📚 A-Level Physics Unit 4: Formula Derivation Mastery from the January 2020 Exam | A-Level物理第四单元：2020年1月真题公式推导精讲
Many high-mark questions in the A-Level Physics Unit 4 paper, especially those seen in the January 2020 sitting, require you not just to recall formulas but to derive them step by step. This article revisits eight classic derivations that frequently appear in Fields and Further Mechanics, building your confidence in applying fundamental principles such as Newton’s laws, conservation of energy, and the definitions of electric and magnetic fields. Each section presents the English explanation followed immediately by its Chinese equivalent so that you can master both the logic and the precise language required for the examination.
A-Level物理第四单元试卷中的许多高分题，尤其是我们在2020年1月试卷中看到的，不仅要求你记住公式，更要求你一步步推导公式。本文回顾了“场与进阶力学”中高频出现的八个经典推导，帮助你建立对基本定律（牛顿定律、能量守恒、电场与磁场定义）的自信运用。每个要点均先提供英文解释，紧接着给出中文对应内容，让你既掌握逻辑，也掌握考试所需的精确语言。
1. Derivation of Centripetal Acceleration a = v²/r | 向心加速度公式 a = v²/r 的推导
Consider an object moving with constant speed v along a circular path of radius r. In a short time interval Δt, the object moves from point A to point B, and the radius turns through a small angle Δθ. The velocity vector changes direction but not magnitude. The two velocity vectors at A and B, both of length v, form an isosceles triangle alongside the displacement triangle made by the two radii. Because the triangles are similar, we can write Δv / v = chord AB / r. For small Δt, the chord length is approximately the arc length vΔt, so Δv / v ≈ vΔt / r, giving Δv / Δt ≈ v² / r. In the limit Δt → 0, this becomes the instantaneous acceleration a = v² / r directed towards the centre of the circle.
考虑一个物体以恒定速率 v 沿半径为 r 的圆周运动。在很短的时间间隔 Δt 内，物体从 A 点运动到 B 点，半径转过一个小角度 Δθ。速度矢量仅改变方向，大小不变。A 点和 B 点的两个速度矢量长度均为 v，它们与两个半径构成的位移三角形形成一对相似三角形。因此可写出 Δv / v = 弦 AB / r。当 Δt 很小时，弦长近似为弧长 vΔt，所以 Δv / v ≈ vΔt / r，从而 Δv / Δt ≈ v² / r。在 Δt → 0 的极限下，可得瞬时加速度 a = v² / r，方向指向圆心。
- Use the similarity between the velocity vector triangle and the radius triangle.
- 利用速度矢量三角形与半径三角形的相似性。
- Approximate the chord as the arc for an infinitesimal angle; the limit gives the exact result.
- 微小角度下将弦长近似为弧长，取极限即得精确结果。
a = v² / r
2. Relationship Between Momentum and Kinetic Energy: p² = 2mEₖ | 动量与动能关系：p² = 2mEₖ 的推导
Kinetic energy Eₖ is defined as ½mv², and momentum p is defined as mv. By squaring the momentum equation we obtain p² = m²v². Multiply the kinetic energy expression by 2m: 2m × (½mv²) = m²v². Therefore, p² = 2mEₖ. This relationship is extremely useful when comparing the momenta of objects that have the same kinetic energy but different masses, or when solving collision problems where you need to link the conservation of momentum to energy considerations.
动能 Eₖ 定义为 ½mv²，动量 p 定义为 mv。将动量方程平方得到 p² = m²v²。将动能表达式乘以 2m：2m × (½mv²) = m²v²。因此 p² = 2mEₖ。当需要比较动能相同但质量不同的物体的动量时，或解决需要将动量守恒与能量相关联的碰撞问题时，这一关系极其有用。
- Start from p = mv and Eₖ = ½mv²; eliminate v.
- 从 p = mv 和 Eₖ = ½mv² 出发，消去 v。
- Always check the units: p² has units kg² m² s⁻², which matches 2mEₖ.
- 务必检查单位：p² 的单位是 kg² m² s⁻²，与 2mEₖ 一致。
p² = 2mEₖ
3. Electric Field Strength as Negative Potential Gradient: E = -dV/dr | 电场强度作为负电势梯度：E = -dV/dr 的推导
From the definition of electric potential, the work done by the field in moving a positive test charge q₀ from point A to point B is q₀(Vₐ – V_B). In a uniform field E, the work done is also force × distance = q₀E × Δr, where Δr is the displacement along the field direction. Equating the two expressions for a small displacement dr gives q₀E dr = -q₀ dV, hence E = -dV/dr. The negative sign indicates that the field points in the direction of decreasing potential. For a radial field around a point charge, we apply the same principle to obtain E = kQ/r² from V = kQ/r.
