OxfordAQA PH03 Key Formula Derivations | 牛津AQA PH03 核心公式推导

📚 OxfordAQA PH03 Key Formula Derivations | 牛津AQA PH03 核心公式推导

Mastering the derivations behind the core equations is essential for success in the OxfordAQA PH03 Unit 3 exam. This unit covers fields and further mechanics, and the mark scheme frequently rewards a clear, logical step‑by‑step derivation of standard results. Below we work through the most important derivations you are expected to know, from centripetal acceleration to transformer ratios.

掌握核心公式背后的推导过程，对于在 OxfordAQA PH03 第三单元考试中取得好成绩至关重要。本单元涵盖力场与进阶力学，评分方案经常给分于对标准结果进行清晰、有逻辑的逐步推导。下面我们将逐一推演你应当熟练掌握的最重要的公式，从向心加速度到变压器变比。

1. Centripetal Acceleration | 向心加速度推导

Consider an object moving at constant speed v in a circle of radius r. In a short time interval Δt, the object moves through a small angle Δθ. The change in velocity Δv points toward the centre. By similar triangles, |Δv| / v = arc length / r = v Δt / r. Therefore the magnitude of acceleration is a = Δv / Δt = v² / r. Since v = rω, we can also write a = ω² r.

考虑一个以恒定速率v在半径为r的圆周上运动的物体。在一个短时间间隔 Δt 内，物体转过一个小角度 Δθ。速度变化量 Δv 指向圆心。由相似三角形可得，|Δv| / v = 弧长 / r = v Δt / r。因此加速度的大小为 a = Δv / Δt = v² / r。由于 v = rω，这也可以写成 a = ω² r。

a = v² / r

a = ω² r

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2. Gravitational Field Strength | 引力场强度推导

Newton’s law of gravitation states that the force between two point masses is F = G M m / r². Gravitational field strength g is defined as the force per unit mass acting on a small test mass. Hence, near a mass M, g = F / m = (G M m / r²) / m = G M / r². This shows the inverse‑square law for a radial field.

牛顿万有引力定律指出，两个质点之间的引力为 F = G M m / r²。引力场强度 g 定义为作用在单位质量检验质量上的力。因此，在质量 M 附近，g = F / m = (G M m / r²) / m = G M / r²。此式展示了辐射状场的平方反比定律。

g = G M / r²

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3. Orbital Speed for a Satellite | 卫星轨道速率推导

For a satellite in a stable circular orbit, the required centripetal force is provided entirely by the gravitational attraction. Equating centripetal force with gravitational force: m v² / r = G M m / r². Cancelling m and one power of r gives v² = G M / r. Taking the square root yields the orbital speed.

对于处于稳定圆形轨道上的卫星，所需的向心力完全由引力提供。令向心力等于引力：m v² / r = G M m / r²。消去 m 并约掉一个 r，得到 v² = G M / r。开平方即得轨道速率。

v = √(G M / r)

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4. Capacitance of a Parallel‑Plate Capacitor | 平行板电容器的电容推导

Start with the uniform electric field between two plates: E = V / d. From Gauss’s law, the field due to a surface charge density σ on one plate is E = σ / (2 ε₀), so the total field between two oppositely charged plates is E = σ / ε₀. Charge Q = σ A, so V = E d = (σ / ε₀) d = (Q / ε₀ A) d. By definition, C = Q / V = ε₀ A / d.

从两板间的匀强电场开始：E = V / d。根据高斯定律，单块板上表面电荷密度 σ 产生的场强为 E = σ / (2 ε₀)，因此两块带相反电荷的板之间的总场强为 E = σ / ε₀。电荷量 Q = σ A，于是 V = E d = (σ / ε₀) d = (Q / ε₀ A) d。根据定义，C = Q / V = ε₀ A / d。

C = ε₀ A / d

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5. Exponential Decay for a Discharging Capacitor | 电容器放电指数衰减推导

During discharge, the current I is the rate of decrease of charge: I = − dq/dt. Also, I = V / R, and V = q / C. Equating: − dq/dt = q / (R C). Separate variables and integrate: ∫ dq / q = − (1 / R C) ∫ dt. This gives ln q = − t / (R C) + constant. Setting q = Q₀ at t = 0 yields the exponential decay law.

