📚 A-Level Physics Unit 4 June 2022: Formula Derivations | A-Level 物理单元4 2022年6月：公式推导
This article walks you through the key derivations that appear – or are directly relevant to – the June 2022 A‑Level Physics Unit 4 exam paper. For each section, the full step‑by‑step reasoning is provided in English first, immediately followed by a matching paragraph in Chinese. The goal is to build deep understanding, not just memorisation.
本文带你逐一推导2022年6月A-Level物理单元4试卷中出现（或与之直接相关）的关键公式。每个小节都先提供完整的英文逐步推导，紧随其后的是对应的中文段落。目标在于建立深刻理解，而非死记硬背。
1. Centripetal Acceleration | 向心加速度
Consider an object moving in a circle of radius r with constant speed v. In a short time Δt, the velocity vector turns through an angle Δθ. The speed v is the arc length divided by time: v = r Δθ / Δt. Since the two velocity vectors have equal magnitude v, the change in velocity Δv points toward the centre, and its magnitude is Δv = v Δθ. Dividing by Δt gives the acceleration a = Δv/Δt = v Δθ/Δt = v × (v/r) = v²/r. Using v = ωr, we also obtain a = ω²r. This acceleration is always directed towards the centre of the circle.
考虑一物体以恒定速率v在半径为r的圆周上运动。在短时间Δt内，速度矢量转过角度Δθ。速率v等于弧长除以时间：v = r Δθ / Δt。由于两个速度矢量大小均为v，速度的变化量Δv指向圆心，其大小为Δv = v Δθ。除以Δt得加速度a = Δv/Δt = v Δθ/Δt = v × (v/r) = v²/r。利用v = ωr，也得到a = ω²r。这个加速度始终指向圆心。
2. Centripetal Force | 向心力
Applying Newton’s second law to an object experiencing centripetal acceleration gives the centripetal force: F = m a = m v²/r = m ω²r. This is not a separate force in nature; it is the net radial force provided by tension, gravity, friction or the normal reaction, depending on the situation. The derivation simply substitutes the expression for centripetal acceleration into ΣF = ma.
对受到向心加速度的物体应用牛顿第二定律即得向心力：F = m a = m v²/r = m ω²r。这并非自然界中一种独立的力；它是根据具体情况由张力、重力、摩擦力或法向反作用力提供的径向合力。推导仅仅是将向心加速度表达式代入ΣF = ma。
3. Momentum and Impulse | 动量与冲量
Newton’s second law in its most general form is F = dp/dt, where p = mv is linear momentum. For a constant net force acting over a time interval Δt, the impulse J = F Δt equals the change in momentum: J = Δp = m(v_f − v_i). This is the impulse–momentum theorem, derived directly by integrating F dt or, for constant force, by rearranging a = Δv/Δt into F = m Δv/Δt and multiplying by Δt.
牛顿第二定律最普遍的形式是F = dp/dt，其中p = mv为线动量。对于在时间间隔Δt内作用的恒定合力，冲量J = F Δt等于动量的变化：J = Δp = m(v_f − v_i)。这就是冲量–动量定理，可直接通过积分F dt导出，或对恒力将a = Δv/Δt整理为F = m Δv/Δt再乘以Δt得到。
4. Electric Field Strength between Parallel Plates | 平行板间的电场强度
For a uniform electric field between two parallel plates separated by distance d and with potential difference V, the work done on a test charge q moving from one plate to the other is W = qV. Also, W = F d = (q E) d. Equating the two expressions gives qV = qEd, hence E = V/d. The direction of the field is from higher to lower potential, perpendicular to the plates.
对于相距d、电势差为V的两平行板间的匀强电场，将试探电荷q从一板移至另一板做功为W = qV。同时W = F d = (q E) d。令两式相等得qV = qEd，因此E = V/d。电场方向从高电势指向低电势，垂直于板面。
5. Capacitance Definition | 电容的定义
Capacitance C is defined as the charge stored per unit potential difference: C = Q / V. This is an empirical formula; no further derivation is needed. However, it is worth noting that for a parallel-plate capacitor, C can also be expressed in terms of geometry: C = ε₀ εᵣ A / d, where A is plate area and d is separation. Combining these we get Q = (ε₀ εᵣ A / d) V.
