📚 A-Level Physics Unit 4: Deriving Key Formulas from the Jan 19 Insert | A-Level 物理 Unit 4：2019年1月公式表关键公式推导
Welcome to this in-depth derivation guide for the essential equations listed in the A-Level Physics Unit 4 formula insert (January 2019). Understanding the origins of these formulas not only deepens your conceptual grasp but also arms you with the ability to tackle unseen problems with confidence. We will walk through step-by-step reasoning for each key relationship.
欢迎阅读这篇深入推导指南，分别呈现 A-Level 物理 Unit 4 公式表（2019年1月）所列的关键方程。理解公式的由来不仅深化概念理解，还能让你自信地应对陌生问题。我们将逐步推演每一个重要关系。
1. Momentum and Impulse | 动量与冲量
ffect of a force acting over a time interval and is defined as the change in momentum: Impulse = Δp = FΔt for a constant force. The relationship follows directly from Newton’s second law F = dp/dt; integrating over time yields the impulse-momentum theorem.
动量 p 定义为质量与速度的乘积：p = mv，是与速度同向的矢量。冲量衡量力在一段时间内作用的效果，定义为动量的变化：冲量 = Δp = FΔt（恒力情况下）。这一关系直接来自牛顿第二定律 F = dp/dt；对时间积分便得到冲量-动量定理。
2. Centripetal Acceleration | 向心加速度
ta;θ. The change in velocity Δv points roughly toward the centre and has magnitude vΔθ. Since angular speed ω = Δθ/Δt, the acceleration magnitude is a = Δv/Δt = vω. Using v = ωr, we obtain the familiar forms a = v2/r = ω2r. This acceleration is always directed towards the centre, hence the term ‘centripetal’.
考虑物体以恒定速率 v 沿半径为 r 的圆周运动。在短时间 Δt 内，速度矢量转过小角 Δθ。速度变化量 Δv 大致指向圆心，其大小为 vΔθ。因为角速度 ω = Δθ/Δt，加速度大小为 a = Δv/Δt = vω。代入 v = ωr，得到熟知的形式 a = v2/r = ω2r。此加速度始终指向圆心，故称之为‘向心’加速度。
3. Coulomb’s Law and Electric Field | 库仑定律与电场
Coulomb’s law gives the force between two point charges: F = kQ1Q2/r2, where k = 1/(4πε0 it positive charge: E = F/q. For a point charge Q, the radial field has magnitude E = kQ/r2 charge and towards a negative charge. In a uniform field between parallel plates, E = V/d.
库仑定律给出两个点电荷之间的作用力：F = kQ1Q2/r2，其中 k = 1/(4πε0)。电场强度 E 定义为单位正电荷所受的力：E = F/q。对于点电荷 Q，其径向电场的大小为 E = kQ/r2，方向背离正电荷而指向负电荷。在平行板间的匀强电场中，E = V/d。
4. Electric Potential around a Point Charge | 点电荷周围的电势
harge from infinity to that point. The force on the test charge q is F = qE, with E = kQ/r2 force is dW = -F dr = -qE dr. Integrating from infinity to a distance r gives V = W/q = ∫∞r (-E dr) = ∫∞r (-kQ/r2 ) = kQ/r. This result shows the potential is positive for positive Q and decreases as 1/r.
电势 V 是指将单位正电荷从无穷远处移到该点时所做的功。试探电荷 q 所受电场力为 F = qE，其中 E = kQ/r2 沿径向。反抗电场力做功 dW = -F dr = -qE dr。从无穷远积分至距离 r 可得 V = W/q = ∫∞r (-E dr) = ∫∞r (-kQ/r2 dr) = kQ/r。结果说明正电荷的电势为正，且以 1/r 递减。
5. Capacitance of a Parallel Plate Capacitor | 平行板电容器的电容
Using Gauss’s law, the uniform electric field between the plates is E = σ/ε0 = Q/(Aε0). The potential difference is V = Ed = Qd/(Aε0). Capacitance is defined as C = Q/V, so substituting gives C = ε0A/d. This formula highlights the geometric dependence: larger area increases capacitance, while greater separation reduces it.
面积为 A、间距为 d 的平行板电容器储存电荷 Q 并产生电压 V。利用高斯定律，两板间的匀强电场为 E = σ/ε0 = Q/(Aε0)。电势差 V = Ed = Qd/(Aε0)。电容定义为 C = Q/V，代入即得 C = ε0A/d。该公式体现出几何依赖性：板面积越大则电容越大，间距越大则电容越小。
6. Energy Stored in a Capacitor | 电容器储存的能量
V dq = (q/C) dq. Integrating from 0 to the final charge Q gives total stored energy W = ∫0Q (q/C) dq = ½ Q2/C. Using C = Q/V, this can be rewritten as W = ½ Q V = ½ C V2. This energy is stored in the electric field between the plates.
