📚 Key Derivations for AQA 9630 PH03 International A-Level Physics | AQA 9630 PH03 国际 A-Level 物理关键推导
The AQA 9630 PH03 unit, 'Fields and their consequences', demands a solid grasp of how fundamental equations arise. According to the v4.2 mark scheme, examiners reward structured derivations that clearly show the logical flow from first principles. This article revisits the core mathematical journeys behind gravitational, electric, and magnetic field formulas, capacitor behaviour, and electromagnetic induction — exactly the steps that secure full marks on 'show that' and long-answer questions.
AQA 9630 PH03 单元“场及其后果”要求考生牢牢掌握基本方程式的来龙去脉。根据 v4.2 评分方案，考官青睐结构清晰的推导过程，能展示从基本原理出发的逻辑链条。本文重温引力场、电场、磁场公式、电容器特性以及电磁感应背后的核心数学旅程——正是“求证”题和长答题斩获满分的步骤。
1. Deriving Gravitational Field Strength g from Newton's Law | 由牛顿定律推导引力场强度 g
Start with Newton' ation: the force between two point masses M and m separated by distance r is F = GMm / r².
从牛顿万有引力定律出发：两个质点 M 和 m 相距 r 时，它们之间的引力为 F = GMm / r²。
gravitational field strength at a point is defined as the force per unit mass: g = F / m.
引力场强度定义为作用在单位质量上的力：g = F / m。
r²) / m = GM / r². This is the magnitude of the field due to a point mass M at distance r.
代入力的表达式，得到 g = (GMm / r²) / m = GM / r²。这就是一个质点 M 在距离 r 处产生的引力场大小。
In radial fields, the direction is always towards the centre of mass M. The mark scheme often expects the step of cancelling m.
在径向场中，方向始终指向质点 M 的中心。评分方案通常要求写出约去 m 的步骤。
2. Derivation of Gravitational Potential V = –GM/r | 引力势 V = –GM/r 的推导
Gravitational potential at a point is the work done per unit mass to bring a small test mass from infinity to that point.
引力势指的是将单位质量从无穷远处移到该点所做的功。
/ x², opposing the displacement. Work done by an external agent against the field is dW = –F dx = –(GMm / x²) dx.
在距离 x 处，测试质量 m 受到的引力为 F = GMm / x²，与位移方向相反。外力反抗引力场做的元功为 dW = –F dx = –(GMm / x²) dx。
Integrate from infinity to r: W = ∫∞ʳ (–GMm / x²) dx = [GMm / x]∞ʳ tial at infinity is zero, and work is done by the field, so the potential is negative: V = –GM / r.
从无穷远积分到 r：W = ∫∞ʳ (–GMm / x²) dx = [GMm / x]∞ʳ = GMm / r。除以质量得到势 V = W / m = GM / r。按约定，无穷远处势为零，实际因引力做正功，势取负值：V = –GM / r。
V = –GM / r
Examiners look for the negative sign and the limits of integration in the mark scheme.
评分者会注意负号以及积分上下限。
3. Deriving Orbital Velocity v = √GM/r | 轨道速度 v = √GM/r 的推导
For a satellite in a circular orbit, the centripetal force is provided entirely by gravitational attraction.
对于做圆周运动的卫星，向心力完全由万有引力提供。
he two forces: GMm / r² = mv² / r, where m is the satellite mass and r the orbital radius.
令二力相等：GMm / r² = mv² / r，其中 m 为卫星质量，r 为轨道半径。
Cancel m and one power of r to obtain v² = GM / r. Hence v = √GM / r.
约去 m 并消去一个 r，得 v² = GM / r，因此 v = √GM / r。
The derivation highlights that orbital speed depends only on the central mass and radius, not on the satellite's mass. The mark scheme rewards explicit mention of the cancellation step.
推导凸显轨道速度仅取决于中心天体的质量和轨道半径，与卫星质量无关。评分方案要求明确展示约分步骤。
4. Electric Field Strength of a Point Charge E = Q/(4πε₀r²) | 点电荷的电场强度
Coulomb's law gives the force between two point charges: F = (1/(4πε₀)) Qq / r².
库仑定律给出两点电荷间的作用力：F = (1/(4πε₀)) Qq / r²。
Electric field strength is force per unit positive charge: E = F / q.
电场强度定义为每单位正电荷所受的力：E = F / q。
lling q yields E = (1/(4πε₀)) Q / r². The direction is radially outward from a positive Q.
代入并约去 q，得到 E = (1/(4πε₀)) Q / r²。若是正电荷 Q，方向沿径向向外。
E = Q / (4πε₀ r²)
ined with F = qE, this forms the basis for many PH03 questions on charged particle motion.
该式与 F = qE 结合，是 PH03 中带电粒子运动许多问题的基础。
5. Electric Potential due to a Point Charge V = Q/(4πε₀r) | 点电荷的电势
Analogous to gravitational potential, electric potential is the work done per unit charge to bring a test charge from infinity to a point distance r from Q.
