A-Level Physics: Key Formula Derivations from June 2018 Paper 2 | A-Level 物理：2018年6月试卷2 关键公式推导

📚 A-Level Physics: Key Formula Derivations from June 2018 Paper 2 | A-Level 物理：2018年6月试卷2 关键公式推导

Many A‑level Physics Paper 2 exams in June 2018 asked students to derive fundamental relationships from first principles. In this article we revisit the essential derivations that appeared – or could easily have appeared – in such papers, covering gravitational fields, circular motion, simple harmonic motion, capacitors and radioactivity. Each derivation is broken into clear steps, helping you master the logic behind the equations you use.

2018年6月的多份A‑level物理试卷2中都出现了对基本公式的推导要求。本文重新梳理了那些已经考过（或极有可能出现）的核心推导，涵盖引力场、圆周运动、简谐运动、电容和放射性。每一个推导都拆解成清晰的步骤，帮你彻底掌握公式背后的逻辑。

1. Deriving Orbital Velocity | 推导轨道速度

For a satellite in a stable circular orbit, the gravitational force provides the necessary centripetal force. Equating the two expressions eliminates the satellite’s mass and yields the orbital speed.

对于做稳定圆周运动的卫星，万有引力提供其所需的向心力。令两力表达式相等即可消去卫星质量，得出轨道速度。

Start with Newton’s law of gravitation and centripetal force: GMm / r² = m v² / r. Cancel the satellite mass m and multiply both sides by r to obtain v² = GM / r.

从万有引力定律和向心力出发：GMm / r² = m v² / r。消去卫星质量 m，两边同乘 r 即得 v² = GM / r。

Taking the square root gives the orbital velocity formula used in many past‑paper questions:

v = √(GM / r)

开平方得到轨道速度公式，历届真题中经常用到：

v = √(GM / r)

2. Deriving Kepler’s Third Law for Circular Orbits | 推导圆轨道下的开普勒第三定律

Kepler’s third law relates the orbital period T to the radius r. By substituting the orbital velocity expression into the period formula, the law emerges directly.

开普勒第三定律联系了轨道周期 T 与半径 r。将轨道速度表达式代入周期公式，即可直接导出该定律。

Orbital period is the circumference divided by speed: T = 2πr / v. Replace v with √(GM / r):

轨道周期等于圆周长除以速度：T = 2πr / v。用 √(GM / r) 替换 v：

T = 2πr / √(GM / r) = 2π √(r³ / GM)

Squaring both sides removes the square root, producing T² = (4π² / GM) r³, i.e. T² ∝ r³. This result is frequently required when analysing satellite data.

两边平方消去根号，得到 T² = (4π² / GM) r³，即 T² ∝ r³。这一结果在分析卫星数据时经常需用。

3. Deriving the Capacitor Discharge Equation | 推导电容器放电方程

One of the most challenging derivations on Paper 2 involves the exponential decay of charge on a capacitor discharging through a resistor. It begins with Kirchhoff’s voltage law and the definition of current.

试卷2中最具挑战性的推导之一是电容器通过电阻放电时电荷的指数衰减。推导从基尔霍夫电压定律和电流定义入手。

Around the loop: V_R + V_C = 0, so V_R = –V_C. Using V_R = IR and V_C = q / C, we have IR = –q / C. But the discharging current reduces the charge on the plates: I = dq/dt (negative sign already accounted for by the minus).

在回路中有：V_R + V_C = 0，所以 V_R = –V_C。代入 V_R = IR 和 V_C = q / C，得 IR = –q / C。但放电电流会减少极板上的电荷：I = dq/dt（负号已被关系中的减号体现）。

Substituting I = dq/dt gives R dq/dt = –q / C. Rearranging: dq/dt = –q / (RC). This first‑order differential equation separates to ∫ (1/q) dq = ∫ –1/(RC) dt.

代入 I = dq/dt 得 R dq/dt = –q / C。整理得 dq/dt = –q / (RC)。这个一阶微分方程分离变量后成为 ∫ (1/q) dq = ∫ –1/(RC) dt。

Integration yields ln q = –t / (RC) + constant. Applying the initial condition q = Q at t = 0 gives the final form:

积分得到 ln q = –t / (RC) + 常数。代入初始条件 t = 0 时 q = Q，得到最终形式：

q = Q e^{–t / (RC)} or V = V₀ e^{–t / (RC)}

q = Q e^{–t / (RC)} 或 V = V₀ e^{–t / (RC)}

4. Deriving Maximum Velocity in Simple Harmonic Motion | 推导简谐运动的最大速度

SHM derivations are common in Paper 2. Starting with the displacement function, differentiation gives the velocity, from which the maximum speed is easily read off.

简谐运动的推导在试卷2中很常见。从位移函数出发，求导可得速度，并直接读出最大速度。

The standard displacement equation is x = A cos(ωt) (assuming starting at maximum displacement). Velocity is the time derivative: v = dx/dt = –Aω sin(ωt).

标准位移方程为 x = A cos(ωt)（假设从最大位移处开始）。速度是位移对时间的导数：v = dx/dt = –Aω sin(ωt)。

The magnitude of v is greatest when |sin(ωt)| = 1. Therefore:

当 |sin(ωt)| = 1 时，v 的模取得最大值。因此：

v_max = Aω

This result can alternatively be derived from energy conservation, equating elastic potential energy at amplitude to kinetic energy at equilibrium.

