📚 A-Level Physics Unit 4 Jan 19 Formula Derivations | A-Level 物理 Unit 4 2019年1月 公式推导精讲
The Edexcel A-Level Physics Unit 4 (WPH04) January 2019 paper frequently tests candidates' ability to derive key formulas from first principles. Understanding these derivations not only secures marks on "show that" questions but also deepens conceptual grasp of further mechanics, fields, and particle physics. This article revisits the essential derivations that appeared in or are highly relevant to that examination session, providing step-by-step walkthroughs with clear reasoning.
Edexcel A-Level 物理 Unit 4 (WPH04) 2019 年 1 月试卷经常考查学生从基本原理出发推导关键公式的能力。掌握这些推导不仅能稳拿 “证明题” 的分数，还能加深对进阶力学、电场磁场与粒子物理的概念理解。本文回顾了该次考试中出现或高度相关的核心推导，提供分步讲解与清晰思路。
1. Deriving Centripetal Acceleration a = v²/r | 推导向心加速度 a = v²/r
the object rotates through an angle Δθ. The velocity vector, which is always tangent to the circle, also rotates by the same angle Δθ. The change in velocity Δv is directed toward the centre and has magnitude Δv ≈ vΔθ (for small angles). The acceleration is a = Δv/Δt = v (Δθ/Δt) = vω, where ω is the angular speed. Since the linear speed v and angular speed are related by v = ωr, we can write ω = v/r. Substituting this gives a = v(v/r) = v²/r. This is the magnitude of the centripetal acceleration.
为推导向心加速度，考虑一物体以恒定速率 v 沿半径为 r 的圆周运动。在很短的时间间隔 Δt 内，物体转过角度 Δθ。速度矢量始终与圆周相切，同样转过角度 Δθ。速度的变化量 Δv 指向圆心，其大小在极小角度下可写为 Δv ≈ vΔθ。加速度 a = Δv/Δt = v (Δθ/Δt) = vω，其中 ω 为角速度。由于线速度与角速度满足 v = ωr，可得 ω = v/r。代入得到 a = v(v/r) = v²/r。这就是向心加速度的大小。
a = v²/r = ω²r
2. Electric Field Strength from Potential Gradient E = -dV/dr | 由电势梯度求电场强度 E = -dV/dr
s the negative gradient of the potential: E = -dV/dr. Differentiating V with respect to r gives dV/dr = -kQ/r², so E = kQ/r², which is consistent with Coulomb' derivation highlights the fundamental link between potential and field, often examined through graphs of V against r.
在点电荷周围的径向电场中，距离 r 处的电势 V = kQ/r，其中 k = 1/(4πε₀)。电场强度 E 是电势的负梯度：E = -dV/dr。对 V 求导可得 dV/dr = -kQ/r²，因此 E = kQ/r²，与库仑定律一致。对于匀强电场，两板之间的电势差为 V、距离为 d 时，场强为恒定值 E = V/d。这一推导突显了电势与电场的基本联系，常通过 V–r 图线进行考查。
E = -dV/dr ⇒ E = kQ/r² (radial); E = V/d (uniform)
3. Energy Stored in a Capacitor E = ½CV² | 电容器储存的能量 E = ½CV²
is dW = V dq = (q/C) dq. Integrating from 0 to the final charge Q gives the total energy stored: W = ∫₀Q (q/C) dq = (1/C)[q²/2]₀Q = Q²/(2C). Since Q = CV, we can write the energy as E = ½CV² or E = ½QV. This energy is stored in the electric field between the plates and can be recovered during discharge.
电容器充电时，两极板间的电势差按 V = q/C 增大。电源送入微小电荷 dq 所做的功为 dW = V dq = (q/C) dq。从 0 积分到最终电荷 Q，得到储存的总能量：W = ∫₀Q (q/C) dq = (1/C)[q²/2]₀Q = Q²/(2C)。利用 Q = CV，能量可以表示为 E = ½CV² 或 E = ½QV。该能量储存在两极板间的电场中，放电时可以释放出来。
E = ½CV² = ½QV = Q²/(2C)
4. Motional EMF: ε = Blv | 动生电动势 ε = Blv
rpendicularly to a uniform magnetic field of strength B with speed v, the flux cut per unit time is given by Φ = BA, where A is the area swept. The area swept in time Δt is ΔA = l v Δt. The rate of change of flux is dΦ/dt = B dA/dt = B l v. According to Faraday' lv. This derivation assumes the conductor, field, and velocity are mutually perpendicular.
