Coulomb's Law Core Points | 库仑定律核心要点

📚 Coulomb’s Law Core Points | 库仑定律核心要点

Coulomb’s law describes the electrostatic force between two point charges. It is one of the most fundamental principles in A-Level physics and a frequent source of exam questions, especially when combined with circular motion, electric fields, or superposition.

库仑定律描述了两个点电荷之间的静电力。它是A-Level物理中最基本的原理之一，也是考试中常见的考点，尤其在结合圆周运动、电场或叠加原理时更是如此。

1. What is Coulomb’s Law? | 什么是库仑定律？

Coulomb’s law states that the magnitude of the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.

库仑定律指出：两个点电荷之间静电力的大小与两电荷电量的乘积成正比，与它们之间距离的平方成反比。

The force acts along the line joining the two charges. If the charges have the same sign, the force is repulsive; if they have opposite signs, the force is attractive.

力的方向沿着两电荷的连线。若两电荷同号，则为斥力；若异号，则为引力。

2. Mathematical Form and Units | 数学形式与单位

In its scalar form, Coulomb’s law is written as:

库仑定律的标量形式为：

F = k Q₁Q₂ / r²

where F is the magnitude of the force in newtons (N), Q₁ and Q₂ are the charges in coulombs (C), r is the separation in metres (m), and k is the Coulomb constant.

其中F是力的大小，单位为牛顿(N)；Q₁和Q₂是电荷量，单位为库仑(C)；r是电荷间的距离，单位为米(m)；k是库仑常数。

In SI units, k is written as:

在国际单位制中，k可写为：

k = 1 / (4πε₀) ≈ 8.99 × 10⁹ N m² C⁻²

where ε₀ is the permittivity of free space, with a value of ε₀ ≈ 8.85 × 10⁻¹² F m⁻¹.

其中ε₀是真空介电常数，其值约为8.85 × 10⁻¹² F m⁻¹。

3. Vector Nature and Direction | 矢量性与方向

Coulomb’s law is a vector equation. When solving problems, you must consider not only the magnitude of the force but also its direction.

库仑定律是矢量方程。解题时不仅要考虑力的大小，还必须考虑力的方向。

In vector form:

矢量形式为：

F₁₂ = k Q₁Q₂ / r₁₂² · r̂₁₂

where r̂₁₂ is a unit vector pointing from charge Q₁ to charge Q₂. The force on Q₂ due to Q₁ is opposite in direction to the force on Q₁ due to Q₂, satisfying Newton’s third law.

其中r̂₁₂是从电荷Q₁指向电荷Q₂的单位矢量。Q₂受到Q₁的力与Q₁受到Q₂的力方向相反，满足牛顿第三定律。

Remember: for unlike charges, the force is attractive, so the arrow on a free-body diagram points towards the other charge. For like charges, it points away.

记住：对于异号电荷，力是吸引力，因此在受力分析图中箭头指向另一个电荷；对于同号电荷，箭头指向远离另一个电荷的方向。

4. Permittivity and Medium | 介电常数与介质

In a vacuum, the constant is k = 1/(4πε₀). If the charges are placed in a different insulating medium, the force is reduced by a factor equal to the relative permittivity εᵣ of that medium.

在真空中，常数k = 1/(4πε₀)。若电荷置于其他绝缘介质中，力会按该介质的相对介电常数εᵣ的大小而减小。

The general expression becomes:

一般情况下，表达式变为：

F = Q₁Q₂ / (4πε₀εᵣr²)

For air, εᵣ ≈ 1.0006, so most A-Level questions assume air is equivalent to a vacuum.

空气的εᵣ约为1.0006，因此大多数A-Level题目将空气视为真空处理。

Be careful: the unit of ε₀ is F m⁻¹ (farads per metre), which is equivalent to C² N⁻¹ m⁻².

注意：ε₀的单位是F m⁻¹（法拉每米），也等价于C² N⁻¹ m⁻²。

5. Comparison with Newton’s Law of Gravitation | 与万有引力定律的比较

Both Coulomb’s law and Newton’s law of gravitation are inverse-square laws. They share similar mathematical structures, but important differences exist.

库仑定律和万有引力定律都是平方反比定律。它们在数学结构上相似，但存在重要区别。

Coulomb’s Law	Newton’s Law of Gravitation
F = k Q₁Q₂ / r²	F = G m₁m₂ / r²
Can be attractive or repulsive	Only attractive
Deals with charges	Deals with masses
k ≈ 8.99 × 10⁹ N m² C⁻²	G ≈ 6.67 × 10⁻¹¹ N m² kg⁻²

A common exam question asks why the gravitational force between two protons is negligible compared with the electrostatic repulsion between them. The answer involves the very small value of G compared with k, and the fact that charges can be large while masses are extremely small.

一个常见的考试问题是：为什么两个质子之间的万有引力与它们之间的静电斥力相比可以忽略不计？答案在于G比k小得多，而且电荷量可以很大，而质量却极其微小。

6. Principle of Superposition | 叠加原理

When more than two charges are present, the net force on any one charge is the vector sum of the forces exerted by each of the other charges individually.

当存在两个以上的电荷时，任一电荷所受的合力等于其他各个电荷单独作用时对该电荷所施力的矢量和。

This is called the principle of superposition. It allows you to solve problems with three or more charges by calculating pairwise forces and then adding them as vectors.

这称为叠加原理。它允许你在处理三个或更多电荷的问题时，先计算每对电荷之间的力，然后再进行矢量求和。

Worked strategy: draw a free-body diagram, label all forces, resolve into perpendicular components if necessary, then add components separately.

