A-Level Physics June 2018 Insert 5 Formula Derivations | A-Level 物理 2018年6月插入页5公式推导

📚 A-Level Physics June 2018 Insert 5 Formula Derivations | A-Level 物理 2018年6月插入页5公式推导

The June 2018 A-Level Physics Unit 5 examination (WPH05) included an insert sheet – often referred to as Insert 5 – containing a concise list of essential formulas, constants and relationships. This article walks through the step‑by‑step derivations of several key equations from that insert, helping students understand the physical principles behind them rather than simply memorising them.

2018年6月的A-Level物理第五单元考试（WPH05）附带了一张插入页——常被称为Insert 5——它列出了必备公式、常数和关系式。本文逐步推导该插入页中的多个关键方程，帮助学生理解其背后的物理原理，而不仅仅是死记硬背。

1. Understanding the Insert 5 Reference Sheet | 理解插入页5参考表

The Insert 5 for Unit 5 (Physics from Creation to Collapse) covered topics such as forced oscillations and resonance, thermal physics, nuclear decay, capacitors, electromagnetism and astrophysics. The derivations below focus on the most conceptually demanding equations that frequently appear in exam questions.

第五单元的Insert 5（从创生到碰撞的物理学）涵盖了受迫振动与共振、热物理、核衰变、电容器、电磁学和天体物理等主题。以下推导将集中讨论那些概念要求较高且常在考题中出现的方程。

2. Deriving the Exponential Decay for Capacitor Discharge | 推导电容器放电的指数衰减

Consider a capacitor of capacitance C charged to an initial voltage V₀, discharging through a resistor R. By definition, Q = C V. The current I in the circuit is the rate of decrease of charge on the capacitor: I = – dQ/dt. Ohm’s law gives V = I R, but during discharge the voltage across the capacitor equals the voltage across the resistor, so V = ( – dQ/dt) R. Substituting V = Q/C yields Q/C = – R dQ/dt. Rearranging gives dQ/dt = – (1/RC) Q.

考虑一个电容C初始充电至电压V₀，通过电阻R放电。根据定义，Q = C V。电路中的电流I等于电容器上电荷减少的速率：I = – dQ/dt。欧姆定律给出V = I R，但在放电过程中电容器两端电压等于电阻两端电压，因此V = ( – dQ/dt) R。代入V = Q/C得到Q/C = – R dQ/dt。整理得 dQ/dt = – (1/RC) Q。

This is a first‑order differential equation whose solution is Q = Q₀ e^–t/RC, where Q₀ is the initial charge. Using Q = C V, the corresponding voltage decay is V = V₀ e^–t/RC. Similarly, the current decays as I = I₀ e^–t/RC.

这是一阶微分方程，其解为 Q = Q₀ e^–t/RC，其中Q₀为初始电荷。利用Q = C V，对应的电压衰减为 V = V₀ e^–t/RC。类似地，电流衰减

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