📚 Formula Derivations in OxfordAQA 9630 PH03 June 2023 | OxfordAQA 9630 PH03 2023年6月公式推导
The June 2023 OxfordAQA Physics Unit 3 (PH03) paper tests students on their ability to derive and apply key formulas in fields, capacitors, electromagnetism, and induction. Mastering these derivations is essential for understanding underlying principles and solving numerical problems effectively. This article walks through the most important derivations that appeared or could appear in such an examination.
2023 年 6 月的 OxfordAQA 物理第三单元（PH03）试卷考查学生推导和应用场、电容、电磁学和感应等关键公式的能力。掌握这些推导对于理解基本原理和有效解决计算题至关重要。本文梳理了该试卷中出现或可能考查的最重要公式推导。
1. Gravitational Field Strength from Newton’s Law | 从万有引力定律推导引力场强度
s law of gravitation for two point masses M and m separated by a distance r: F = G M m / r2. By definition, g = F / m, so substituting gives g = G M / r2. This shows that the field strength decreases with the square of the distance from the centre of mass M.
引力场强度 g 定义为置于场中的小检验质量每单位质量所受的力。从两个点质量 M 和 m 相距 r 的牛顿万有引力定律出发：F = G M m / r2。根据定义，g = F / m，代入得到 g = G M / r2。这表明场强随距离质量 M 中心的距离平方而衰减。
g = G M / r2
2. Gravitational Potential | 引力势的推导
ng a test mass from infinity to that point. For a radial field, the force is F = G M m / r2 placement dr is dW = F dr. Integrating from infinity to r: V = - ∫(infinity to r) (G M / r2) dr = -[ -G M / r ]∞r gative sign indicates that work is done by the field when a mass is brought from infinity.
某点的引力势 V 是把单位检验质量从无穷远移到该点外力所做的功。在径向场中，力为 F = G M m / r2。克服此力移动无穷小位移 dr 所做的功为 dW = F dr。从无穷远积分到 r：V = - ∫(无穷远到 r) (G M / r2) dr = -[ -G M / r ]∞r = -G M / r。负号表示将质量从无穷远移近时场做正功。
V = - G M / r
3. Coulomb’s Law and Electric Field Strength | 库仑定律与电场强度
Coulomb’s law gives the force between two point charges Q and q: F = k Q q / r2, where k = 1 / (4πε0 field strength E is defined as the force per unit positive charge, so E = F / q = k Q / r2. Thus the radial electric field around a point charge follows an inverse square law analogous to gravity.
库仑定律给出两点电荷 Q 和 q 之间的作用力：F = k Q q / r2，其中 k = 1 / (4πε0)。电场强度 E 定义为单位正电荷所受的力，因此 E = F / q = k Q / r2。因而点电荷周围的径向电场遵循与引力类似的平方反比律。
E = k Q / r2 = Q / (4πε0 r2)
4. Electric Potential due to a Point Charge | 点电荷的电势
sitive test charge from infinity to a point. The force on the test charge is F = k Q q / r2. The work done by an external agent is dW = -F dr. Integrating yields V = - ∫∞r (k Q / r2) dr = k Q / r. This potential is positive for a positive source charge Q.
电势 V 是将单位正检验电荷从无穷远移到某点外力所做的功。检验电荷受到的力为 F = k Q q / r2。外力做功 dW = -F dr。积分得 V = - ∫∞r (k Q / r2) dr = k Q / r。当源电荷 Q 为正时，该电势为正。
V = k Q / r = Q / (4πε0 r)
5. Energy Stored in a Capacitor | 电容器储存的能量
ded to add charge dq at a potential v: dW = v dq. Since v = q / C, integration gives W = ∫0Q (q / C) dq = ½ Q2 / C. Using Q = C V, we obtain two equivalent forms: E = ½ C V2 = ½ Q V.
当电容为 C 的电容器被充电至电势差 V 时，储存的能量可通过对微小充电过程做功来推导：在电势 v 时添加电荷 dq 所做的功 dW = v dq。由于 v = q / C，积分得 W = ∫0Q (q / C) dq = ½ Q2 / C。代入 Q = C V 得到两种等价形式：E = ½ C V2 = ½ Q V。
E = ½ C V2 = ½ Q V = ½ Q2 / C
6. Capacitor Discharge Equation | 电容器放电方程
During discharge through a resistor R, Kirchhoff’s voltage law gives VC = R I, where VC hus q / C = - R dq/dt. Rearranging: dq/q = - dt / (R C). Integrating from initial charge Q0 to q: ∫ dq/q = - ∫ dt / (R C), gives ln(q / Q0) = - t / (R C). Exponentiating both sides yields the exponential decay law.