该结果也可从能量守恒推导——令振幅处的弹性势能等于平衡位置处的动能。

5. Deriving Centripetal Acceleration | 推导向心加速度

A vector‑based derivation of a = v² / r (or a = ω²r) is often awarded several marks. The key is to consider the small angle approximation for two velocity vectors.

基于矢量的 a = v² / r（或 a = ω²r）的推导常能获得高分。关键是对两个速度矢量使用小角度近似。

Imagine a particle moving from point P to Q through a small angle Δθ in time Δt. The change in velocity Δv has magnitude Δv ≈ v Δθ (since the two velocity vectors form an isosceles triangle with small vertex angle).

设想一个质点在 Δt 时间内经小角度 Δθ 从 P 运动到 Q。速度变化量 Δv 的大小为 Δv ≈ v Δθ（因为两个速度矢量构成一个夹角很小的等腰三角形）。

The distance travelled along the arc is Δs = r Δθ ≈ v Δt, so Δθ = v Δt / r. The magnitude of acceleration is a = Δv / Δt ≈ (v · v Δt / r) / Δt = v² / r.

沿圆弧运动的距离为 Δs = r Δθ ≈ v Δt，所以 Δθ = v Δt / r。加速度的大小为 a = Δv / Δt ≈ (v · v Δt / r) / Δt = v² / r。

Using v = ωr, the alternative form a = ω²r is immediately obtained.

代入 v = ωr，立即得到另一形式 a = ω²r。

6. Deriving Energy Stored in a Capacitor | 推导电容器储存的能量

Energy stored by a capacitor is not simply QV because the voltage rises as charge builds up. The derivation uses the work done while moving an infinitesimal charge dq against the growing potential difference.

电容器储存的能量不能简单写成 QV，因为充电过程中电压随电荷积累而升高。该推导借助移动微小电荷 dq 以克服逐渐增大的电势差所做的功。

Work done to add a small charge dq when the potential difference is V = q / C is dW = V dq = (q / C) dq.

当电势差为 V = q / C 时，移动微小电荷 dq 所做的功为 dW = V dq = (q / C) dq。

Total work, and hence energy, is the integral from 0 to Q: E = ∫₀^Q (q / C) dq = (1 / C) [½ q²]₀^Q.

总功即储存能量，是对 0 到 Q 的积分：E = ∫₀^Q (q / C) dq = (1 / C) [½ q²]₀^Q。

Evaluating the definite integral gives the familiar three forms:

计算定积分得到三个常见形式：

E = ½ QV = ½ CV² = ½ Q² / C

7. Deriving the Magnetic Force on a Current‑Carrying Wire | 推导载流直导线所受磁力

The formula F = BIL sinθ can be derived from the Lorentz force on individual moving charges. This links the macroscopic force to the microscopic behaviour of electrons.

公式 F = BIL sinθ 可从单个运动电荷的洛伦兹力导出，从而将宏观作用力与电子的微观行为联系起来。

Each charge q moving with drift velocity v experiences a force F_q = B q v sinθ. In a wire of length L with n charge carriers per unit volume, the total number of moving charges is N = n A L (where A is the cross‑sectional area).

每个以漂移速度 v 运动的电荷 q 受力 F_q = B q v sinθ。对长度为 L、单位体积载流子数为 n 的导线，运动电荷总数为 N = n A L（A 为截面积）。

The total force on the wire is F = N B q v sinθ = (n A L) B q v sinθ. Recognise that the current I is the rate of charge flow: I = n A v q.

导线所受总力为 F = N B q v sinθ = (n A L) B q v sinθ。注意到电流 I 是电荷流动的速率：I = n A v q。

Substituting n A v q with I yields F = (I) × B L sinθ, i.e. F = B I L sinθ. This neatly explains why the force is proportional to both current and length.

将 n A v q 替换为 I 即得 F = (I) × B L sinθ，即 F = B I L sinθ。这清晰地解释了为何磁力与电流和长度均成正比。

8. Deriving Half‑Life from the Decay Equation | 从衰变方程推导半衰期

Radioactive decay is a random process, but the mathematical relationship between half‑life T_½ and decay constant λ is deterministic. Starting from the exponential decay law, the derivation is straightforward and often examined.

放射性衰变是一种随机过程，但半衰期 T_½ 与衰变常数 λ 之间的数学关系却是确定的。从指数衰变律出发，推导直接且常见于考试。

The number of undecayed nuclei obeys N = N₀ e^–λt. After one half‑life, N = N₀ / 2. Substitute and cancel N₀: ½ = e^{–λ T_½}.

未衰变原子核数遵循 N = N₀ e^–λt。经历一个半衰期后，N = N₀ / 2。代入并消去 N₀：½ = e^{–λ T_½}。

Taking natural logarithms of both sides gives ln(½) = –λ T_½, and since ln(½) = –ln2 we obtain:

两边取自然对数得 ln(½) = –λ T_½，再利用 ln(½) = –ln2 得到：

T_½ = ln 2 / λ

This equation appears routinely when calculating half‑lives from activity measurements or decay‑constant data.

在通过活度测量或衰变常数数据计算半衰期时，这一方程会反复出现。

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A-Level Physics: Key Formula Derivations from June 2018 Paper 2 | A-Level 物理：2018年6月试卷2 关键公式推导