的功为 W = ∫∞ʳ −F dr = ∫∞ʳ −(GMm/r²) dr。计算积分得 W/m = [GM/r]∞ʳ = GM/r − 0，但由于引力是吸引力，引力场本身做功为负，最终得到 V = −GM/r。许多学生处理符号时犯了错误。
5. Electric Field Strength and Potential Gradient | 电场强度与电势梯度的关系
The exam exposed confusion between uniform field formulas and the general relation. For a uniform electric field between parallel plates, E = V/d is derived from the work done moving a charge q through a potential difference V: work = qV. Work is also force × distance: Fd = qV. Since F = qE, we get E = V/d. More generally, the field strength is the negative potential gradient. In a uniform field this reduces to E = −ΔV/Δx. In radial fields, E = kQ/r² can be obtained from V = kQ/r by differentiating: E = −dV/dr = kQ/r². Candidates needed to show that E = V/d is a special case, not the definition.
考试暴露出学生在匀强电场公式与一般关系上的混淆。对于平行板间的匀强电场，E = V/d 可通过移动电荷 q 经过电势差 V 所做的功来推导：功 = qV。功也等于力乘距离：Fd = qV。因为 F = qE，可得 E = V/d。更一般地，电场强度是电势梯度的负值。在匀强电场中，这简化为 E = −ΔV/Δx。在径向电场中，由 V = kQ/r 求导可得 E = −dV/dr = kQ/r²。考生需要证明 E = V/d 只是一个特例，而非定义。
6. Capacitor Discharge Equation | 电容器放电方程推导
Many marks were lost on the exponential decay derivation because students started with the solution without showing the differential equation. For a capacitor C discharging through a resistor R, the potential difference V across the capacitor equals the p.d. across the resistor, so V = IR. The charge on the capacitor Q = CV, and the current I is the rate of decrease of charge: I = −dQ/dt. Substituting gives Q/C = −R dQ/dt, hence dQ/dt = −Q/(RC). This is a first‑order differential equation. Solving by separation of variables: dQ/Q = −dt/(RC). Integrating gives ln Q = −t/(RC) + constant. Using initial condition Q = Q₀ at t = 0 leads to Q = Q₀ e⁻⁽ᵗ/ᴿᶜ⁾. Since V ∝ Q, the same form applies for V and I.
许多考生在指数衰减的推导上丢分，因为他们直接写出解，而没有展示微分方程。对于通过电阻 R 放电的电容 C，电容器两端的电势差 V 等于电阻两端的电压，因此 V = IR。电容器的电荷量 Q = CV，电流 I 是电荷减少的速率：I = −dQ/dt。代入得 Q/C = −R dQ/dt，因此 dQ/dt = −Q/(RC)。这是一个一阶微分方程。用分离变量法求解：dQ/Q = −dt/(RC)。积分得 ln Q = −t/(RC) + 常数。利用初始条件 t=0 时 Q=Q₀，可得 Q = Q₀ e⁻⁽ᵗ/ᴿᶜ⁾。由于 V ∝ Q，V 和 I 也具有相同形式的解。
7. Deriving the Exponential Law of Radioactive Decay | 放射性衰变指数规律的推导
The examiner’s report highlighted that a significant number of students could not justify N = N₀ e⁻⁽λᵗ⁾ from the fundamental assumption that decay rate is proportional to the number of nuclei present. The activity A = −dN/dt = λN, where λ is the decay constant. This leads to dN/dt = −λN. Separating variables: dN/N = −λ dt. Integrating gives ln N = −λt + constant. With N = N₀ at t = 0, we obtain N = N₀ e⁻⁽λᵗ⁾. The half‑life T₁/₂ is found by setting N = N₀/2, giving ln(1/2) = −λ T₁/₂, so T₁/₂ = ln2/λ.