由电势的定义可知，静电场将正检验电荷 q₀ 从 A 点移至 B 点所做的功为 q₀(Vₐ – V_B)。在匀强电场 E 中，该功也等于力乘以距离，即 q₀E × Δr，其中 Δr 是沿电场方向的位移。对于微小位移 dr，令两式相等得到 q₀E dr = -q₀ dV，因此 E = -dV/dr。负号表示电场指向电势下降的方向。对于点电荷周围的径向场，我们应用同一原理即可从 V = kQ/r 得到 E = kQ/r²。
- Relate work done by the field to potential difference and to force × distance.
- 将电场力做功与电势差及力乘距离相联系。
- The negative sign is essential: E is directed from high to low potential.
- 负号至关重要：E 的方向从高电势指向低电势。
E = -dV/dr
4. Capacitor Discharge Equation Q = Q₀ e^{-t/RC} | 电容器放电公式 Q = Q₀ e^{-t/RC} 的推导
During discharge, the current I = -dQ/dt (negative because Q decreases). By Kirchhoff’s voltage law, the potential difference across the capacitor Q/C equals the potential difference across the resistor IR, so Q/C = IR. Substituting I gives Q/C = -R dQ/dt. Rearranging: dQ/dt = -Q/(RC). This is a first-order differential equation, which can be solved by separating variables: ∫(1/Q) dQ = -∫(1/RC) dt. Integrating yields ln Q = -t/RC + constant. Applying the initial condition Q = Q₀ at t = 0, we obtain ln(Q/Q₀) = -t/RC, and hence Q = Q₀ e^{-t/RC}.
放电过程中，电流 I = -dQ/dt（负号表示 Q 随时间减少）。根据基尔霍夫电压定律，电容器两端的电势差 Q/C 等于电阻两端的电势差 IR，因此 Q/C = IR。代入 I 的表达式得到 Q/C = -R dQ/dt。重新排列得 dQ/dt = -Q/(RC)。这是一个一阶微分方程，可通过分离变量法求解：∫(1/Q) dQ = -∫(1/RC) dt。积分后得到 ln Q = -t/RC + 常数。代入初始条件 t = 0 时 Q = Q₀，可得 ln(Q/Q₀) = -t/RC，从而 Q = Q₀ e^{-t/RC}。
- Use I = -dQ/dt and the loop equation Q/C = IR.
- 应用 I = -dQ/dt 与回路方程 Q/C = IR。
- Recognise that RC is the time constant τ of the circuit.
- 注意 RC 是电路的时间常数 τ。
Q = Q₀ e^{-t/RC}
5. Energy Stored by a Capacitor: E = ½QV = ½CV² = ½Q²/C | 电容器储存的能量：E = ½QV = ½CV² = ½Q²/C 的推导
When a small amount of charge dq is moved through a potential difference V, the work done is V dq. For a capacitor, V = q/C, where q is the instantaneous charge. Therefore, the incremental work dW = (q/C) dq. To charge the capacitor from 0 to a final charge Q, we integrate: W = ∫₀ᴼ (q/C) dq = [q²/(2C)]₀ᴼ = Q²/(2C). Since Q = CV, this stored energy can also be written as ½CV² or ½QV. This derivation shows that half the energy supplied by the battery is stored, the other half being dissipated in the charging circuit.
当微小电荷 dq 在电势差 V 下移动时，所做的功为 V dq。对于电容器，V = q/C，其中 q 为瞬时电荷量。因此，微量功 dW = (q/C) dq。为将电容器从 0 充至最终电荷 Q，进行积分：W = ∫₀ᴼ (q/C) dq = [q²/(2C)]₀ᴼ = Q²/(2C)。由于 Q = CV，储存的能量也可写为 ½CV² 或 ½QV。这一推导表明电池提供的能量只有一半被储存，另一半在充电回路中耗散。
- dW = V dq is the fundamental starting point.
- 基本出发点是 dW = V dq。
- The area under the V–Q graph for a capacitor is a triangle, giving ½QV directly.
- 电容器的 V–Q 图线下面积为三角形，直接得出 ½QV。
E = ½CV² = ½QV = Q²/(2C)
6. Magnetic Force on a Moving Charge and Radius of Circular Path: r = mv/(Bq) | 运动电荷的磁力及圆周运动半径：r = mv/(Bq) 的推导
A charged particle of charge q moving with velocity v perpendicular to a uniform magnetic field of flux density B experiences a magnetic force F = Bqv (from F = BIl and I = q/t). This force is always perpendicular to the velocity, so it acts as a centripetal force. Equating the magnetic force to the centripetal force: Bqv = mv²/r. Solving for the radius r gives r = mv/(Bq). The period T of the circular motion, found from v = 2πr/T, results in T = 2πm/(Bq), which is independent of the particle’s speed – a key concept in cyclotron theory.