在放电过程中，电流 I 等于电荷的减少速率：I = − dq/dt。同时，I = V / R，且 V = q / C。联立得：− dq/dt = q / (R C)。分离变量并积分：∫ dq / q = − (1 / R C) ∫ dt。由此得到 ln q = − t / (R C) + 常数。代入 t = 0 时 q = Q₀，即得指数衰减规律。

q = Q₀ e^{− t / R C}

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6. Radius of a Charged Particle in a Magnetic Field | 磁场中带电粒子的轨道半径推导

For a particle of charge q moving at speed v perpendicular to a uniform magnetic field B, the magnetic force acts as a centripetal force: B q v = m v² / r. Rearranging for r gives the radius of the circular path.

对于电荷量为 q、以速度 v 垂直于匀强磁场 B 运动的粒子，磁力充当向心力：B q v = m v² / r。重新整理即得圆形路径的半径。

r = m v / (B q)

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7. EMF Induced in a Moving Conductor | 运动导体中的感应电动势推导

Consider a straight conductor of length L moving at speed v perpendicular to a magnetic field B. The flux cut per second is B × area swept out per second = B L v. By Faraday’s law, the magnitude of the induced emf is equal to the rate of change of flux linkage, which for a single conductor is ε = B L v.

考虑一根长度为 L 的直导体，以速度 v 垂直于磁场 B 运动。每秒切割的磁通量为 B × 每秒扫过的面积 = B L v。根据法拉第定律，感应电动势的大小等于磁链的变化率，对于单根导体，有 ε = B L v。

ε = B L v

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8. The Ideal Transformer Equation | 理想变压器公式推导

An ideal transformer assumes 100 % efficiency. The rate of change of flux is the same in both primary and secondary coils. From Faraday’s law, the induced emf per turn is equal for each coil. Thus Vₚ / Nₚ = Vₛ / Nₛ, which rearranges to the familiar voltage ratio. Power conservation then gives Iₚ Vₚ = Iₛ Vₛ, leading to the current ratio.

理想变压器假设效率为 100%。初级线圈和次级线圈中磁通量的变化率相同。由法拉第定律，每匝线圈的感应电动势相等。因此 Vₚ / Nₚ = Vₛ / Nₛ，重新整理即得我们熟悉的电压比公式。再由功率守恒 Iₚ Vₚ = Iₛ Vₛ，即可得出电流比关系。

Vₚ / Vₛ = Nₚ / Nₛ

Iₚ / Iₛ = Nₛ / Nₚ

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9. Gravitational Potential | 引力势的推导

Gravitational potential V is defined as the work done per unit mass in bringing a small test mass from infinity to a point. The gravitational force varies with distance, so we integrate F · dr = G M m / r² dr from ∞ to r. The work done per unit mass is − G M / r, giving the standard expression for potential in a radial field.

引力势 V 定义为将单位质量的检验质量从无穷远处移动到某点所做的功。引力随距离变化，因此我们对 F · dr = G M m / r² dr 从 ∞ 到 r 积分。单位质量对应的功为 − G M / r，从而得到辐射状场中势的标准表达式。

V = − G M / r

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10. Time Constant of an RC Circuit | RC电路的时间常数推导

The time constant τ is defined from the exponential decay equation. From q = Q₀ e^{− t / R C}, when t = R C, the exponent becomes −1, so q = Q₀ e⁻¹ ≈ 0.37 Q₀. Thus τ = R C represents the time for the charge to fall to about 37 % of its initial value, and is a key parameter in timing circuits.

时间常数 τ 由指数衰减方程定义而来。由 q = Q₀ e^{− t / R C}，当 t = R C 时，指数变为 −1，因此 q = Q₀ e⁻¹ ≈ 0.37 Q₀。因此 τ = R C 表示电荷衰减至初始值约 37% 所需的时间，是定时电路中的一个关键参数。

τ = R C

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11. Escape Velocity | 逃逸速度的推导

To escape a planet’s gravitational field, an object must have sufficient kinetic energy to overcome the gravitational potential energy at the surface. Equating ½ m v²_esc = G M m / r (ignoring sign), the mass m cancels, giving v²_esc = 2 G M / r. Taking the square root yields the escape velocity.

要脱离行星的引力场，物体必须具有足够的动能以克服其表面的引力势能。令 ½ m v²_esc = G M m / r（忽略符号），质量 m 可消去，得到 v²_esc = 2 G M / r。开平方即得逃逸速度。

v_esc = √(2 G M / r)

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