📚 Deriving Key Formulas in IAL Physics Unit 4 (Jan 2021 Examiner Report) | 基于2021年1月考官报告：IAL物理第四单元核心公式推导
The January 2021 examiner report for International A-Level Physics Unit 4 highlighted a recurring weakness: candidates often recalled formulas correctly but could not derive them from fundamental principles. This inability to show logical step‑by‑step derivations cost marks, particularly in “show that” questions and in explaining the physics behind the equations. Understanding derivations deepens conceptual grasp and builds the confidence needed to apply formulas in unfamiliar contexts. The following article revisits the most important derivations mentioned in the examiner report, presenting each one with clear reasoning and paired English‑Chinese explanations.
2021年1月的国际A-Level物理第四单元考官报告揭示了一个反复出现的弱点：考生虽然能正确记忆公式，却无法从基本原理出发进行推导。这种缺乏分步逻辑推导的能力在“证明”类题目和解释公式背后物理原理的题目中造成失分。理解推导过程能加深概念掌握，增强将公式应用于陌生情景的信心。本文基于考官报告，重新梳理最重要的公式推导，每个推导均提供清晰的推理过程和中英文对照讲解。
1. Centripetal Acceleration: a = v² / r and a = ω²r | 向心加速度的推导
The examiner noted that many students failed to relate the change in velocity vector to the centripetal acceleration when an object moves through a small angle Δθ at constant speed. A common mistake was treating the change in speed rather than the change in velocity.
考官指出，许多学生未能将速度矢量的变化与物体以恒定速率转过小角度Δθ时的向心加速度联系起来。常见错误是只考虑速率的变化而忽略了速度矢量方向的改变。
Consider an object moving at constant speed v around a circle of radius r. In a short time Δt, it moves through angle Δθ and its velocity vector turns by Δθ. The change in velocity Δv is directed toward the centre and has magnitude v Δθ for small Δθ.
考虑一个物体以恒定速率 v 绕半径为 r 的圆周运动。在短时间 Δt 内，它转过角度 Δθ，速度矢量方向同样改变 Δθ。速度变化量 Δv 指向圆心，且当 Δθ 很小时其大小为 v Δθ。
Since arc length = r Δθ = v Δt, we have Δθ = v Δt / r. Substituting into Δv = v Δθ gives Δv = v (v Δt / r) = v² Δt / r. Acceleration magnitude is a = Δv / Δt = (v² Δt / r) / Δt.
由于弧长 = r Δθ = v Δt，可得 Δθ = v Δt / r。代入 Δv = v Δθ 得到 Δv = v (v Δt / r) = v² Δt / r。加速度大小 a = Δv / Δt = (v² Δt / r) / Δt。
a = v² / r
Using v = ω r gives the alternative form a = ω² r. The examiner stressed that this derivation must be understood, not just stated.
利用 v = ω r 可得另一形式 a = ω² r。考官强调，考生必须理解此推导过程，而不能仅仅背诵结果。
2. Electric Field Strength of a Point Charge: E = kQ / r² | 点电荷电场强度的推导
Candidates frequently wrote E = F / q without showing how it leads to the Coulombic form. The examiner advised starting from the definition of electric field strength and Coulomb’s law.
考生经常写出 E = F / q，却未展示如何由此导出库仑形式的表达式。考官建议从电场强度定义和库仑定律出发进行推导。
By definition, electric field strength E is the force per unit positive test charge: E = F / q₀. For a point charge Q, the force on a test charge q₀ placed at distance r is given by Coulomb’s law: F = k Q q₀ / r².
根据定义，电场强度 E 是单位正检验电荷所受的力：E = F / q₀。对于点电荷 Q，放置在距离 r 处的检验电荷 q₀ 所受的力由库仑定律给出：F = k Q q₀ / r²。
Substituting and cancelling q₀ yields the magnitude of the field due to Q alone.
代入并消去 q₀ 即得到仅由 Q 产生的场强大小。
E = kQ / r²
The direction is radially outward from a positive Q. The examiner commented that many lost marks by forgetting the conditions: the formula applies to a point charge and to external points of a spherically symmetric charge distribution.
方向从正电荷 Q 沿径向向外。考官评论说，很多考生因忘记适用条件而失分：此公式适用于点电荷及球对称电荷分布的外部点。
3. Electric Potential of a Point Charge: V = kQ / r | 点电荷电势的推导
Examiners observed confusion between electric potential and potential energy. Deriving V = kQ / r from the work done against the electric field clarifies the relationship.
