📚 A-Level Physics June 2018 Examiner Report: Deriving the Capacitor Discharge Equation | 2018年6月A-Level物理考官报告：推导电容放电方程
In the June 2018 A-Level Physics examination, many candidates struggled with the derivation of the exponential decay equation for a discharging capacitor. The examiner’s report highlighted that students often memorised the final formula V = V₀ e–t/RC without understanding the underlying calculus. This article revisits the derivation step by step, addresses common mistakes, and provides clear guidance to help you master one of the most important mathematical processes in the electricity module.
在2018年6月的A-Level物理考试中，许多考生在推导电容放电的指数衰减方程时遇到了困难。考官报告指出，学生往往只记住了最终公式 V = V₀ e–t/RC，却不理解其背后的微积分过程。本文将逐步重新推导这一公式，指出常见错误，并为你提供清晰的指导，帮助你掌握电学模块中最重要的数学推导之一。
1. Background: Why Derivation Matters | 背景：为什么推导很重要
Examiners frequently set questions that require you to show how V = V₀ e–t/RC arises from basic principles. Simply quoting the formula earns no marks in a ‘show that’ question. The derivation tests your ability to connect circuit theory with a simple first-order differential equation, a skill that appears across many A-Level topics.
考官经常设置需要你展示 V = V₀ e–t/RC 如何由基本原理推导而来的问题。在 ‘show that’ 类问题中，直接引用公式是不得分的。这一推导考察你将电路理论与简单一阶微分方程联系起来的能力，这是一种跨越多项A-Level主题的核心技能。
2. Starting Point: Relationship Between Charge, Voltage and Current | 起点：电荷、电压与电流的关系
Consider a capacitor of capacitance C connected in series with a resistor R. At time t = 0, the capacitor carries an initial charge Q₀ and the switch is closed to allow discharge. By definition, capacitance C = Q/V, so at any instant the p.d. across the capacitor is V = Q/C. The current I in the circuit is the rate of flow of charge leaving the capacitor: I = –dQ/dt (negative because Q decreases).
考虑一个电容为 C 的电容器与电阻 R 串联。在 t = 0 时，电容器带有初始电荷 Q₀，合上开关开始放电。根据定义，电容 C = Q/V，因此在任意时刻电容器两端的电压为 V = Q/C。电路中的电流 I 是电荷离开电容器的速率：I = –dQ/dt（负号表示 Q 在减少）。
3. Applying Kirchhoff’s Voltage Law | 应用基尔霍夫电压定律
During discharge there is no external e.m.f., so the sum of p.d.s around the loop is zero: V_R + V_C = 0. Using Ohm’s law V_R = IR and V_C = Q/C, we get IR + Q/C = 0. Substituting I = –dQ/dt gives –R(dQ/dt) + Q/C = 0, which rearranges to dQ/dt = –Q/(RC). This is the differential equation governing the discharge.
放电过程中没有外部电动势，因此回路中的电压之和为零：V_R + V_C = 0。利用欧姆定律 V_R = IR 和 V_C = Q/C，得到 IR + Q/C = 0。代入 I = –dQ/dt 得到 –R(dQ/dt) + Q/C = 0，整理后即得 dQ/dt = –Q/(RC)。这就是控制放电过程的微分方程。
4. Separating the Variables | 分离变量
To solve dQ/dt = –Q/(RC), we collect all terms involving Q on one side and t on the other: (1/Q) dQ = –1/(RC) dt. This step is crucial; many candidates incorrectly leave a negative sign off or forget to invert Q. Take care to keep the minus sign on the right-hand side.
为了求解 dQ/dt = –Q/(RC)，我们将所有包含 Q 的项移到一边，包含 t 的项移到另一边：(1/Q) dQ = –1/(RC) dt。这一步至关重要；许多考生会错误地遗漏负号或忘记将 Q 取倒数。务必把负号保留在右侧。
5. Integrating Both Sides | 两边积分
Integrate the left side with respect to Q and the right side with respect to t: ∫ (1/Q) dQ = ∫ –1/(RC) dt. This yields ln Q = –t/(RC) + K, where K is the constant of integration. A common mistake is to forget the integration constant, which then prevents the correct application of initial conditions.
左边对 Q 积分，右边对 t 积分：∫ (1/Q) dQ = ∫ –1/(RC) dt。得到 ln Q = –t/(RC) + K，其中 K 是积分常数。一个常见错误是忘记积分常数，这会使随后无法正确应用初始条件。
6. Determining the Constant of Integration | 确定积分常数
At t = 0, the charge on the capacitor is Q = Q₀. Substituting these values into ln Q = –t/(RC) + K gives ln Q₀ = 0 + K, so K = ln Q₀. Therefore the equation becomes ln Q = –t/(RC) + ln Q₀. Using the subtraction rule for logs, ln Q – ln Q₀ = ln(Q/Q₀) = –t/(RC).
