📚 PH03 International A-Level Physics: Key Formula Derivations for Electric Fields, Capacitors and Magnetic Fields | PH03国际A-Level物理：电场、电容器与磁场核心公式推导
In the International A-Level Physics PH03 unit, commonly covering 'Fields and their consequences', students are expected not only to apply key formulas but also to understand how they are derived from fundamental principles. This article walks you through the step-by-step derivations of essential equations related to electric fields, capacitance, RC circuits, and the motion of charged particles in magnetic fields. Mastering these derivations deepens your conceptual understanding and equips you with the confidence to tackle both calculation and explanation questions in the 2023 examination series.
在国际A-Level物理PH03单元（通常涵盖“场及其影响”）中，学生不仅需要应用核心公式，更要理解这些公式如何从基本原理推导而来。本文将逐步展示电场、电容、RC电路以及带电粒子在磁场中运动等重要公式的推导过程。掌握这些推导能够深化你的概念理解，让你在2023年考试中从容应对计算和解释类题目。
1. Coulomb's Law and the Electric Constant | 库仑定律与电场常量
Coulomb's law states that the magnitude of the electrostatic force F between two point charges Q and q separated by a distance r is directly proportional to the product of the charges and inversely proportional to the square of the distance. In SI units this is expressed with the electric constant ε0 (permittivity of free space):
库仑定律指出，两个点电荷 Q 和 q 相距 r 时，静电作用力的大小与电荷量的乘积成正比，与距离的平方成反比。在国际单位制中，用电场常量 ε0（真空介电常数）表示为：
F = (1 / (4πε0)) · (Qq / r2)
The constant 1/(4πε0) is often written as k, where k ≈ 8.99 × 109 N m2 C−2. The inverse‑square nature of the law arises from the geometry of field lines radiating uniformly in three dimensions.
常量 1/(4πε0) 常写作 k，其值约为 8.99 × 109 N m2 C−2。平方反比规律源于电场线在三维空间中均匀辐射的几何特性。
2. Electric Field Strength of a Point Charge | 点电荷的电场强度
By definition, the electric field strength E at a point is the force per unit positive charge placed at that point: E = F/q. Substituting F from Coulomb's law for a source charge Q and a test charge q gives the field strength at a distance r from Q:
根据定义，某点的电场强度 E 等于该点单位正电荷所受的力：E = F/q。将库仑定律中源电荷 Q 和检验电荷 q 受的力 F 代入，即得距离 Q 为 r 处的电场强度：
E = (1 / (4πε0)) · (Q / r2)
This expression reveals that the field is independent of the test charge and points radially outward for a positive Q. Its magnitude falls off as 1/r2, which is a hallmark of a radial field.
该表达式表明电场与检验电荷无关，且对正 Q 沿径向向外。场的大小按 1/r2 减弱，这是径向场的标志性特征。
3. Electric Potential due to a Point Charge | 点电荷的电势
Electric potential V all test charge from infinity to that point. The work done against the electric field is ∫F·dr. For a point charge Q, integrating the force from ∞ to r gives:
电势 V 定义为将单位正电荷从无穷远处移到该点所做的功。克服电场力所做的功为 ∫F·dr。对点电荷 Q，从无穷远积分到 r 可得：
V = W/q = (1 / (4πε0)) · (Q / r)
Note that V is a scalar quantity and decreases as 1/r. The zero of potential is taken at infinity, which makes the potential around a positive charge positive everywhere in free space.
注意 V 是标量，按 1/r 规律减小。电势零点取在无穷远处，这使得正电荷周围的电势在自由空间中处处为正。
4. Uniform Electric Field and Potential Difference | 匀强电场与电势差
In a uniform electric field, such as that between two parallel plates, the field lines are straight, parallel, and equally spaced. Moving a charge q through a potential difference ΔV requires work W = qΔV. This work is also equal to the force F = qE multiplied by the displacement d parallel to the field: W = qEd. Equating the two expressions yields:
在匀强电场中（例如两平行板之间），电场线是笔直、平行且等间距的。将电荷 q 移动通过电势差 ΔV 需要做功 W = qΔV。该功也等于电场力 F = qE 乘以沿场方向的位移 d：W = qEd。令两式相等得：
E = ΔV / d
This relationship is exceptionally useful for linking the field strength between parallel plates to the applied voltage and plate separation.
这一关系在联系平行板间的电场强度与外加电压及板间距时极为有用。
5. Capacitance of a Parallel-Plate Capacitor | 平行板电容器的电容
Capacitance C is defined as the ratio of charge stored to the potential difference: C = Q/V. For an ideal parallel-plate capacitor of plate area A and separation d, a uniform electric field exists between the plates. Using Gauss's law, the field due to the surface charge density σ = Q/A is E = σ/ε0 = Q/(Aε0). The potential difference is V = Ed = Qd/(Aε0). Hence the capacitance is:
电容 C 定义为存储的电荷量与电势差之比：C = Q/V。对于极板面积为 A、间距为 d 的理想平行板电容器，板间存在匀强电场。利用高斯定律，由面电荷密度 σ = Q/A 产生的电场为 E = σ/ε0 = Q/(Aε0)。电势差 V = Ed = Qd/(Aε0)。因此电容为：
C = ε0A / d
With a dielectric material of relative permittivity εr inserted, the capacitance becomes C = εrε0A/d. The derivation highlights why capacitance increases with plate area and decreases with separation.
