📚 A-Level Physics: Key Formula Derivations from June 2018 Paper 4 | A-Level 物理：2018年6月卷4核心公式推导
Understanding the derivations of key equations is essential for tackling Paper 4 questions, which often require you to start from fundamental principles and build up to the final result. This article walks through several important derivations that appeared in or are related to the June 2018 Paper 4 exam, explaining each step clearly using standard notation.
理解核心方程的推导对于应对卷4的问题至关重要，这类问题常常要求你从基本原理出发，逐步构建出最终结果。本文回顾了与2018年6月卷4相关的几个重要推导，并使用标准符号对每一步进行了清晰的解释。
1. Derivation of 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 Δt, it moves from point A to point B, subtending an angle Δθ. The velocity vector changes direction from vA to vB, both of magnitude v.
考虑一个物体以恒定速率 v 在半径为 r 的圆周上运动。在很短的时间 Δt 内，它从点 A 运动到点 B，转过的角度为 Δθ。速度矢量从 vA 变为 vB，两者大小均为 v。
The change in velocity Δv can be represented as the base of an isosceles triangle with sides v and included angle Δθ. For small Δθ, the magnitude of Δv is approximately vΔθ. Since Δθ = (arc length)/r = vΔt/r, we have:
速度的变化量 Δv 可以用一个等腰三角形的底边表示，腰长为 v，夹角为 Δθ。当 Δθ 很小时，Δv 的大小近似为 vΔθ。又因为 Δθ = (弧长)/r = vΔt/r，于是得到：
Δv ≈ v * (vΔt/r) = v² Δt / r
The acceleration towards the centre is the rate of change of velocity: a = Δv/Δt, so:
向心加速度是速度的变化率：a = Δv/Δt，因此：
a = v² / r
Using v = rω, we also obtain a = rω². This acceleration is always directed towards the centre of the circle.
利用 v = rω，也可得到 a = rω²。这个加速度始终指向圆心。
2. Kepler’s Third Law from Newton’s Law of Gravitation | 从牛顿引力定律推导开普勒第三定律
For a planet of mass m orbiting a star of mass M in a circular path of radius r, the gravitational force provides the centripetal force:
对于质量为 m 的行星绕质量为 M 的恒星做半径为 r 的圆周运动，引力提供向心力：
GMm / r² = m v² / r
Cancel m and substitute v = 2πr/T, where T is the orbital period:
消去 m，并代入 v = 2πr/T，其中 T 为轨道周期：
GM / r² = (4π² r² / T²) / r ⇒ GM / r² = 4π² r / T²
Rearrange to isolate T²:
整理得到 T²：
T² = (4π² / GM) * r³
Since 4π²/GM is constant for a given star, T² is proportional to r³. This is Kepler’s third law.
对于给定的恒星，4π²/GM 为常数，因此 T² 与 r³ 成正比。这就是开普勒第三定律。
3. Gravitational Potential Energy | 引力势能推导
The gravitational potential energy U of two point masses M and m separated by distance r is defined as the work done by an external agent to bring m from infinity to that point without acceleration. The force on m is GMm/x² towards M, so the work done against gravity is:
两个质点 M 和 m 相距 r 时的引力势能 U 定义为外力将 m 从无穷远处无加速地移到该点所做的功。m 受到的引力为 GMm/x²，方向指向 M，因此克服引力所做的功为：
W = ∫∞r (GMm/x²) dx = GMm [-1/x]∞r = -GMm/r
The negative sign indicates that gravity does positive work as the separation decreases, so the system’s potential energy decreases. Thus:
负号表明随着距离减小，引力做正功，系统的势能减小。因此：
U = -GMm / r
4. Capacitance of a Parallel-Plate Capacitor | 平行板电容器的电容
For two parallel plates of area A separated by distance d, with vacuum between them, we start with Gauss’s law to find the electric field. The field due to one plate with surface charge density σ is σ/(2ε₀). Between the plates, the fields add to give a uniform field:
对于面积为 A、间距为 d、真空介质的平行板，我们从高斯定律出发求电场。一个带表面电荷密度 σ 的极板产生的场为 σ/(2ε₀)。在两板之间，电场叠加形成匀强电场：
E = σ/ε₀ = Q/(ε₀A)
The potential difference V between the plates is V = E d, so:
两极板间的电势差 V = E d，因此：
V = (Q d)/(ε₀A)
Capacitance C = Q/V, hence:
根据电容定义 C = Q/V，得到：
C = ε₀A / d
If a dielectric of relative permittivity εr is inserted, C = εrε₀A/d.
