📚 Key Formula Derivations for OxfordAQA PH05 (Unit 5 Physics) | 牛津AQA物理PH05 (单元5) 关键公式推导
This article dissects the most essential formula derivations that appear across the OxfordAQA International A-Level Physics Unit 5 (PH05) mark scheme from June 2023. Understanding these derivations not only prepares you for direct ‘show that’ questions but also deepens your grasp of fields, particles, and nuclear physics. We will go through each derivation step by step, using clear notation and linking every step to the underlying physical principle.
本文深度解析牛津AQA国际A-Level物理单元5（PH05）2023年6月评分标准中最核心的公式推导。掌握这些推导不仅能应对直接的”证明”类考题，更能加深你对场、粒子和核物理的理解。我们将逐步推演每个公式，使用清晰符号并将每一步与基本原理挂钩。
1. Gravitational Field Strength from Newton’s Law | 从万有引力定律推导引力场强度
The gravitational field strength g at a distance r from a point mass M is defined as the gravitational force per unit mass. Starting with Newton’s law of gravitation, F = GMm/r², we divide both sides by the test mass m. This directly yields g = F/m = GM/r². The direction of g is radially towards the centre of the mass M.
引力场强度 g 定义为点质量 M 距离 r 处单位质量所受的引力。从牛顿万有引力定律 F = GMm/r² 出发，两边同时除以检验质量 m，直接得到 g = F/m = GM/r²。g 的方向沿径向指向质量 M 的中心。
g = GM / r²
This derivation is fundamental because it shows how the inverse-square law of gravitation translates into a field that varies as 1/r². A common exam question asks candidates to show that the gravitational field strength near the Earth’s surface is approximately 9.81 N kg⁻¹ by substituting M for Earth’s mass and r for Earth’s radius.
这一推导之所以基础，是因为它展示了引力的平方反比定律如何转化为一个随 1/r² 变化的场。考试中常要求考生代入地球质量和半径，证明地表附近的引力场强度约为 9.81 N kg⁻¹。
2. Gravitational Potential Energy: U = -GMm / r | 引力势能：U = -GMm / r
Gravitational potential energy at a separation r is defined as the work done by an external force in bringing a mass m from infinity to that point without acceleration. The gravitational force is F = GMm/r² directed radially inward. To move m slowly from infinity to r, the external force must act opposite to the displacement, so the work done is W = ∫ (from ∞ to r) −(GMm/r²) dr. Integrating gives W = [GMm/r] from ∞ to r = GMm/r − 0 = GMm/r. Wait, careful: the integral of −1/r² is +1/r. So we get W = GMm(1/r − 1/∞) = GMm/r. However, potential energy is usually taken as negative because the external agent does negative work (the field does positive work). Indeed, the correct definition gives U = −GMm/r because we integrate the force exerted by the field in the direction of displacement. Formal derivation: work done by the gravitational field when moving m from infinity to r is W_field = ∫∞r (−GMm/r²) dr = [GMm/r]∞r = GMm/r. Since change in potential energy ΔU = −W_field, we have U(r) − U(∞) = −GMm/r. Setting U(∞)=0 gives U(r) = −GMm/r. The negative sign indicates that the mass is bound.
引力势能定义为将质量 m 从无穷远缓慢移到距离 r 处外力所做的功。引力为 F = GMm/r²，方向径向向内。要缓慢移动，外力须与位移方向相反，因此做功 W = ∫ (从 ∞ 到 r) −(GMm/r²) dr。积分得到 W = [GMm/r]₍∞₎⁽r⁾ = GMm/r。但通常势能取负值，因为实际上场力做正功，外力做负功。正规推导：引力场将 m 从 ∞ 移至 r 所做功为 W_field = ∫∞r (−GMm/r²) dr = [GMm/r]∞r = GMm/r。由于势能变化 ΔU = −W_field，有 U(r) − U(∞) = −GMm/r。取 U(∞)=0，得 U(r) = −GMm/r。负号表示质量处于束缚态。
U = − GMm / r
This result is crucial for understanding escape velocity and total energy in orbits. In the mark scheme, you may need to show that the total energy of a satellite is −GMm/(2r) using the virial theorem or by combining kinetic and potential energy.