考官发现考生易混淆电势与电势能。从反抗电场力做功出发推导 V = kQ / r 可澄清两者关系。
Electric potential V at a point is the work done per unit positive test charge in bringing it from infinity to that point. The force on q₀ due to Q is F = k Q q₀ / x² at separation x, acting outward. To bring q₀ from infinity to r with constant speed, an external force must do work against the repulsive electric force (for like charges).
电势 V 定义为将单位正检验电荷从无穷远处移至该点外力所做的功。距离 Q 为 x 时，检验电荷 q₀ 受的力为 F = k Q q₀ / x²，方向沿径向向外。为将 q₀ 从无穷远匀速移至 r，外力必须反抗电场力做功（对于同号电荷）。
Work done W = ∫ from ∞ to r F dx = ∫ (k Q q₀ / x²) dx from ∞ to r = k Q q₀ [ −1/x ] from ∞ to r = k Q q₀ (1/r). Dividing by q₀ gives V.
外力做的功 W = ∫ 从 ∞ 到 r F dx = ∫ (k Q q₀ / x²) dx 从 ∞ 到 r = k Q q₀ [ −1/x ] 从 ∞ 到 r = k Q q₀ (1/r)。除以 q₀ 即得 V。
V = kQ / r
For a negative Q, the potential is negative, and work is done by the field. The examiner recommended practising this integration to secure the concept.
对于负的 Q，电势为负，且电场力做正功。考官建议通过练习该积分来巩固概念。
4. Energy Stored in a Capacitor: W = ½ QV = ½ CV² | 电容器储存能量的推导
A notable error in the exam was assuming the energy was simply QV, ignoring the fact that the p.d. across a capacitor builds up linearly with charge. The correct derivation uses the average p.d.
考试中一个明显的错误是认为能量就是简单的 QV，忽略了电容器两端电势差随电荷量线性增加的事实。正确的推导需要使用平均电势差。
When a small charge dq is transferred onto a capacitor at potential v, the work done is dW = v dq. Since v = q / C, dW = (q / C) dq. The total energy stored is the integral from 0 to Q.
当微小电荷量 dq 在电势差 v 下充入电容器时，做功为 dW = v dq。因 v = q / C，故 dW = (q / C) dq。总储存能量为从 0 到 Q 的积分。
W = ∫₀Q (q / C) dq = 1/(2C) [q²]₀Q = Q² / (2C). Substituting Q = CV yields W = ½ CV² = ½ QV. Graphically, it is the area under the V–Q line.
W = ∫₀Q (q / C) dq = 1/(2C) [q²]₀Q = Q² / (2C)。代入 Q = CV 即得 W = ½ CV² = ½ QV。从图形上看，它就是 V–Q 直线下方的面积。
W = ½ QV = ½ CV²
The examiner pointed out that when a capacitor is charged through a resistor, half the energy supplied by the battery is dissipated; understanding the derivation helps explain this.
考官指出，当电容器通过电阻充电时，电池提供的能量有一半被消耗；理解上述推导有助于解释这一现象。
5. Magnetic Force on a Moving Charge: F = Bqv | 运动电荷在磁场中所受力的推导
Many candidates simply quoted F = Bqv without linking it to the force on a current‑carrying conductor. The examiner stressed the importance of deriving it from F = BIl and the microscopic current model.
许多考生只是直接写出 F = Bqv 而不联系通电导线所受的安培力。考官强调了从 F = BIl 和微观电流模型推导该式的重要性。
Consider a straight conductor of length l carrying current I in a uniform magnetic field B at right angles. The magnetic force is F = BIl. Inside the conductor, the current is due to moving charge carriers: I = n q v A, where n is the number density of carriers, q is the charge on one carrier, v is the drift velocity, and A is cross‑sectional area.
考虑一段长度为 l 的直导线，载有电流 I 并垂直放置于匀强磁场 B 中。磁场力为 F = BIl。在导线内部，电流由运动载流子形成：I = n q v A，其中 n 为载流子数密度，q 为单个载流子的电荷量，v 为漂移速度，A 为横截面积。
Substituting I gives F = B (n q v A) l. The total number of moving charges in the length l is N = n A l. Therefore the force per charge is F / N = B q v l / l, provided the velocity is perpendicular to B.