一个电荷量为 q 的粒子以速度 v 垂直于磁通密度 B 的匀强磁场运动时，受到磁力 F = Bqv（由 F = BIl 与 I = q/t 导出）。该力始终与速度垂直，因此充当向心力。令磁力等于向心力：Bqv = mv²/r。解出轨道半径 r 得 r = mv/(Bq)。由 v = 2πr/T 可得圆周运动的周期 T = 2πm/(Bq)，该周期与粒子速度无关——这是回旋加速器理论中的关键概念。
- Start from F = Bqv and note its direction from Fleming’s left-hand rule.
- 从 F = Bqv 出发，并用弗莱明左手定则判定方向。
- The condition ‘perpendicular’ is vital; if velocity has a parallel component, the path becomes helical.
- “垂直”条件至关重要；若速度有平行分量，轨迹将变为螺旋线。
r = mv / (Bq)
7. Faraday’s Law of Electromagnetic Induction: ε = -N dΦ/dt | 法拉第电磁感应定律：ε = -N dΦ/dt 的推导
Faraday discovered that a changing magnetic flux Φ through a coil induces an electromotive force (emf). The flux through a single turn is Φ = BA cos θ. For a coil of N turns, the flux linkage is NΦ. By analysing the work done when a conductor moves in a magnetic field, one finds that the induced emf equals the rate of change of flux linkage: ε = -N dΦ/dt. The negative sign embodies Lenz’s law: the induced current opposes the change in flux that produced it. This law can be derived from the Lorentz force on electrons in the conductor moving through the field, reinforcing the unity of electromagnetic theory.
法拉第发现，通过线圈的变化磁通量 Φ 会感应出电动势。单匝线圈的磁通量为 Φ = BA cos θ。对于 N 匝线圈，磁链为 NΦ。通过分析导体在磁场中移动时所做的功，可以得出感应电动势等于磁链的变化率：ε = -N dΦ/dt。负号体现了楞次定律：感应电流的方向总是阻碍引起它的磁通量变化。这一定律也可从运动导体中电子所受的洛伦兹力导出，体现了电磁理论的统一性。
- Flux linkage = NΦ; a change is needed to produce an emf.
- 磁链 = NΦ；产生电动势需要磁通量发生变化。
- Lenz’s law is a statement of energy conservation; without the minus sign, a perpetual motion machine would be possible.
- 楞次定律实质上是能量守恒的体现；若没有负号，永动机将成为可能。
ε = -N dΦ/dt
8. Transformer EMF Equation and Turns Ratio Vₛ/Vₚ = Nₛ/Nₚ | 变压器电动势方程与匝数比 Vₛ/Vₚ = Nₛ/Nₚ 的推导
An ideal transformer assumes no flux leakage and perfect magnetic coupling. The primary coil produces a changing magnetic flux that entirely links the secondary coil. From Faraday’s law, the induced emf in the primary is Vₚ = Nₚ dΦ/dt, and in the secondary Vₛ = Nₛ dΦ/dt. Since the rate of change of flux is the same for both coils, dividing the two equations yields Vₛ/Vₚ = Nₛ/Nₚ. For an ideal transformer with 100% efficiency, the power input equals the power output: Vₚ Iₚ = Vₛ Iₛ, which leads to Iₚ/Iₛ = Nₛ/Nₚ. This shows how step-up transformers trade current for voltage.
理想变压器假设无磁漏且完全磁耦合。初级线圈产生的变化磁通量全部耦合到次级线圈。根据法拉第定律，初级感应电动势 Vₚ = Nₚ dΦ/dt，次级 Vₛ = Nₛ dΦ/dt。由于两线圈的磁通量变化率相同，两式相除得 Vₛ/Vₚ = Nₛ/Nₚ。对于效率为100%的理想变压器，输入功率等于输出功率：Vₚ Iₚ = Vₛ Iₛ，于是有 Iₚ/Iₛ = Nₛ/Nₚ。这表明升压变压器是以电流换取电压。
- Assumption: dΦ/dt is identical in both coils; this requires a closed iron core.
- 假设：两线圈中 dΦ/dt 完全相同，这需要闭合铁芯。
- Real transformers have losses due to eddy currents, hysteresis and winding resistance.
- 实际变压器因涡流、磁滞和绕组电阻而存在损耗。
Vₛ / Vₚ = Nₛ / Nₚ
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