这一结果是理解逃逸速度和轨道总能量的关键。在评分标准中，你可能需要结合动能和势能证明卫星的总能量为 −GMm/(2r)。
3. Kepler’s Third Law: T² ∝ r³ | 开普勒第三定律：T² ∝ r³ 的推导
For a planet or satellite in a circular orbit, the centripetal force is provided by gravity: GMm/r² = mv²/r. The orbital speed v can be expressed as v = 2πr/T, where T is the period. Substituting gives GMm/r² = m(2πr/T)² / r = 4π²mr/T². Cancel m and rearrange: GM/r² = 4π²r/T² → T² = (4π²/GM) r³. This shows T² is proportional to r³, with the constant depending only on the central mass M.
对于沿圆轨道运行的行星或卫星，向心力由引力提供：GMm/r² = mv²/r。轨道速率 v 可表示为 v = 2πr/T，其中 T 为周期。代入得 GMm/r² = m(2πr/T)² / r = 4π²mr/T²。约去 m 并整理：GM/r² = 4π²r/T² → T² = (4π²/GM) r³。这表明 T² 与 r³ 成正比，比例常数仅取决于中心质量 M。
T² = (4π² / GM) r³
This derivation appears frequently in the PH05 paper. If the orbit is elliptical, the same relationship holds with r replaced by the semi-major axis a. You may be asked to derive the constant from given data or predict the period of a satellite.
此推导在 PH05 卷中频繁出现。若轨道为椭圆，只需将 r 替换为半长轴 a，关系仍成立。考题可能要求从给定数据推导常数，或预测卫星周期。
4. Uniform Electric Field: E = V / d | 匀强电场：E = V / d
For a uniform electric field between two parallel plates separated by distance d with potential difference V, the work done on a positive test charge q moving from one plate to the other is W = qV. This work also equals the force qE multiplied by distance d, so qEd = qV. Cancel q to get E = V/d. The direction of the field is from high to low potential.
在相距 d、电势差为 V 的两块平行板之间，匀强电场对正检验电荷 q 从一板移到另一板做功 W = qV。此功也等于力 qE 乘以距离 d，故 qEd = qV。约去 q 得 E = V/d。场方向为由高电势指向低电势。
E = V / d
This relationship is often combined with F = qE and motion equations. In the mark scheme, you might derive E from a graph of V against d or explain why E is uniform. The SI unit V m⁻¹ is equivalent to N C⁻¹.
此关系常与 F = qE 和运动学方程结合。评分标准中可能要求根据 V-d 图像求 E，或解释 E 为何均匀。单位 V m⁻¹ 与 N C⁻¹ 等价。
5. Electric Field of a Point Charge: E = kQ / r² | 点电荷电场：E = kQ / r²
Coulomb’s law gives the force between two point charges: F = kQq/r², where k = 1/(4πε₀). The electric field strength E at a distance r from Q is defined as the force per unit positive charge: E = F/q. Replacing F gives E = kQ/r². This field radiates outward for a positive Q and inward for a negative Q.
库仑定律给出两点电荷间作用力：F = kQq/r²，其中 k = 1/(4πε₀)。距离 Q 为 r 处的电场强度 E 定义为单位正电荷所受的力：E = F/q。代入 F 得 E = kQ/r²。正电荷的场沿径向向外，负电荷向内。
E = (1 / 4πε₀) × Q / r²
This inverse-square law mirrors the gravitational case. It is essential for superposition problems and for deriving the relationship between field and potential (E = −dV/dr). Note that the mark scheme may expect you to compare gravitational and electric fields.