类比引力势，电势是把单位正电荷从无穷远移到距离 Q 为 r 的点所做的功。
ge q at distance x is F = (1/(4πε₀)) Qq / x². External agent works against the field: dW = –F dx = –(1/(4πε₀)) Qq/x² dx.
测试电荷 q 在距离 x 处受力 F = (1/(4πε₀)) Qq / x²。外力反抗电场做功：dW = –F dx = –(1/(4πε₀)) Qq/x² dx。
Integrate from ∞ to r: W = ∫∞ʳ –(1/(4πε₀)) Qq/x² dx = (1/(4πε₀)) Qq / r. Dividing by q gives V = W / q = (1/(4πε₀)) Q / r.
从∞积分到 r：W = ∫∞ʳ –(1/(4πε₀)) Qq/x² dx = (1/(4πε₀)) Qq / r。除以 q 得到 V = W / q = (1/(4πε₀)) Q / r。
V = Q / (4πε₀ r)
The mark scheme often awards marks for correctly stating the integral and handling the sign.
评分方案常对正确写出积分式并处理符号给予分数。
6. Capacitance and Energy Stored E = ½QV = ½CV² | 电容储能公式推导
e C = Q / V. The potential difference V across the plates builds up as charge accumulates.
对电容器，电容 C = Q / V。极板间电势差 V 随电荷积累而增加。
one plate to the other requires work dW = V dq = (q/C) dq, because V = q/C at any instant.
移动微小电荷 dq 从一个极板到另一个极板需要做功 dW = V dq = (q/C) dq，因为在任意时刻 V = q/C。
Total work done to charge from 0 to Q is W = ∫₀Q (q/C) dq = [q²/(2C)]₀Q = Q²/(2C).
从 0 充电到 Q 的总功为 W = ∫₀Q (q/C) dq = [q²/(2C)]₀Q = Q²/(2C)。
: E = ½ QV = ½ C V². This derivation relies on the direct proportionality between V and q.
利用 Q = CV，该功以能量形式储存：E = ½ QV = ½ C V²。推导依赖 V 与 q 的正比关系。
E = ½QV = ½CV² = Q²/(2C)
7. Derivation of Capacitor Discharge Equation Q = Q₀ e⁻ᵗ/ᴿᶜ | 电容放电方程推导
Consider an RC series circuit with no cell. Kirchhoff's voltage law gives VR + VC = 0, so IR + Q/C = 0.
考虑无电源的 RC 串联电路。基尔霍夫电压定律给出 VR + VC = 0，即 IR + Q/C = 0。
Current is the rate of discharge, I = –dQ/dt (negative as Q decreases). Substituting: –R dQ/dt + Q/C = 0 → dQ/dt = –Q/(RC).
电流是放电速率，I = –dQ/dt（因为 Q 减小而加负号）。代入得：–R dQ/dt + Q/C = 0 → dQ/dt = –Q/(RC)。
Separate variables: 1/Q dQ = –1/(RC) dt. Integrate both sides: ln Q = –t/(RC) + constant.
分离变量：1/Q dQ = –1/(RC) dt。两侧积分：ln Q = –t/(RC) + 常数。
At t = 0, Q = Q₀, so ln Q₀ = constant. Therefore ln Q – ln Q₀ = –t/(RC) → Q = Q₀ e⁻ᵗ/ᴿᶜ.
当 t = 0 时 Q = Q₀，所以 ln Q₀ = 常数。因此 ln Q – ln Q₀ = –t/(RC) → Q = Q₀ e⁻ᵗ/ᴿᶜ。
Q = Q₀ e⁻ᵗ/ᴿᶜ
V and I decay similarly: V = V₀ e⁻ᵗ/ᴿᶜ, I = I₀ e⁻ᵗ/ᴿᶜ. The mark scheme expects the correct sign for I and the separation of variables.
电压和电流也按类似规律衰减：V = V₀ e⁻ᵗ/ᴿᶜ, I = I₀ e⁻ᵗ/ᴿᶜ。评分方案要求 I 的符号正确以及变量分离步骤。
8. Time Constant τ = RC and Half-life | 时间常数 τ = RC 与半衰期
its of time (Ω × F = s) and is called the time constant τ. At t = τ, Q = Q₀ e⁻¹ ≈ 0.37 Q₀.
乘积 RC 具有时间单位（Ω × F = s），称为时间常数 τ。在 t = τ 时，Q = Q₀ e⁻¹ ≈ 0.37 Q₀。
The half-life t₁/₂ is found by setting Q = Q₀/2: Q₀/2 = Q₀ e⁻ᵗ/ᴿᶜ → 1/2 = e⁻ᵗ/ᴿᶜ → ln(1/2) = –t/RC → t₁/₂ = RC ln 2.