当 t = 0 时，电容器上的电荷为 Q = Q₀。将这些值代入 ln Q = –t/(RC) + K，得到 ln Q₀ = 0 + K，因此 K = ln Q₀。于是方程变为 ln Q = –t/(RC) + ln Q₀。利用对数运算法则，ln Q – ln Q₀ = ln(Q/Q₀) = –t/(RC)。
7. Exponentiating to Obtain Q(t) | 取指数得到 Q(t)
To remove the natural logarithm, we exponentiate both sides: Q/Q₀ = e–t/RC, hence Q = Q₀ e–t/RC. Since V = Q/C at any time, multiplying both sides by 1/C gives V = V₀ e–t/RC, where V₀ = Q₀/C. This is the standard exponential decay of voltage across a discharging capacitor.
为消去自然对数，两边同时取指数：Q/Q₀ = e–t/RC，因此 Q = Q₀ e–t/RC。由于任意时刻 V = Q/C，两边同乘以 1/C 得到 V = V₀ e–t/RC，其中 V₀ = Q₀/C。这就是放电电容器两端电压的标准指数衰减公式。
8. Alternative Derivation Using Current Directly | 直接利用电流的替代推导
Some syllabi prefer to start from the relationship for current through a capacitor: I = C (dV/dt). During discharge, the same current flows through the resistor, so V = IR gives I = V/R. Equating the two expressions for current yields C (dV/dt) = –V/R (negative because voltage is decreasing). Rearranging gives dV/dt = –V/(RC), leading by the same separation and integration steps to V = V₀ e–t/RC.
一些教学大纲倾向于从电容器的电流关系 I = C (dV/dt) 开始。放电时，同样的电流流过电阻，因此 V = IR 给出 I = V/R。让两个电流表达式相等得到 C (dV/dt) = –V/R（负号是因为电压在下降）。整理后得到 dV/dt = –V/(RC)，经相同的分离变量和积分步骤最终导出 V = V₀ e–t/RC。
9. Common Mistakes Highlighted in the Examiner Report | 考官报告中强调的常见错误
The June 2018 examiner report noted several recurring errors: (i) omitting the negative sign when relating I to dQ/dt; (ii) writing I = +dQ/dt and then adding a negative sign later without justification; (iii) forgetting to include the integration constant and thus obtaining ln Q = –t/RC directly; (iv) incorrectly manipulating logarithms when moving from ln Q to Q; and (v) failing to state clearly the transition from Q to V using V = Q/C.
2018年6月的考官报告指出了几类反复出现的错误：(i) 在将 I 与 dQ/dt 关联时遗漏负号；(ii) 写为 I = +dQ/dt，随后不加解释地添加负号；(iii) 忘记包含积分常数，从而直接得到 ln Q = –t/RC；(iv) 从 ln Q 转化为 Q 时对数运算错误；(v) 未能清晰地使用 V = Q/C 完成从 Q 到 V 的过渡。
10. The Meaning of the Time Constant RC | 时间常数 RC 的含义
From V = V₀ e–t/RC, when t = RC, the voltage drops to V₀ e–1 ≈ 0.37 V₀. Thus the time constant τ = RC represents the time taken for the voltage to fall to about 37% of its initial value. In a graphical analysis, τ can also be found from the intercept of a ln V vs t graph, where the gradient is –1/RC. Examiners expect you to be able to interpret these features both algebraically and graphically.
由 V = V₀ e–t/RC 可知，当 t = RC 时，电压降至 V₀ e–1 ≈ 0.37 V₀。因此时间常数 τ = RC 表示电压下降到初始值约37%所需的时间。在图像分析中，τ 也可通过 ln V 对 t 直线的截距求得，其斜率为 –1/RC。考官要求你能够从代数和图像两个方面解释这些特征。
11. Practical Relevance and Experimental Verification | 实际意义与实验验证
The exponential discharge can be investigated by charging a capacitor through a known resistor and recording the p.d. at regular intervals using a voltmeter or data logger. Plotting ln V against t should produce a straight line with negative gradient. Deviations from linearity often indicate a faulty capacitor or incorrect resistance value. Examiners may ask you to describe such an experiment and link the gradient to the time constant.
指数放电过程可以通过将一个电容器通过已知电阻放电，并用电压表或数据记录器定期记录电压来进行研究。绘制 ln V 对 t 的图像应得到一条斜率为负的直线。偏离线性的情况通常表明电容器故障或电阻值不正确。考官可能会要求你描述这一实验，并将斜率与时间常数联系起来。
12. Summary and Exam Tips | 总结与考试技巧
To gain full marks on a derivation question, always: show the sign convention clearly, write the differential equation before integrating, include the constant of integration, use initial conditions to find it, apply correct log rules, and convert to the required variable (Q or V). Practise writing the derivation in a logical sequence without skipping steps. The examiners value clear, structured mathematical reasoning far more than a half-remembered formula.
要在推导题中获得满分，应始终做到：清晰地展示符号规定，在积分前写出微分方程，包含积分常数，利用初始条件求出该常数，正确使用对数运算法则，并将结果转换为所要求的变量（Q 或 V）。练习按逻辑顺序写出推导过程，不要跳步。考官更看重清晰、结构化的数学推理，而非一知半解地背出的公式。
Published by TutorHao | Physics Revision Series | aleveler.com
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