解题策略：画受力分析图，标出所有力，必要时分解为互相垂直的分量，然后分别对分量求和。

7. Point Charges and Limitations | 点电荷与局限性

Coulomb’s law applies strictly to point charges — objects whose size is much smaller than the distance between them. If charges are distributed over extended objects, the force cannot be calculated simply using the centre-to-centre distance unless the charge distribution is spherically symmetric.

库仑定律严格适用于点电荷——即物体的大小远小于它们之间距离的情况。如果电荷分布在有限大小的物体上，除非电荷分布具有球对称性，否则不能简单使用质心到质心的距离来计算力。

For a uniformly charged sphere, the external force can be treated as if all the charge were concentrated at the centre. Inside a conducting sphere, however, the electric field is zero, but this is usually covered in the electric fields topic.

对于均匀带电球体，外部受力可视为所有电荷集中在球心时的情况。然而，在导体球内部，电场为零，但这通常在电场专题中讨论。

Another limitation: Coulomb’s law becomes inaccurate at extremely small distances (below atomic scale) where quantum effects dominate, and at very high speeds where relativistic effects matter.

另一个局限性：在极小的距离下（小于原子尺度），量子效应占主导，库仑定律不再精确；在极高速度下，相对论效应也会影响其适用性。

8. Worked Examples | 例题解析

Example 1: Two point charges, Q₁ = +4.0 μC and Q₂ = -2.0 μC, are separated by a distance of 0.30 m in a vacuum. Calculate the magnitude of the electrostatic force between them.

例题1：两个点电荷Q₁ = +4.0 μC和Q₂ = -2.0 μC在真空中相距0.30 m。求它们之间的静电力大小。

Solution:

解答：

F = k|Q₁Q₂| / r² = (8.99 × 10⁹)(4.0 × 10⁻⁶)(2.0 × 10⁻⁶) / (0.30)²

First, 4.0 × 10⁻⁶ × 2.0 × 10⁻⁶ = 8.0 × 10⁻¹² C². Then F = (8.99 × 10⁹ × 8.0 × 10⁻¹²) / 0.09 = 0.0719 / 0.09 ≈ 0.80 N.

首先，4.0 × 10⁻⁶ × 2.0 × 10⁻⁶ = 8.0 × 10⁻¹² C²。然后F = (8.99 × 10⁹ × 8.0 × 10⁻¹²) / 0.09 = 0.0719 / 0.09 ≈ 0.80 N。

Since the charges have opposite signs, the force is attractive. The magnitude is 0.80 N.

由于两电荷异号，力为吸引力。大小为0.80 N。

Example 2: Three charges are placed at the vertices of an equilateral triangle of side 0.20 m. Each charge has magnitude 2.0 μC; two are positive and one is negative. Find the magnitude of the net force on the negative charge.

例题2：三个电荷分别置于边长0.20 m的等边三角形顶点。每个电荷大小为2.0 μC；两个为正，一个为负。求负电荷所受合力的大小。

Solution: The two positive charges each exert an attractive force on the negative charge of magnitude:

解答：两个正电荷各自对负电荷施加引力，大小为：

F = k(2.0 × 10⁻⁶)² / (0.20)² = (8.99 × 10⁹ × 4.0 × 10⁻¹²) / 0.04 = 0.90 N

The two forces are directed along the sides of the triangle towards the positive charges. Since the triangle is equilateral, the angle between the two force vectors is 60°. The resultant magnitude is:

这两个力分别沿三角形边指向正电荷。由于是等边三角形，两个力矢量之间的夹角为60°。合力大小为：

F_net = √(F² + F² + 2F²cos60°) = √(3F²) = F√3 ≈ 1.56 N

So the net force is approximately 1.56 N directed along the bisector of the angle opposite the negative charge.

因此合力约为1.56 N，方向沿负电荷所对角平分线方向。

9. Common Exam Pitfalls | 常见考试陷阱

Forgetting to convert microcoulombs (μC) to coulombs (C). Always multiply by 10⁻⁶.
Forgetting to convert microcoulombs (μC) to coulombs (C)。务必乘以10⁻⁶。
Using the distance between charges instead of the square of the distance in the denominator.
在分母中使用电荷间距离而不是距离的平方。
Ignoring the vector nature of force; when calculating net force, you must add forces as vectors, not as scalars.
忽略力的矢量性；计算合力时必须按矢量相加，而不是标量相加。
Using the centre-to-centre distance for non-point, non-spherical charge distributions.
对非点电荷、非球对称电荷分布直接使用中心到中心的距离。
Confusing ε₀ with εᵣ; remember εᵣ is the relative permittivity, a dimensionless factor.
混淆ε₀与εᵣ；记住εᵣ是相对介电常数，是无量纲的比值。
Neglecting the sign of charges when only the magnitude of force is asked.
当只要求力的大小时忽略电荷的正负号。

10. Summary | 总结

Coulomb’s law gives the electrostatic force between point charges: F = kQ₁Q₂/r², with k = 1/(4πε₀). The force is an inverse-square law, acts along the line joining the charges, and obeys the principle of superposition. In a medium, use ε₀εᵣ in the denominator. When solving problems, always pay attention to units, vector directions, and the conditions under which the law applies.

库仑定律给出了点电荷之间的静电力：F = kQ₁Q₂/r²，其中k = 1/(4πε₀)。该力遵循平方反比规律，作用在两电荷连线上，并满足叠加原理。在介质中，分母使用ε₀εᵣ。解题时，务必注意单位、矢量方向以及定律的适用条件。

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Coulomb’s Law Core Points | 库仑定律核心要点