📚 Electric Fields and Capacitance Formula Derivations | 电场与电容公式推导
In this revision article, we systematically derive the essential formulas for electric fields and capacitance as required by the Oxford AQA International A-Level Physics specification. Understanding these derivations not only helps you memorize the equations but also deepens your insight into the underlying principles of electrostatics and circuit elements. Let's work through the key relationships step by step.
在这篇复习文章中，我们将系统地推导 Oxford AQA 国际 A-Level 物理考试要求的电场与电容核心公式。理解这些推导不仅有助于记忆方程，还能加深你对静电学和电路元件基本原理的理解。让我们逐步掌握这些关键关系。
1. Coulomb's Law and Electric Field Strength of a Point Charge | 库仑定律与点电荷的电场强度
Coulomb' trength E is the force per unit positive charge experienced by a small test charge q placed in the field: E = F/q. Substituting Coulomb' point charge Q at a distance r is E = kQ / r², directed radially outward if Q is positive.
库仑定律指出，两个距离为 r 的点电荷 Q 和 q 之间的力 F = kQq / r²，其中 k = 1/(4πε₀)。根据定义，电场强度 E 是放置在电场中的小检验电荷 q 所受到的每单位正电荷的力：E = F/q。代入库仑定律得 E = (kQq / r²) / q = kQ / r²。因此，点电荷 Q 在距离 r 处产生的电场强度大小为 E = kQ / r²，若 Q 为正，方向沿径向向外。
E = kQ / r²
2. Electric Potential due to a Point Charge | 点电荷的电势
rk done against the electric field force is given by integrating F dr from infinity to r: W = ∫[∞, r] (kQq / r²) dr. Solving this integral, W = kQq [ -1/r ] from ∞ to r = kQq / r. Therefore, V = W/q = kQ / r. This shows that the potential due to a point charge is inversely proportional to distance.
电势 V 定义为将单位正电荷从无穷远处移动到某点所做的功。克服电场力所做的功由积分 F dr 从无穷远到 r 给出：W = ∫[∞, r] (kQq / r²) dr。计算积分，W = kQq [ -1/r ] 从 ∞ 到 r = kQq / r。因此，V = W/q = kQ / r。这表明点电荷产生的电势与距离成反比。
V = kQ / r
3. Uniform Electric Field between Parallel Plates | 平行板间的匀强电场
m one plate to the other is W = qV. This same work is also given by the force F = qE multiplied by the distance d: W = qEd. Equating the two expressions, qV = qEd, which simplifies to E = V/d. The field is uniform and directed from the positive plate to the negative plate.
对于两块带有异种电荷、间距为 d 且电势差为 V 的平行板，将电荷 q 从一块板移动到另一块板所做的功为 W = qV。同样的功也可表示为力 F = qE 乘以距离 d：W = qEd。令两式相等，得 qV = qEd，简化即 E = V/d。该电场为匀强场，方向从正极板指向负极板。
E = V / d
4. Capacitance and the Parallel Plate Capacitor | 电容与平行板电容器
aration d, we use the uniform field relation V = Ed and the field due to plates E = σ/ε₀, where σ = Q/A is the surface charge density. Substituting, V = (σ/ε₀) d = (Q/(Aε₀)) d. Then C = Q/V = Q / [Qd/(Aε₀)] = ε₀ A / d. This derivation ignores edge effects. With a dielectric of relative permittivity εr, capacitance becomes C = ε₀ εr A / d.
电容 C 定义为储存的电荷 Q 与电势差 V 的比值：C = Q/V。对于极板面积为 A、间距为 d 的平行板电容器，利用匀强电场关系 V = Ed 和极板产生的电场 E = σ/ε₀，其中 σ = Q/A 为面电荷密度。代入得 V = (σ/ε₀) d = (Q/(Aε₀)) d。于是 C = Q/V = Q / [Qd/(Aε₀)] = ε₀ A / d。此推导忽略边缘效应。若存在相对介电常数为 εr 的电介质，电容变为 C = ε₀ εr A / d。
C = ε₀ A / d
C = ε₀ εr A / d
5. Energy Stored in a Capacitor | 电容器储存的能量
the potential is v is dW = v dq. Since v = q/C, dW = (q/C) dq. Integrating from 0 to Q gives total work W = ∫[0,Q] (q/C) dq = (1/2) Q² / C = ½ QV. Using Q = CV, we also obtain W = ½ CV², and W = ½ Q² / C. This energy is stored as electric potential energy in the field between the plates.
为电容器充电时，必须克服逐渐增大的电势差移动电荷而做功。当电势为 v 时，增加电荷微量 dq 所需的微功为 dW = v dq。因 v = q/C，所以 dW = (q/C) dq。从 0 到 Q 积分得总功 W = ∫[0,Q] (q/C) dq = (1/2) Q² / C = ½ QV。利用 Q = CV，还可得到 W = ½ CV² 和 W = ½ Q² / C。该能量以极板间电场的电势能形式储存。
E = ½ QV = ½ CV² = ½ Q² / C
6. RC Charging and Discharging Equations | RC 充放电方程
t with a resistor R and capacitor C connected to a DC supply of emf V₀, applying Kirchhoff's voltage law gives V₀ = vR + vC. Using vR = iR and i = dq/dt, and vC al equation has the solution for the charge on the capacitor while charging: q = Q₀ (1 - e-t/RC During discharging with no supply, the equation becomes 0 = R dq/dt + q/C, giving q = Q₀ e-t/RC. The current and voltage follow similar exponential decays.
在电阻 R 与电容 C 串联且连接至直流电源（电动势 V₀）的电路中，应用基尔霍夫电压定律得 V₀ = vR + vC。利用 vR = iR 且 i = dq/dt，以及 vC = q/C，可得 V₀ = R dq/dt + q/C。这个一阶微分方程在充电时的电荷解为：q = Q₀ (1 - e-t/RC)，其中 Q₀ = CV₀。时间常数 τ = RC。在无电源的放电过程中，方程变为 0 = R dq/dt + q/C，解得 q = Q₀ e-t/RC。电流和电压遵循类似的指数衰减规律。
q = Q₀ (1 - e-t/RC)
q = Q₀ e-t/RC
7. Capacitors in Series and Parallel | 电容器的串联与并联
For capacitors in parallel, they share the same potential difference V. The total charge Qtotal = Q₁ + Q₂ + ... = C₁V + C₂V + ... = (C₁ + C₂ + ...)V. Hence the equivalent capacitance Cₚ
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