若插入相对介电常数为 εr 的电介质，则 C = εrε₀A/d。
5. Discharge of a Capacitor through a Resistor | 电容器通过电阻放电
When a capacitor of capacitance C charged to initial charge Q₀ discharges through a resistor R, the current I = -dQ/dt (negative because charge decreases). Kirchhoff’s voltage law gives:
当电容为 C、初始电荷为 Q₀ 的电容器通过电阻 R 放电时，电流 I = -dQ/dt（负号表示电荷减少）。根据基尔霍夫电压定律：
Q/C – IR = 0 ⇒ Q/C + R dQ/dt = 0
Separate variables and integrate:
分离变量并积分：
dQ/Q = -dt/(RC) ⇒ ∫Q₀Q dQ/Q = -∫0t dt/(RC)
ln(Q/Q₀) = -t/(RC) ⇒ Q = Q₀ e-t/(RC)
The time constant τ = RC determines how fast the discharge occurs; after t = RC, the charge falls to Q₀/e.
时间常数 τ = RC 决定了放电的快慢；当 t = RC 时，电荷降为 Q₀/e。
6. Energy Stored by a Capacitor | 电容器储存的能量
To charge a capacitor from uncharged to a final charge Q, work must be done against the increasing potential difference. At an instant when the charge is q, the p.d. is v = q/C. The work dW done to add a small charge dq is:
为了将电容器从零电荷充电至最终电荷 Q，必须克服不断增大的电势差做功。在电荷为 q 的瞬间，p.d. 为 v = q/C。增加微小电荷 dq 所做的功 dW 为：
dW = v dq = (q/C) dq
Total work = ∫₀Q (q/C) dq = ½ Q²/C. This work is stored as electric potential energy:
总功 = ∫₀Q (q/C) dq = ½ Q²/C。这部分功以电势能的形式储存：
U = ½ QV = ½ CV² = ½ Q²/C
where V is the final potential difference.
其中 V 为最终的电势差。
7. Ideal Gas Pressure Equation | 理想气体压强方程
Consider N molecules of an ideal gas in a cubic box of side L. A molecule with velocity component ux along the x-axis collides elastically with a wall, reversing its x-component and imparting an impulse 2mux. The time between collisions with the same wall is 2L/ux, so the average force on the wall from one molecule is:
考虑边长为 L 的立方体盒子中 N 个理想气体分子。一个沿 x 轴的速度分量为 ux 的分子与器壁发生弹性碰撞，其 x 分量反向，施加的冲量为 2mux。与同一器壁两次碰撞的时间间隔为 2L/ux，因此一个分子对器壁的平均作用力为：
F1 = (2mux) / (2L/ux) = mux² / L
Summing over all N molecules, the total force F = (m/L) Σ ux². The mean square x-component is
对所有 N 个分子求和，总力 F = (m/L) Σ ux²。x 方向速度分量均方值
F = (m/L) N (⅓
Pressure p = F/A = F/L², and L³ = V, so:
压强 p = F/A = F/L²，且 L³ = V，所以：
p = F/L² = (m N
Introducing density ρ = Nm/V, we obtain:
代入密度 ρ = Nm/V，得到：
p = ⅓ ρ
This links a macroscopic quantity (pressure) to microscopic particle motion.
这一公式将宏观量（压强）与微观粒子运动联系起来。
8. Bohr Model Energy Levels | 玻尔模型能级推导
In the Bohr model of hydrogen, the electron orbits the nucleus in allowed circular orbits where angular momentum is quantized: mvr = nħ, with n = 1,2,3… and ħ = h/2π. The centripetal force is provided by the Coulomb attraction:
在氢原子的玻尔模型中，电子在允许的圆形轨道上运行，其角动量是量子化的：mvr = nħ，其中 n = 1,2,3…，ħ = h/2π。向心力由库仑引力提供：
ke²/r² = mv²/r, where k = 1/(4πε₀)
From mv²/r = ke²/r², we get mv² = ke²/r. The total energy E = kinetic + potential = ½ mv² – ke²/r = -½ ke²/r. Using the quantization condition v = nħ/(mr), we find r:
由 mv²/r = ke²/r² 得 mv² = ke²/r。总能量 E = 动能 + 势能 = ½ mv² – ke²/r = -½ ke²/r。利用量子化条件 v = nħ/(mr)，求出 r：
m (nħ/(mr))² = ke²/r ⇒ n²ħ²/(mr) = ke² ⇒ r = n²ħ²/(mke²) = n² a₀
where a₀ is the Bohr radius. Substituting r into E:
其中 a₀ 是玻尔半径。将 r 代入 E：
En = -½ ke² / r = – (mk²e⁴)/(2ħ²) * 1/n²
Evaluating constants gives En = -13.6 eV / n². This derivation shows how quantized angular momentum leads to discrete energy levels.
代入常数计算可得 En = -13.6 eV / n²。这一推导展示了量子化的角动量如何导致分立的能级。
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