代入 I 可得 F = B (n q v A) l。在长度 l 内运动电荷的总数为 N = n A l。因此在速度与 B 垂直的条件下，每个电荷所受的力为 F / N = B q v l / l。
F = B q v
When the velocity makes an angle θ with the field, the component v⊥ = v sin θ is used: F = B q v sin θ. The derivation underlines why the force is always perpendicular to both B and v.
当速度与磁场成 θ 角时，使用垂直分量 v⊥ = v sin θ：F = B q v sin θ。该推导揭示了为何该力总是垂直于 B 和 v。
6. Induced EMF in a Rotating Coil: ε = B A N ω sin ωt | 旋转线圈感应电动势的推导
The report noted that students struggled with deriving the alternating EMF equation from Faraday’s law, often forgetting that flux changes as cos ωt and that EMF depends on the rate of change.
报告指出，考生难以从法拉第定律推导交变电动势公式，常常忘记磁通量按 cos ωt 变化且电动势取决于变化率。
For a flat coil of N turns, each of area A, rotating with angular speed ω in a uniform magnetic field B, the flux linkage Φ at time t is Φ = B A N cos ωt, where ωt is the angle between the field and the normal to the coil.
对于匝数为 N、每匝面积为 A 的平面线圈，在匀强磁场 B 中以角速度 ω 旋转，t 时刻的磁链为 Φ = B A N cos ωt，其中 ωt 为磁场方向与线圈法线之间的夹角。
According to Faraday-Lenz law, the induced EMF ε is the negative rate of change of flux linkage: ε = − dΦ / dt. Using the derivative of cos ωt gives d(cos ωt)/dt = − ω sin ωt.
根据法拉第-楞次定律，感应电动势 ε 等于磁链的负变化率：ε = − dΦ / dt。利用 cos ωt 的导数 d(cos ωt)/dt = − ω sin ωt。
Therefore ε = − B A N (− ω sin ωt) = B A N ω sin ωt.
因此 ε = − B A N (− ω sin ωt) = B A N ω sin ωt。
ε = B A N ω sin ωt
The peak EMF is ε₀ = B A N ω. The examiner advised clearly showing the differentiation step rather than jumping to the final answer.
峰值电动势为 ε₀ = B A N ω。考官建议清晰展示求导步骤，而不是直接跳到最终答案。
7. Impulse and Change in Momentum: F Δt = Δp | 冲量与动量变化的推导
Even though this appears in earlier units, the examiner found that Unit 4 candidates often could not derive the impulse–momentum relationship from Newton’s second law in its force‑equals‑rate‑of‑change‑of‑momentum form.
尽管此内容出现在更早的单元中，考官发现第四单元的考生仍常常不能从牛顿第二定律的“力等于动量变化率”形式推导出冲量-动量关系。
Newton’s second law can be written as F = dp / dt, where p = m v is the momentum. If the mass is constant, F = m dv/dt = m a, but the momentum form is more general.
牛顿第二定律可以写作 F = dp / dt，其中 p = m v 为动量。若质量不变，则 F = m dv/dt = m a，但动量形式更具普遍性。
For a constant net force acting over a short time interval Δt, the integral gives F Δt = Δp, where Δp = final momentum − initial momentum. This is the impulse I.
对于在短时间间隔 Δt 内作用的恒定合外力，积分得到 F Δt = Δp，其中 Δp = 末动量 − 初动量。这就是冲量 I。
F Δt = Δp
The examiner stressed that the area under a force–time graph represents the impulse and equals the change in momentum. Candidates must be able to link this to collision problems.
考官强调，力-时间图下的面积代表冲量，且等于动量的变化量。考生必须能将其与碰撞问题联系起来。
8. Radioactive Decay Law: N = N₀ e⁻ˡᵗ | 放射性衰变定律的推导
While the exponential law is often stated, the examiner reported that few students could derive it from the assumption that the activity is proportional to the number of undecayed nuclei, A = λN, combined with A = − dN/dt.
虽然指数定律经常被直接引用，考官报告称很少有学生能从“活度与未衰变原子核数成正比，A = λN”以及 A = − dN/dt 这一前提推导出该定律。
If N undecayed nuclei are present, the activity A = − dN/dt = λN, where λ is the decay constant. Separating variables gives (1/N) dN = − λ dt.