这一平方反比规律与引力场类似，对叠加原理问题以及推导场与电势的关系 (E = −dV/dr) 至关重要。评分标准可能期望你比较引力场与电场。
6. Energy Stored in a Capacitor: E = ½QV = ½CV² | 电容器储能：E = ½QV = ½CV²
When a capacitor of capacitance C is charged to a potential difference V, the charge q and p.d. v are related by q = Cv at any instant. The work done in adding an infinitesimal charge dq is dW = v dq = (q/C) dq. Integrating from 0 to Q gives total work W = ∫0Q (q/C) dq = [q²/(2C)]0Q = Q²/(2C). Since Q = CV, this can be written as ½CV² or ½QV. All three forms are equivalent.
当电容 C 充电至电势差 V 时，任意时刻电荷 q 与电压 v 满足 q = Cv。添加微小电荷 dq 所做功为 dW = v dq = (q/C) dq。从 0 积分到 Q 得总功 W = ∫0Q (q/C) dq = [q²/(2C)]0Q = Q²/(2C)。借助 Q = CV，可改写为 ½CV² 或 ½QV。三种形式等价。
Estored = ½ Q V = ½ C V² = ½ Q² / C
This derivation is often required when explaining how energy is transferred in RC circuits. The mark scheme may ask you to find the energy stored from a Q–V graph (area under line).
解释 RC 电路能量转移时常常需要此推导。评分标准可能要求根据 Q–V 图线下的面积求储存的能量。
7. Charged Particle in a Magnetic Field: r = mv / (Bq) | 带电粒子在磁场中的运动半径
A charged particle of mass m, charge q, moving with speed v perpendicular to a uniform magnetic field B experiences a magnetic force FB = Bqv (sin 90° = 1). This force acts as a centripetal force: Bqv = mv²/r. Solving for radius r gives r = mv/(Bq). The period of circular motion T = 2πr/v = 2πm/(Bq), which is independent of speed – a useful fact for mass spectrometers and cyclotrons.
质量为 m、电荷为 q 的粒子以速度 v 垂直于匀强磁场 B 运动时，受到磁力 FB = Bqv (sin 90° = 1)。该力提供向心力：Bqv = mv²/r。解出半径 r = mv/(Bq)。圆周运动周期 T = 2πr/v = 2πm/(Bq)，与速度无关——这一点对质谱仪和回旋加速器非常有用。
r = m v / (B q)
T = (2π m) / (B q)
In the PH05 exam, you may need to combine this with an energy equation (e.g., the particle was accelerated through a potential difference, qV = ½mv²) to find r in terms of V, B, m, and q.
在 PH05 考试中，你可能需要结合能量方程（例如经电势差加速后 qV = ½mv²）将 r 表达为 V、B、m 和 q 的函数。
8. Faraday’s Law and Motional EMF: ε = Blv | 法拉第定律与动生电动势：ε = Blv
Faraday’s law states that the induced emf ε in a circuit equals the rate of change of magnetic flux linkage: ε = −N dΦ/dt. For a straight conductor of length l moving with velocity v perpendicular to a uniform magnetic field B, the area swept per unit time is ΔA/Δt = lv. The change in flux ΔΦ = B ΔA = Blv Δt. Thus, dΦ/dt = Blv. For a single conductor, N=1, and the magnitude of induced emf is ε = Blv. The direction is given by Lenz’s law or Fleming’s right-hand rule.
法拉第定律指出回路中感应电动势 ε 等于磁通量变化率的负值：ε = −N dΦ/dt。对于长 l 的直导体以速度 v 垂直于匀强磁场 B 运动，单位时间内扫过的面积 ΔA/Δt = lv。磁通量变化 ΔΦ = B ΔA = Blv Δt。因此 dΦ/dt = Blv。对于单根导体，N=1，感应电动势大小为 ε = Blv。方向由楞次定律或弗莱明右手定则确定。
ε = B l v
This derivation is a cornerstone for understanding electric generators. The mark scheme often asks students to derive ε = Blv from first principles using ΔΦ and Δt.