插入相对介电常数为 εr 的电介质后，电容变为 C = εrε0A/d。该推导揭示了电容随极板面积增大而增大、随间距增大而减小的原因。
6. Energy Stored in a Capacitor | 电容器储存的能量
Charging a capacitor involves transferring charge from one plate to the other, which requires work against the growing potential difference. When a small amount of charge dq is moved at an instantaneous potential difference v = q/C, the work done is dW = v dq = (q/C) dq. Integrating from q = 0 to the final charge Q gives the total stored energy:
给电容器充电实质上是将电荷从一个极板转移到另一个极板，这一过程需要克服逐渐增大的电势差做功。当瞬时电势差为 v = q/C 时，移动微小电荷 dq 所做的功为 dW = v dq = (q/C) dq。从 q = 0 积分至最终电荷 Q，可得总储存能量：
W = ½ Q2/C = ½ QV = ½ CV2
The factor ½ appears because the average potential difference during charging is half the final value. This energy resides in the electric field between the plates.
因子 ½ 出现的原因是充电过程中的平均电势差仅为最终值的一半。该能量储存在极板间的电场中。
7. RC Discharge Circuit: Differential Equation | RC放电电路：微分方程
Consider a charged capacitor of capacitance C connected in series with a resistor R, with no battery in the loop. At any instant, the potential difference across the capacitor q/C equals the p.d. across the resistor iR. Applying Kirchhoff's voltage rule, q/C + iR = 0. Since current is the rate of discharge, i = −dq/dt (negative because charge decreases), we obtain:
考虑一个已充电的电容器 C 与电阻 R 串联，回路中没有电池。任意时刻电容器两端的电势差 q/C 等于电阻两端的电势差 iR。根据基尔霍夫电压定律，q/C + iR = 0。由于电流是放电速率，i = −dq/dt（负号表示电荷减少），可得：
R (dq/dt) + q/C = 0
This is a first-order linear differential equation that encapsulates the behaviour of the discharging capacitor.
这是一个一阶线性微分方程，概括了放电电容器的行为。
8. Solving the RC Discharge Equation | 求解RC放电方程
Separating variables gives dq/q = −dt/(RC). Integrating both sides leads to ln q = −t/(RC) + constant. At t = 0, q = Q0 (the initial charge), so the constant is ln Q0. Exponentiating yields the well-known exponential decay:
分离变量得 dq/q = −dt/(RC)。两边积分得 ln q = −t/(RC) + 常数。当 t = 0 时，q = Q0（初始电荷），故常数为 ln Q0。指数化后得到著名的指数衰减：
q = Q0 e−t/RC
The product RC It has units of seconds and represents the time taken for the charge to fall to 1/e (≈ 37%) of its initial value.
乘积 RC 称为时间常数 τ，其单位为秒，表示电荷衰减至初始值 1/e（约37%）所需的时间。
9. Half-life of RC Discharge | RC放电的半衰期
The half-life T1/2 is the time required for the charge on the capacitor to drop to half its initial value. Setting q = Q0/2 in the decay equation: Q0/2 = Q0 e−T1/2/RC. Taking natural logarithms gives ln(½) = −T1/2/RC, and since ln(½) = −ln 2, we find:
半衰期 T1/2 是电容器电荷量降至初始值一半所需的时间。在衰减方程中令 q = Q0/2：Q0/2 = Q0 e−T1/2/RC。取自然对数得 ln(½) = −T1/2/RC，而 ln(½) = −ln 2，故有：
T1/2 = RC ln 2 ≈ 0.693 RC
Like the time constant, the half-life is independent of the initial charge and is a characteristic of the circuit itself.
与时间常数一样，半衰期与初始电荷无关，是电路本身的特征参数。
10. Circular Motion of a Charged Particle in a Magnetic Field | 带电粒子在磁场中的圆周运动
When a charged particle of charge q moves with velocity v perpendicular to a uniform magnetic field of flux density B, it experiences a magnetic force F = qvB (Lorentz force). This force is always perpendicular to the velocity, providing the centripetal force required for circular motion: mv2/r = qvB. Rearranging gives the radius of the circular path:
当一个电荷量为 q 的带电粒子以速度 v 垂直于磁感应强度为 B 的匀强磁场运动时，它会受到磁力 F = qvB（洛伦兹力）。该力始终垂直于速度方向，提供圆周运动所需的向心力：mv2/r = qvB。整理可得圆周路径的半径：
r = mv / (qB)
The period of one complete revolution T is the circumference divided by speed: T = 2πr/v. Substituting the expression for r makes the period independent of velocity:
旋转一周的周期 T 等于周长除以速率：T = 2πr/v。代入 r 的表达式后发现周期与速度无关：
T = 2πm / (qB)
This principle is exploited in cyclotrons and mass spectrometers; particles with larger speeds orbit with larger radii, but all take exactly the same time to complete a cycle.
这一原理被应用于回旋加速器和质谱仪；速度较大的粒子以较大的半径运行，但所有粒子完成一周的时间完全相同。
Published by TutorHao | Physics Revision Series | aleveler.com
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