若存在 N 个未衰变原子核，活度 A = − dN/dt = λN，其中 λ 为衰变常量。分离变量得 (1/N) dN = − λ dt。
Integrating both sides: ∫ (1/N) dN = ∫ − λ dt → ln N = − λ t + constant. At t = 0, N = N₀, so constant = ln N₀. Therefore ln N = ln N₀ − λ t → ln (N/N₀) = − λ t.
两边积分：∫ (1/N) dN = ∫ − λ dt → ln N = − λ t + 常数。t = 0 时，N = N₀，故常数 = ln N₀。因此 ln N = ln N₀ − λ t → ln (N/N₀) = − λ t。
Exponentiating gives N = N₀ e⁻ˡᵗ. The half‑life T₁/₂ is obtained by setting N = N₀/2, giving T₁/₂ = ln 2 / λ.
指数化即得 N = N₀ e⁻ˡᵗ。令 N = N₀/2 可得到半衰期 T₁/₂ = ln 2 / λ。
N = N₀ e^(-λ t)
The examiner advised practicing the derivation to avoid confusing the decay constant with half‑life when solving structured problems.
考官建议通过练习推导来避免在求解结构化问题时混淆衰变常量和半衰期。
9. Mass Defect and Binding Energy: E = Δm c² | 质量亏损与结合能的推导
The concept of mass‑energy equivalence caused difficulties when candidates were asked to explain why the mass of a nucleus is less than the sum of its nucleons. A clear derivation from the Einstein equation bridges the gap.
当要求解释原子核的质量为何小于其核子质量之和时，质能等价概念给考生带来了困难。从爱因斯坦方程出发进行清晰推导可以弥补这一缺口。
The total mass of separated nucleons is greater than the mass of the assembled nucleus. The mass defect Δm is the difference: Δm = Z mₚ + (A−Z) mₙ − M_nucleus.
分离核子的总质量大于组合后原子核的质量。质量亏损 Δm 就是这个差值：Δm = Z mₚ + (A−Z) mₙ − M_核。
According to Einstein, mass and energy are equivalent: E = m c². The binding energy is the energy released when the nucleus is formed, equivalent to the mass defect: B.E. = Δm c².
根据爱因斯坦的理论，质量与能量等价：E = m c²。结合能是原子核形成时释放的能量，等价于质量亏损：B.E. = Δm c²。
B.E. = Δm c²
The examiner noted that unit conversions between atomic mass units u and MeV/c² must be shown. Often, 1 u = 931.5 MeV/c² is used, so multiplying Δm in u by 931.5 gives B.E. in MeV.
考官指出必须展示原子质量单位 u 与 MeV/c² 之间的单位换算。通常用 1 u = 931.5 MeV/c²，因此用 u 表示的 Δm 乘以 931.5 即可得到以 MeV 为单位的结合能。
10. Ideal Transformer Equation: εₛ / εₚ = Nₛ / Nₚ | 理想变压器公式的推导
Examiners found that students often wrote the turns‑ratio equation without justifying it using Faraday’s law. The derivation depends on the same changing magnetic flux linking both coils.
考官发现，学生经常写出匝数比公式却没有用法拉第定律定性地证明它。其推导依赖于同一变化磁通同时交链两个线圈这一事实。
For an ideal transformer, the alternating current in the primary coil creates a changing flux Φ in the iron core. This same flux passes through each turn of both coils. By Faraday’s law, the induced EMF per turn is the same for primary and secondary.
对于理想变压器，初级线圈中的交变电流在铁芯中产生变化的磁通量 Φ。同一磁通穿过两个线圈的每一匝。根据法拉第定律，每匝感应电动势对于初级和次级是相同的。
Thus εₚ / Nₚ = εₛ / Nₛ = rate of change of flux. Rearranging gives the relationship.
因此 εₚ / Nₚ = εₛ / Nₛ = 磁通变化率。整理即得所需关系。
εₛ / εₚ = Nₛ / Nₚ
Assuming 100% efficiency, power in equals power out: Iₚ εₚ = Iₛ εₛ, which leads to Iₚ / Iₛ = Nₛ / Nₚ. The examiner emphasized stating the assumptions (no flux leakage, negligible resistance) to gain full marks.
假设效率100%，输入功率等于输出功率：Iₚ εₚ = Iₛ εₛ，从而得到 Iₚ / Iₛ = Nₛ / Nₚ。考官强调必须陈述假设条件（无磁通泄漏、电阻可忽略）才能获得满分。
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