当导体的运动切割磁力线时会产生动生电动势。对于一根长度为 l 的直导体，垂直于磁感应强度为 B 的匀强磁场并以速度 v 运动，单位时间内切割的磁通量为 Φ = BA，其中 A 为扫过的面积。在 Δt 时间内扫过的面积 ΔA = l v Δt。磁通量变化率 dΦ/dt = B dA/dt = B l v。根据法拉第定律，感应电动势的大小 ε = dΦ/dt，因此 ε = Blv。这个推导要求导体、磁场与速度三者互相垂直。
ε = Blv
5. Radius of Circular Path in a Magnetic Field: r = mv/(Bq) | 磁场中回旋半径 r = mv/(Bq)
the velocity. This force acts as a centripetal force, causing circular motion. Equating the magnetic force to the centripetal force required for circular motion gives qvB = mv²/r. Solving for the radius r yields r = mv/(qB). The period of revolution T = 2πr/v = 2πm/(qB), which is independent of the particle's speed. This derivation is the basis for understanding mass spectrometers and cyclotrons.
一个带电粒子垂直于匀强磁场运动时，会受到始终与速度垂直的洛伦兹力 F = qvB。这个力充当向心力，使粒子做匀速圆周运动。令洛伦兹力等于圆周运动所需的向心力：qvB = mv²/r。解出回旋半径 r = mv/(qB)。运动周期 T = 2πr/v = 2πm/(qB)，与粒子速率无关。这一推导是理解质谱仪和回旋加速器工作的基础。
r = mv/(qB) ; T = 2πm/(qB)
6. Deflection of a Charged Particle in a Uniform Electric Field | 匀强电场中带电粒子的偏转
c force gives an acceleration a = eE/m upward. In the horizontal direction, the motion is uniform, so the time spent between the plates is t = L/v₀. The vertical deflection y is obtained from y = ½ a t² = ½ (eE/m) (L/v₀)² = (eE L²)/(2m v₀²). The angle of emergence θ satisfies tan θ = vy/vx ation is frequently combined with subsequent free flight to a screen, forming the basis of an electron deflection tube.
考虑电子以水平初速度 v₀ 射入长度为 L 的平行板匀强电场中。竖直方向的电场力产生加速度 a = eE/m。水平方向为匀速运动，故电子在极板间的时间 t = L/v₀。竖直偏转距离 y = ½ a t² = ½ (eE/m) (L/v₀)² = (eE L²)/(2m v₀²)。出射角度 θ 满足 tan θ = vy/vx = (a t)/v₀ = (eE L)/(m v₀²)。这一推导常常结合后续无场区的匀速直线运动，形成电子偏转管的工作基础。
y = eE L²/(2m v₀²) ; tan θ = eE L/(m v₀²)
7. Relating Kinetic Energy and Momentum | 动能与动量的关系
Kinetic energy and momentum are linked through the object's mass and speed. Starting from the definitions Ek = ½mv² and p = mv, we can express v = p/m. Substituting into the energy formula gives Ek = ½ m (p/m)² = p²/(2m). Conversely, momentum can be written as p = √2mE_k. These relations are particularly useful in elastic collision problems and when comparing the motion of particles with identical kinetic energies but different masses.
动能和动量通过物体的质量和速率相互关联。从定义式 Ek = ½mv² 和 p = mv 出发，可写出 v = p/m。将之代入动能公式得 Ek = ½ m (p/m)² = p²/(2m)。反过来，动量可以表示为 p = √2mE_k。这些关系在处理弹性碰撞问题以及比较具有相同动能但质量不同的粒子运动时格外有用。
Ek = p²/(2m) ; p = √2mE_k
8. Deriving the RC Discharge Equation Q = Q₀ e-t/RC | RC 放电方程 Q = Q₀ e-t/RC 的推导
For a capacitor discharging through a resistor, Kirchhoff's voltage law gives VC = VR ifferential equation is solved by separating variables: dQ/Q = -dt/(RC). Integrating both sides gives ln Q = -t/(RC) + constant. Applying the initial condition Q = Q₀ at t = 0 leads to Q = Q₀ e-t/RC. The time constant τ = RC is the time for the charge to fall to Q₀/e (about 37%). Similar derivations apply to voltage and current during charging and discharging.
对于电容器通过电阻器放电的情况，由基尔霍夫电压定律可得 VC = VR，即 Q
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