半衰期通过令 Q = Q₀/2 求得：Q₀/2 = Q₀ e⁻ᵗ/ᴿᶜ → 1/2 = e⁻ᵗ/ᴿᶜ → ln(1/2) = –t/RC → t₁/₂ = RC ln 2。
τ = RC, t₁/₂ = RC ln 2
These relationships often appear in PH03 questions linking time constants to graphical analysis of exponential decay.
上述关系常在 PH03 试题中结合指数衰减图像分析出现。
9. Faraday's Law and Induced emf ε = –N dΦ/dt | 法拉第定律与感应电动势推导
Faraday discovered that an emf is induced when the magnetic flux Φ through a coil changes. The magnitude is proportional to the rate of change of flux.
法拉第发现当穿过线圈的磁通量 Φ 发生变化时，会感应出电动势。其大小与磁通量变化率成正比。
For a coil of N turns, flux linkage is NΦ. The induced emf is ε = – d(NΦ)/dt = –N dΦ/dt. The negative sign reflects Lenz's law.
对于 N 匝线圈，磁链为 NΦ。感应电动势 ε = – d(NΦ)/dt = –N dΦ/dt。负号体现了楞次定律。
This empirical law can be derived from energy conservation: moving a magnet in a coil requires work against the induced magnetic forces, converting mechanical energy to electrical energy.
这条经验定律可从能量守恒导出：移动磁铁进出线圈需抵抗感应磁场力做功，机械能转化为电能。
A cos θ, so a change in B, A, or θ induces an emf. The mark scheme rewards stating both the law and the linking of minus sign to Lenz's law.
在匀强磁场中，Φ = BA cos θ，因此 B、A 或 θ 的变化都会感应出电动势。评分方案要求既陈述定律，也把负号与楞次定律关联。
10. emf Induced in a Rotating Coil ε = BANω sin(ωt) | 旋转线圈中的感应电动势
Consider a rectangular coil of area A, N turns, rotating with angular speed ω in a uniform magnetic field B.
考虑面积为 A、N 匝的矩形线圈在匀强磁场 B 中以角速度 ω 旋转。
e field and the normal to the coil is θ = ωt. The flux through the coil is Φ = BA cos(ωt).
在 t 时刻，磁场方向与线圈法线夹角 θ = ωt。穿过线圈的磁通量 Φ = BA cos(ωt)。
Apply Faraday's law: ε = –N dΦ/dt = –N × BA × (–ω sin(ωt)) = BANω sin(ωt). Peak emf ε₀ = BANω.
应用法拉第定律：ε = –N dΦ/dt = –N × BA × (–ω sin(ωt)) = BANω sin(ωt)。峰值电动势 ε₀ = BANω。
ε = ε₀ sin(ωt), ε₀ = BANω
Examiners expect the chain rule differentiation of cos(ωt) and recognition that the induced emf varies sinusoidally.
考官期望对 cos(ωt) 使用链式法则求导，并识别感应电动势呈正弦变化。
11. Force on a Moving Charge F = BQv (Derivation from F = BIl) | 运动电荷受力 F = BQv 的推导
r experiences a force F = BIl sin θ in a magnetic field. Current is due to moving charges.
载流导体在磁场中受力 F = BIl sin θ。电流源自电荷的运动。
q, current I = nAvq. Substituting into F = BIl gives F = B (nAvq) l = BQ_total v, where Qtotal is the total moving charge.
对于长度为 l 的导体，漂移速度为 v，每单位体积有 n 个电荷 q，电流 I = nAvq。代入 F = BIl 得 F = B (nAvq) l = BQ_total v，其中 Qtotal 为总移动电荷。
ply F = BQv. This derivation clarifies the link between macroscopic and microscopic force.
对于以速度 v 垂直于 B 运动的单个电荷 Q，磁场力即为 F = BQv。该推导理清了宏观力与微观力间的联系。
F = BQv (with v ⟂ B)
12. Radius of Curvature for a Charged Particle in a Magnetic Field | 带电粒子在磁场中的回转半径
ndicular to a uniform B experiences a force F = BQv, which provides the centripetal force.
电荷 Q 以速度 v 垂直射入匀强磁场 B 中时，受到力 F = BQv，此力充当向心力。
Set BQv = mv² / r, where m is the mass and r the radius of the circular path.
令 BQv = mv² / r，其中 m 为质量，r 为圆周轨道半径。
rrange: r = mv / (BQ). This equation is fundamental for mass spectrometers and cyclotrons.
约去一个 v 并整理得：r = mv / (BQ)。此公式是质谱仪和回旋加速器的基本原理。
r = mv / (BQ)
The derivation must show the force balance and justify that the velocity stays perpendicular to B for uniform circular motion.
推导过程需展示力平衡，并说明因速度始终垂直于 B 而维持匀速圆周运动。
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