考官报告指出，相当多的学生无法从衰变速率与现存核子数成正比这一基本假设出发，推导出 N = N₀ e⁻⁽λᵗ⁾。活度 A = −dN/dt = λN，其中 λ 为衰变常量。由此得 dN/dt = −λN。分离变量：dN/N = −λ dt。积分得 ln N = −λt + 常数。代入 t=0 时 N=N₀，可得 N = N₀ e⁻⁽λᵗ⁾。半衰期 T₁/₂ 可通过令 N=N₀/2 求得，于是 ln(1/2) = −λ T₁/₂，故 T₁/₂ = ln2/λ。
8. Radius of Curvature for a Charged Particle in a Magnetic Field | 带电粒子在磁场中圆周运动半径推导
Questions requiring the derivation of r = mv/(Bq) were often rushed, with candidates skipping the force direction justification. A charged particle of charge q moving with velocity v perpendicular to a uniform magnetic field B experiences a magnetic force F = Bqv (using Fleming’s left‑hand rule). This force is perpendicular to velocity, providing the centripetal force for circular motion. Hence Bqv = mv²/r. Cancelling one power of v gives r = mv/(Bq). A common error was failing to state that the force acts as the centripetal force. For non‑perpendicular motion, the component v⊥ must be used.
要求推导 r = mv/(Bq) 的题目往往处理得很仓促，考生跳过了力的方向论证。带电粒子电荷为 q，以速度 v 垂直于匀强磁场 B 运动，受到的磁力为 F = Bqv（用弗莱明左手定则判断方向）。该力始终垂直于速度，提供圆周运动所需的向心力。因此 Bqv = mv²/r。消去一个 v 即得 r = mv/(Bq)。常见错误是未指明磁力充当向心力。若非垂直入射，则需要使用速度的垂直分量 v⊥。
9. Energy Stored in a Capacitor | 电容器储存能量的推导
The PH04 report mentioned that many could not explain why E = ½CV², resorting to memorised formula. Consider charging a capacitor from zero to final charge Q. At an intermediate stage, when the charge is q, the potential difference is v = q/C. To add a small amount of charge dq, the work done is v dq = (q/C) dq. Total work W = ∫₀⁰ (q/C) dq = [q²/(2C)]₀⁰ = Q²/(2C). Since Q = CV, W = ½CV². This energy is stored in the electric field between plates. Candidates lost marks by omitting the integration steps.
PH04 报告提到，许多人无法解释为何 E = ½CV²，只能死记公式。考虑将电容器从零充电至最终电荷 Q。中间阶段电荷量为 q 时，电势差为 v = q/C。移动微小电荷 dq 需做功 v dq = (q/C) dq。总功 W = ∫₀⁰ (q/C) dq = [q²/(2C)]₀⁰ = Q²/(2C)。因为 Q = CV，所以 W = ½CV²。这一能量储存在两极板之间的电场中。考生若省略积分步骤就会丢分。
10. Linking Gravitational Potential and Escape Velocity | 引力势与逃逸速度的关联推导
Escape velocity derivation appeared indirectly in PH04, and examiners were strict about starting from energy conservation. To escape a planet’s gravitational field from its surface, a mass m must have kinetic energy at least equal to the magnitude of its gravitational potential energy at the surface. Potential energy U = −GMm/R, so the minimum kinetic energy required is ½mv² = GMm/R, giving v = √(2GM/R). Alternatively, using g = GM/R², we can write v = √(2gR). Students were expected to note that escape speed does not depend on the direction of launch (as long as it’s not straight into the ground).
逃逸速度的推导在PH04中间接出现，考官对于从能量守恒出发的要求非常严格。要挣脱行星引力场而从其表面逃逸，质量 m 所需的动能至少须等于其在表面引力势能的大小。势能 U = −GMm/R，因此所需的最小动能满足 ½mv² = GMm/R，得出 v = √(2GM/R)。利用 g = GM/R²，也可写成 v = √(2gR)。学生需要指出，逃逸速度与发射方向无关（只要不是直接撞向地面）。
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