此推导是理解发电机的基础。评分标准常要求学生从 ΔΦ 和 Δt 出发推导 ε = Blv。
9. Radioactive Decay Law: N = N₀ e⁻λt | 放射性衰变定律：N = N₀ e⁻λt
The activity A of a radioactive sample is proportional to the number of undecayed nuclei N: A = −dN/dt = λN, where λ is the decay constant. Rearranging: dN/N = −λ dt. Integrate both sides: ∫ dN/N = −λ ∫ dt → ln N = −λt + C. At t=0, N = N₀, so C = ln N₀. Thus ln(N/N₀) = −λt, which gives N = N₀ e⁻λt. The corresponding activity decays as A = A₀ e⁻λt.
放射性样品的活度 A 与未衰变核数 N 成正比：A = −dN/dt = λN，λ 为衰变常量。重排得 dN/N = −λ dt。两边积分：∫ dN/N = −λ ∫ dt → ln N = −λt + C。t=0 时 N = N₀，故 C = ln N₀。于是 ln(N/N₀) = −λt，得 N = N₀ e⁻λt。相应地，活度衰变规律为 A = A₀ e⁻λt。
N = N₀ e−λ t
This derivation is frequently examined by asking for half-life: set N = N₀/2, so ½ = e⁻λt₁/₂, leading to t₁/₂ = ln2 / λ. You may need to show these steps explicitly.
此推导常通过半衰期考察：令 N = N₀/2，得 ½ = e⁻λt₁/₂，故 t₁/₂ = ln2 / λ。你可能需要明确展示这些步骤。
10. Mass–Energy Equivalence: E = Δm c² | 质能等价：E = Δm c² 在结合能中的应用
Einstein’s mass–energy relation states that a mass Δm is equivalent to energy E = Δm c². In nuclear physics, the mass of a nucleus is always less than the sum of the masses of its individual protons and neutrons. This mass defect Δm is converted into binding energy that holds the nucleus together. The binding energy per nucleon is a crucial indicator of stability. For a nucleus with Z protons and (A−Z) neutrons, Δm = Z mₚ + (A−Z) mₙ − Mnucleus, and binding energy = Δm c². In exam derivations, you will often use atomic masses and account for electrons.
爱因斯坦质能关系指出质量 Δm 等效于能量 E = Δm c²。在核物理中，原子核的实际质量总小于其单个质子与中子质量之和。这一质量亏损 Δm 转化为将核子束缚在一起的结合能。平均结合能（每核子）是稳定性的关键指标。对于含有 Z 个质子和 (A−Z) 个中子的核，Δm = Z mₚ + (A−Z) mₙ − Mnucleus，结合能 = Δm c²。考试推导中常采用原子质量并修正电子影响。
Binding energy = Δm c²
The mark scheme often expects you to calculate Δm in unified atomic mass units u, then convert using 1 u = 931.5 MeV/c² to find energy in MeV. A typical ‘show that’ question involves verifying the energy released in a nuclear reaction via mass difference.
评分标准通常期望你以原子质量单位 u 计算 Δm，再利用 1 u = 931.5 MeV/c² 换算得到以 MeV 为单位的能量。典型的“证明”题会通过质量差验证核反应释放的能量。
These ten derivations form the backbone of the OxfordAQA PH05 formula toolkit. Each one requires not only mathematical manipulation but also a confident command of the underlying physical concepts. Practice writing them out step by step, and you will be fully prepared for the derivation marks on your June 2023-style assessment.
上述十个推导构成了牛津AQA PH05 公式工具包的脊梁。每一组不仅要求数学操作，还需要对背后物理概念的自信掌握。逐步练写它们，你就能为2023年6月风格考试中的推导分值做好充分准备。
Published by TutorHao | Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply