📚 A-Level Physics Unit 4 Insert Jan 20 Formula Derivations | A-Level 物理 Unit 4 2020年1月插页公式推导
Physics at A2 level demands a deep understanding of how key equations are derived — not just how to apply them. The formula sheet provided in the January 2020 Unit 4 examination (often referred to as the Insert) lists the essential relationships for further mechanics, fields, and oscillations. This article walks through the derivations of these equations, linking them to fundamental principles such as Newton’s laws, conservation of energy, and calculus. By working through these derivations, you will build a more robust knowledge that is essential for high marks in A-Level Physics.
A2阶段的物理不仅要求会运用公式，更需要理解这些关键方程是如何推导出来的。2020年1月Unit 4考试中提供的公式表（通常称为“插页”）列出了进阶力学、场和振动的重要关系式。本文将逐步推导这些公式，将其与牛顿定律、能量守恒及微积分等基本原理联系起来。通过掌握这些推导过程，你将构建更扎实的知识体系，这是在A-Level物理中取得高分的关键。
1. Angular Speed and Linear Speed | 角速度与线速度
An object moving in a circle of radius r sweeps out an angle Δθ (in radians) in time Δt. By definition, angular speed ω = Δθ/Δt. The arc length travelled is s = rΔθ. Dividing both sides by Δt gives v = r ω, where v is the linear speed. This relationship is fundamental for all circular motion problems.
一个物体在半径为r的圆周上运动时，在时间Δt内转过的角度为Δθ（单位为弧度）。根据定义，角速度ω = Δθ/Δt。物体经过的弧长s = rΔθ。两边同时除以Δt即得v = r ω，其中v为线速度。这个关系是所有圆周运动问题的基础。
2. Centripetal Acceleration and Force | 向心加速度与向心力
Consider a particle moving with constant speed v in a circle. In a short time δt, the velocity vector changes direction by a small angle δθ. The magnitude of the change in velocity is approximately v δθ, so the acceleration is a = v δθ/δt = v ω. Substituting ω = v/r gives a = v²/r. Using rω² yields a = r ω². According to Newton’s second law, the centripetal force is F = m a = m v²/r = m r ω².
考虑一个以恒定速率v做圆周运动的质点。在短时间δt内，速度矢量方向改变一个小角度δθ。速度变化量的大小约为v δθ，因此加速度a = v δθ/δt = v ω。代入ω = v/r得到a = v²/r。使用rω²同样可得a = r ω²。根据牛顿第二定律，向心力F = m a = m v²/r = m r ω²。
3. Newton’s Law of Gravitation and Field Strength | 万有引力定律与引力场强度
Newton’s law of gravitation states that the force between two point masses M and m separated by distance r is F = G M m / r². The gravitational field strength g at a point is defined as the force per unit mass, g = F/m. Substituting the force expression yields g = G M / r², which shows that g decreases with the square of the distance from the centre of a spherical mass.
万有引力定律指出，两个相距r的质点M和m之间的引力为F = G M m / r²。引力场强度g定义为每单位质量所受的力，即g = F/m。代入力的表达式即得g = G M / r²，表明引力场强度随到球形质量中心的距离平方而减弱。
4. Gravitational Potential Energy and Potential | 引力势能与引力势
Gravitational potential V at a point is the work done per unit mass in bringing a small test mass from infinity to that point. The force varies with distance, so integration is needed: V = – ∫∞→r (G M / x²) dx = – G M / r. The negative sign indicates that the potential decreases as one approaches the mass. Potential energy of a mass m is then U = m V = – G M m / r.
某点的引力势V是将单位质量从无穷远处移动到该点所做的功。由于引力随距离变化，需要积分：V = – ∫∞→r (G M / x²) dx = – G M / r。负号表示随着靠近质量，势能减小。质量为m的物体的引力势能为U = m V = – G M m / r。
5. Electric Field Strength for a Point Charge | 点电荷的电场强度
Coulomb’s law gives the force between two point charges Q and q as F = k Q q / r², where k = 1/(4π ε₀). Electric field strength E is the force per unit positive charge, E = F/q. Hence, E = k Q / r². The field radiates outwards for a positive source charge, and its magnitude follows an inverse‑square law.
库仑定律给出两个点电荷Q与q之间的力为F = k Q q / r²，其中k = 1/(4π ε₀)。电场强度E是单位正电荷所受的力，E = F/q，因此E = k Q / r²。对于正源电荷，电场向外辐射，其大小遵循平方反比律。
6. Electric Potential due to a Point Charge | 点电荷带来的电势
Similar to gravitational potential, the electric potential V at a distance r from a point charge Q is the work done per unit charge in bringing a test charge from infinity to that point. Integrating the field gives V = k Q / r. Unlike the gravitational case, the sign of V depends on the sign of Q. The potential energy of a charge q placed in this region is W = q V = k Q q / r.
与引力势类似，距离点电荷Q为r处的电势V是将单位正电荷从无穷远移动到该点所做的功。对电场积分可得V = k Q / r。与引力情况不同，V的正负取决于Q的符号。将电荷q放入该区域时，其电势能为W = q V = k Q q / r。
7. Capacitance and Energy Stored in a Capacitor | 电容与电容器储能
Capacitance C is defined as the ratio of charge stored to potential difference, C = Q / V. As a capacitor charges from 0 to Q, the potential difference rises proportionally. The work done in adding a small increment of charge dq is dW = v dq, where v = q / C. Integrating from 0 to Q gives total energy E = ∫₀Q (q/C) dq = ½ Q² / C. Using Q = C V, this can be written as E = ½ C V² = ½ Q V.
电容C定义为储存的电荷与电势差之比，C = Q / V。电容器从0充电至Q时，电势差成正比上升。增加小量电荷dq所做的功为dW = v dq，其中v = q / C。从0到Q积分得到总能量E = ∫₀Q (q/C) dq = ½ Q² / C。利用Q = C V，可改写为E = ½ C V² = ½ Q V。
8. Exponential Discharge of a Capacitor | 电容器的指数放电
When a capacitor discharges through a resistor R, the current is I = – dQ/dt. By Kirchhoff’s voltage law, Q/C = I R. Substituting for I gives Q/C = – R dQ/dt. Rearranging yields the differential equation dQ/dt = – Q/(RC). Separating variables and integrating leads to Q = Q₀ e^(-t/RC), where Q₀ is the initial charge. The product RC is the time constant τ.
当电容器通过电阻R放电时，电流I = – dQ/dt。根据基尔霍夫电压定律，Q/C = I R。代入I得到Q/C = – R dQ/dt。整理后得到微分方程dQ/dt = – Q/(RC)。分离变量并积分可得Q = Q₀ e^(-t/RC)，其中Q₀为初始电荷。乘积RC即为时间常数τ。
9. Simple Harmonic Motion: Acceleration and Displacement | 简谐运动：加速度与位移
Many oscillating systems, such as a mass on a spring or a simple pendulum (for small angles), obey Hooke’s law F = -k x. Newton’s second law gives m a = -k x, so a = – (k/m) x. Defining ω² = k/m, we obtain the characteristic SHM equation a = – ω² x. The solution is sinusoidal: x = A cos(ωt + φ).
许多振动系统，如弹簧上的重物或单摆（在小角度下），都遵循胡克定律F = -k x。由牛顿第二定律得m a = -k x，所以a = – (k/m) x。定义ω² = k/m，即得到简谐运动的特征方程a = – ω² x。其解为正弦函数：x = A cos(ωt + φ)。
10. Velocity in Simple Harmonic Motion | 简谐运动中的速度
Using the displacement equation x = A sin(ωt) (or cosine), velocity is the first derivative: v = dx/dt = A ω cos(ωt). Since cos(ωt) = ±√(1 – sin²(ωt)) = ±√(1 – x²/A²), we obtain v = ± ω √(A² – x²). This expresses velocity as a function of displacement and shows that maximum speed v_max = ω A occurs at the equilibrium position.
利用位移方程x = A sin(ωt)（或余弦形式），速度是其一阶导数：v = dx/dt = A ω cos(ωt)。因为cos(ωt) = ±√(1 – sin²(ωt)) = ±√(1 – x²/A²)，可得到v = ± ω √(A² – x²)。这给出了速度随位移变化的函数，表明最大速率v_max = ω A出现在平衡位置。
11. Energy Transformations in SHM | 简谐运动中的能量转换
In the absence of damping, the total mechanical energy of an SHM system remains constant. Kinetic energy is K = ½ m v² = ½ m ω² (A² – x²). The elastic potential energy for a spring system is U = ½ k x² = ½ m ω² x². Summing them gives total energy E_total = ½ m ω² A², which depends only on the amplitude. This expression is often used to relate maximum kinetic or potential energy.
在没有阻尼的情况下，简谐运动系统的总机械能保持不变。动能为K = ½ m v² = ½ m ω² (A² – x²)。对弹簧系统，弹性势能为U = ½ k x² = ½ m ω² x²。两者相加得总能量E_total = ½ m ω² A²，仅取决于振幅。这个表达式常用于联系最大动能或最大势能。
12. Magnetic Force on a Moving Charge | 运动电荷在磁场中所受的力
Experiments show that the magnetic force on a charge q moving with velocity v in a magnetic field B is given by F = q v × B. For a straight conductor of length L carrying a current I, the force is F = B I L sinθ, where θ is the angle between the field and the current. Deriving this from the microscopic force: the charge passing a point in time t is q = I t, and drift velocity v = L/t. Substituting gives F = B (I t) (L/t) sinθ = B I L sinθ. This relation is vital for motor effect calculations.
实验表明，以速度v在磁场B中运动的电荷q所受磁力为F = q v × B。对于长度为L、通有电流I的直导线，其受力为F = B I L sinθ，其中θ是磁场与电流方向之间的夹角。从微观力推导：在时间t内通过某点的电荷为q = I t，漂移速度v = L/t。代入得F = B (I t) (L/t) sinθ = B I L sinθ。此关系对于电动机效应的计算至关重要。
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