给电容器充电需要做功，因为每增加微小电荷 dq 都要反抗不断升高的电压 V = q/C。做功 dW = V dq = (q/C) dq。从 0 积分到最终电荷 Q，得总储能 W = ∫0Q (q/C) dq = ½ Q2/C。利用 C = Q/V，可改写为 W = ½ Q V = ½ C V2。这部分能量储存在两极板间的电场中。
7. Magnetic Force on a Current-Carrying Wire | 电流在磁场中的受力
Bqv. For a steady current, the total moving charge in the wire is q = It, and the drift velocity relates to length by L = v t. Combining, the total force is F = B (It) v = B I (v t) = B I L. If the wire is at an angle θ to the field, the perpendicular component gives F = B I L sinθ. This is the foundation for motor effect calculations.
当长度为 L、载有电流 I 的导线垂直于匀强磁场 B 放置时，每个载流子受到磁力 F = Bqv。对于稳恒电流，导线中运动的总电荷 q = It，且漂移速度与长度的关系为 L = v t。综合得总力 F = B (It) v = B I (v t) = B I L。若导线与磁场成 θ 角，则垂直分量给出 F = B I L sinθ。这是电动机效应计算的基础。
8. Magnetic Flux and Faraday’s Law | 磁通量与法拉第定律
face. Faraday’s law states that the induced emf in a coil is equal to the negative rate of change of flux linkage: ε = -N dΦ/dt. The negative sign embodies Lenz’s law, which says the induced current opposes the change producing it. A common derivation for a straight wire moving perpendicularly through a field gives the motional emf ε = B L v, consistent with the flux cutting picture.
通过某一面的磁通量 Φ 定义为 Φ = B A cosθ，其中 θ 是磁感线与面法线的夹角。法拉第定律指出，线圈中感生电动势等于磁通匝链数变化率的负值：ε = -N dΦ/dt。负号体现了楞次定律——感生电流总在阻碍产生它的变化。对于直导线垂直切割磁感线的常见推导给出动生电动势 ε = B L v，与磁通切割图像一致。
9. The Transformer Equation | 变压器公式
An ideal transformer operates with two coils wound on a common iron core so that the changing flux from the primary coil links fully with the secondary coil. The rate of flux change is the same for both, so the induced emf per turn is identical: εp/Np = εs/Ns. Since the terminal voltages are nearly equal to the induced emfs, we get Vp/Vs = Np/Ns. Assuming 100% efficiency, power conservation gives Vp Ip = Vs Is, leading to Is/Ip = Np/Ns.
理想变压器在共同铁芯上绕有两个线圈，使初级线圈的变化磁通完全交链到次级线圈。两线圈磁通变化率相同，因而每匝感生电动势相等：εp/Np = εs/Ns。由于端电压近似等于感生电动势，可得 Vp/Vs = Np/Ns。假设效率100%，功率守恒给出 Vp Ip = Vs Is，继而导出 Is/Ip = Np/Ns。
10. Elastic Collisions in One Dimension | 一维弹性碰撞
In a perfectly elastic collision, both momentum and kinetic energy are conserved. For two objects of masses m1 and m2 with initial velocities u1 and u2, the conservation laws are: m1u1 + m2u2 = m1v1 + m2v2 and ½ m1u12 + ½ m2u22 = ½ m1v12 + ½ m2v22. When the target is initially stationary (u2 = 0), solving simultaneously yields v1 = (m1 - m2)u1/(m1 + m2) and v2 = (2 m1)u1/(m1 + m2). These special cases are frequently tested in particle physics and mechanics problems.
完全弹性碰撞中，动量和动能均守恒。设有质量 m1 和 m2 的两物体，初速度分别为 u1 和 u2，守恒律为：m1u1 + m2u2 = m1v1 + m2v2 以及 ½ m1u12 + ½ m2u22 = ½ m1v12 + ½ m2v22。当靶最初静止 (u2 = 0)，联立解得 v1 = (m1 - m2)u1/(m1 + m2) 和 v2 = (2 m1)u1/(m1 + m2)。这些特殊情况常在粒子物理与力学问题中考查。
11. Radioactive Decay Law and Half-Life | 放射性衰变定律与半衰期
derlying differential equation is dN/dt = -λN. Separating variables and integrating gives ln(N/N0) = -λt, or N = N0 e-λt. The half-life T½ is the time taken for half the nuclei to decay; setting N = N0/2 yields e-λT½ = ½, so T½ = ln2/λ. The same form applies to activity and mass.
放射性样品的活度与尚未衰变的原子核数 N 成正比：A = λN，其中 λ
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