在通过电阻 R 放电时，基尔霍夫电压定律给出 VC = R I，其中 VC = q / C，I = - dq/dt。因此 q / C = - R dq/dt。整理得 dq/q = - dt / (R C)。从初始电荷 Q0 积分到 q：∫ dq/q = - ∫ dt / (R C)，得到 ln(q / Q0) = - t / (R C)。两边取指数即得到指数衰减规律。
q = Q0 e-t / RC
Differentiating gives the current: I = - dq/dt = (Q0 / R C) e-t / RC = I0 e-t / RC, where I0 The time constant τ = R C is the time for the charge to fall to 1/e of its initial value.
对时间求导得到电流：I = - dq/dt = (Q0 / R C) e-t / RC = I0 e-t / RC，其中 I0 为初始电流。时间常数 τ = R C 是电荷衰减到初始值 1/e 所需的时间。
7. Force on a Charged Particle in a Magnetic Field | 磁场对运动电荷的作用力
force given by the Lorentz force law: F = B q v. The direction is given by Fleming’s left‑hand rule. This force provides the centripetal force for circular motion: B q v = m v2 / r.
电荷量为 q 的粒子以速度 v 垂直于均匀磁场 B 运动时，受到洛伦兹力：F = B q v。方向由左手定则确定。该力提供圆周运动所需的向心力：B q v = m v2 / r。
F = B q v
8. Radius of Circular Motion in a Magnetic Field | 磁场中圆周运动半径
From the equality of magnetic force and centripetal force, B q v = m v2 he magnetic flux density and charge. The period T = 2π r / v = 2π m / (B q) is independent of speed, which is the basis of a cyclotron.
由磁力等于向心力 B q v = m v2 / r，可解出圆周轨迹半径：r = m v / (B q)。这表明半径随动量 m v 增大而增大，与磁通量密度和电荷成反比。周期 T = 2π r / v = 2π m / (B q) 与速度无关，这是回旋加速器的基础。
r = m v / (B q)
9. Faraday’s Law of Electromagnetic Induction | 法拉第电磁感应定律
x Φ per turn, flux linkage = N Φ. The law is expressed as ε = - N dΦ / dt. The minus sign reflects Lenz’s law: the induced emf opposes the change in flux.
法拉第定律指出，电路中感应电动势的大小等于磁链的变化率。对于 N 匝线圈，每匝磁通量为 Φ，则磁链 = N Φ。定律表达为 ε = - N dΦ / dt。负号反映了楞次定律：感应电动势阻碍磁通量的变化。
ε = - N (dΦ / dt)
= - N ΔΦ / Δt. The law can be applied to a moving rod, a rotating coil, or a transformer.
若磁通均匀变化，则可简化为 ε = - N ΔΦ / Δt。此定律可用于运动导杆、旋转线圈或变压器。
10. Transformer Voltage Equation | 变压器电压方程
An ideal transformer assumes no flux leakage and no power loss. The primary coil with Np turns produces a changing flux Φ in the core that links the secondary coil with Ns turns. By Faraday’s law, the primary emf is εp = Np dΦ/dt, and the secondary emf is εs = Ns dΦ/dt. Dividing gives the turns ratio equation. For an ideal transformer, the terminal voltages approximate the induced emfs, so Vs / Vp = Ns / Np.
理想变压器假设无磁漏和无功率损耗。原边 Np 匝产生变化磁通 Φ 于铁芯中，该磁通链接着 Ns 匝的副边线圈。由法拉第定律，原边电动势 εp = Np dΦ/dt，副边电动势 εs = Ns dΦ/dt。两式相除得到匝数比方程。对于理想变压器，端电压近似等于感应电动势，因此 Vs / Vp = Ns / Np。
Vs / Vp = Ns / Np
Combined with conservation of energy (Pp = Ps) we also get Ip Vp = Is Vs, leading to Is / Ip = Np / Ns.
结合能量守恒（Pp = Ps），还可得 Ip Vp = Is Vs，从而 Is / Ip = Np / Ns。
11. Summary and Exam Technique | 总结与考试技巧
In the PH03 exam, derivation questions often require a logical sequence of steps: state the fundamental law, define quantities, perform algebraic manipulation, and finally quote the formula with units. Always mention the sign conventions where appropriate (e.g. gravitational potential, Lenz’s law). Use diagrams if needed to support your reasoning, but the written derivation must be self‑contained.
在 PH03 考试中，推导题通常需要合乎逻辑的步骤顺序：陈述基本定律，定义物理量，进行代数运算，最后写出带单位的公式。适当时务必提及符号约定（如引力势、楞次定律）。如需要可配合草图辅助推理，但文字推导本身必须完整自洽。
Published by TutorHao | Physics Revision Series | aleveler.com
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