电容C定义为每单位电势差所储存的电荷量：C = Q / V。这是一个经验公式，无需进一步推导。不过值得注意的是，对于平行板电容器，C也可用几何参数表达：C = ε₀ εᵣ A / d，其中A为板面积，d为间距。结合两者可得Q = (ε₀ εᵣ A / d) V。
6. Energy Stored in a Capacitor | 电容器储存的能量
As a capacitor charges through a resistor, the p.d. v across it rises from 0 to V while a small charge dq is moved onto the plates. The work done by the battery in moving dq is dW = v dq. Using Q = C v, the total energy stored is E = ∫₀ᵝ v dq = ∫₀ᵝ (q/C) dq = ½ Q²/C. Substituting Q = C V gives the three common forms: E = ½ Q V = ½ C V² = ½ Q² / C.
电容器通过电阻充电时，其两端电压v从0升至V，同时微小电荷dq被移至极板。电源移动dq所做的功为dW = v dq。利用Q = C v，总储存能量为E = ∫₀ᵝ v dq = ∫₀ᵝ (q/C) dq = ½ Q²/C。代入Q = C V得到三种常用形式：E = ½ Q V = ½ C V² = ½ Q² / C。
7. Discharging a Capacitor – Time Constant | 电容器放电 – 时间常数
For a capacitor discharging through a resistor R, Kirchhoff’s voltage law gives v_C + v_R = 0, or Q/C + I R = 0. Since I = dQ/dt (but dQ is negative because charge decreases), we write dQ/dt = −Q/(RC). Solving this differential equation yields Q = Q₀ e^(−t/RC). The product RC has units of time and is called the time constant τ. After t = τ, the charge falls to about 37% of its initial value.
电容器通过电阻R放电时，基尔霍夫电压定律给出v_C + v_R = 0，即Q/C + I R = 0。因为I = dQ/dt（由于电荷减少，dQ为负），可写作dQ/dt = −Q/(RC)。解此微分方程得Q = Q₀ e^(−t/RC)。乘积RC具有时间量纲，称为时间常数τ。经过t = τ后，电荷降至初始值的约37%。
8. Magnetic Force on a Moving Charge | 运动电荷在磁场中的受力
Experimentally, the force on a charge q moving with velocity v in a magnetic field B is given by F = q v B sin θ, where θ is the angle between v and B. In vector form: F = q (v × B). For a current-carrying conductor of length L, replacing q v by I L gives F = B I L sin θ. This is derived by considering the conduction electrons drifting with speed v_d: the total charge passing a point in time t is I t, so the magnetic force is F = B (I t) v_d / t = B I (v_d t) / t = B I L.
实验表明，电荷q以速度v在磁场B中运动时所受的力为F = q v B sin θ，其中θ为v与B的夹角。矢量形式为F = q (v × B)。对于长度为L的载流导体，将q v替换为I L即得F = B I L sin θ。推导时考虑以漂移速度v_d运动的传导电子：在时间t内通过某点的总电荷为I t，于是磁场力F = B (I t) v_d / t = B I (v_d t) / t = B I L。
9. Faraday’s Law of Electromagnetic Induction | 法拉第电磁感应定律
Faraday’s law states that the induced e.m.f. ε in a circuit equals the negative rate of change of magnetic flux linkage: ε = − d(NΦ)/dt. If a single loop of area A rotates in a uniform magnetic field B, the flux Φ = B A cos θ, with θ = ω t for constant angular speed. Then ε = − d/dt (B A cos ω t) = B A ω sin ω t. The negative sign embodies Lenz’s law: the induced current opposes the change in flux.
法拉第定律指出，回路中感应电动势ε等于磁链变化率的负值：ε = − d(NΦ)/dt。若面积为A的单匝线圈在匀强磁场B中旋转，磁通量Φ = B A cos θ，对于匀角速度有θ = ω t。则ε = − d/dt (B A cos ω t) = B A ω sin ω t。负号体现了楞次定律：感应电流反抗磁通量的变化。
10. Transformer Equation | 变压器方程
For an ideal transformer, the rate of change of flux in the primary and secondary coils is the same because they share a common core. The induced e.m.f. in each coil is proportional to its number of turns: ε_p = −N_p dΦ/dt, ε_s = −N_s dΦ/dt. Dividing the two gives ε_s / ε_p = N_s / N_p. Assuming negligible power losses, V_p I_p = V_s I_s, hence I_s / I_p = N_p / N_s. This links the ratios of voltages and currents.
对于理想变压器，由于初级和次级线圈共享同一铁芯，磁通量变化率相同。每匝线圈的感应电动势与其匝数成正比：ε_p = −N_p dΦ/dt，ε_s = −N_s dΦ/dt。两式相除得ε_s / ε_p = N_s / N_p。假设功率损耗可忽略，V_p I_p = V_s I_s，故I_s / I_p = N_p / N_s。从而建立起电压比与电流比之间的联系。
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