Electric Force on Charges in an Electric Field | A-Level 物理：电荷在电场中的受力分析

📚 Electric Force on Charges in an Electric Field | A-Level 物理：电荷在电场中的受力分析

In A-Level Physics, analysing the force experienced by a charge placed in an electric field is a fundamental skill. It connects the ideas of field strength, potential difference, and Newton’s laws of motion, and underpins many exam questions involving charged particles, parallel plates, and electric fields.

在 A-Level 物理中，分析电荷在电场中所受的力是一项核心技能。它将电场强度、电势差与牛顿运动定律联系起来，也是回答涉及带电粒子、平行板电容器和电场分布等许多考题的基础。

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1. What Is an Electric Field? | 什么是电场？

An electric field is a region of space in which an electric charge experiences a force. The field is a vector quantity, meaning it has both magnitude and direction. The direction of the field is defined as the direction of the force on a positive test charge placed in the field.

电场是电荷在其中会受到电场力作用的空间区域。电场是矢量，既有大小又有方向。电场方向定义为放置在电场中的正检验电荷所受力的方向。

For a uniform electric field, such as the field between two parallel charged plates, the field lines are parallel and equally spaced. This indicates that the electric field strength is constant in that region.

对于匀强电场，例如两块平行带电金属板之间的电场，电场线是平行且等间距的。这表示该区域内电场强度处处相等。

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2. Electric Field Strength E | 电场强度 E

Electric field strength E is defined as the force per unit positive charge experienced by a small test charge placed at a point in the field. Mathematically, it is written as:

电场强度 E 的定义是：放置在电场中某一点的小检验电荷所受到的电场力与其电荷量的比值。其数学表达式为：

E = F / Q

where F is the electric force in newtons (N), Q is the charge in coulombs (C), and E is the electric field strength in newtons per coulomb (N C⁻¹), which is equivalent to volts per metre (V m⁻¹).

其中 F 是电场力，单位为牛顿（N）；Q 是电荷量，单位为库仑（C）；E 是电场强度，单位为牛顿每库仑（N C⁻¹），也等价于伏特每米（V m⁻¹）。

Rearranging this equation gives the electric force experienced by a charge Q in a field of strength E:

将该式变形，可以得到电荷 Q 在电场强度为 E 的电场中所受电场力为：

F = Q E

This simple equation is the most important starting point when analysing the motion of a charged particle in an electric field.

这个简单公式是分析带电粒子在电场中运动的起点，也是考查频率最高的知识点之一。

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3. Direction of the Electric Force | 电场力的方向

The direction of the electric force on a charge depends on the sign of the charge. A positive charge experiences a force in the same direction as the electric field, while a negative charge experiences a force in the opposite direction to the electric field.

电荷在电场中受力的方向取决于电荷的正负。正电荷所受电场力的方向与电场方向相同，而负电荷所受电场力的方向与电场方向相反。

For example, if an electric field points to the right, a proton will accelerate to the right, while an electron will accelerate to the left.

例如，如果电场方向指向右方，质子将向右加速，而电子将向左加速。

It is important to distinguish between the electric field direction and the electric force direction. They are the same only for positive charges.

务必区分电场方向与电场力方向：只有在正电荷的情况下，两者方向才一致。

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4. Uniform Electric Field Between Parallel Plates | 平行板之间的匀强电场

Two parallel conducting plates connected to a direct current supply produce a uniform electric field between them. The field lines run from the positive plate to the negative plate, and the field strength is constant in the central region.

两块平行导体板与直流电源连接后，两板之间会产生匀强电场。电场线从正极板指向负极板，在中间区域电场强度恒定。

If the separation between the plates is d and the potential difference is V, the electric field strength is given by:

若两板间距为 d，电压为 V，则电场强度为：

E = V / d

This equation is extremely useful in exam problems because it links the electric field strength to quantities that can be easily controlled and measured. The force on a charge between the plates then becomes:

这个公式非常实用，因为它将电场强度与容易控制和测量的物理量联系起来。于是，两板间电荷所受的电场力为：

F = Q V / d

This force is constant throughout the region between the plates, assuming edge effects are ignored.

在忽略边缘效应的情况下，该电场力在两板之间的整个区域内恒定不变。

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5. Motion of a Charged Particle in a Uniform Field | 带电粒子在匀强电场中的运动

When a charged particle enters a uniform electric field, its motion depends on its initial velocity and the direction of the electric force relative to that velocity. If the particle starts from rest or moves parallel to the field, its motion is one-dimensional and uniformly accelerated.

当带电粒子进入匀强电场时，其运动取决于初速度方向以及电场力与初速度之间的夹角。若粒子从静止开始运动，或初速度方向与电场方向平行，则粒子做一维匀加速直线运动。

Using Newton’s second law, the acceleration of the particle is:

由牛顿第二定律，粒子的加速度为：

a = F / m = Q E / m

where m is the mass of the particle. Since F is constant in a uniform field, a is also constant, and the standard kinematic equations can be applied to determine the particle’s velocity and displacement over time.

其中 m 为粒子的质量。由于匀强电场中 F 恒定，因此加速度 a 也恒定，可以使用运动学公式求解粒子的速度和位移随时间的变化。

If a charge is projected perpendicular to the field lines, its trajectory becomes a parabola. The electric force provides a constant acceleration perpendicular to its initial velocity, similar to projectile motion under gravity.

如果带电粒子垂直于电场线射入，其运动轨迹为抛物线。电场力提供垂直于初速度方向的恒定加速度，类似于重力场中的抛体运动。

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6. Work Done and Energy Changes | 电场力做功与能量变化

When a charge moves through an electric field, the electric force may do work on the charge, changing its kinetic energy. In a uniform field, the work done by the electric force when a charge Q moves a distance x in the direction of the field is:

电荷在电场中移动时，电场力可能对电荷做功，从而改变其动能。在匀强电场中，电荷 Q 沿电场方向移动距离 x 时，电场力做的功为：

W = F x = Q E x

Since E x equals the potential difference between the two points, the work done can also be written as:

由于 E x 等于两点之间的电势差，因此做功也可以写为：

W = Q V

This energy relationship is often used together with the kinetic energy equation to find the speed of a charged particle accelerated from rest through a potential difference V:

这个能量关系常与动能定理联用，用于求解带电粒子从静止开始经过电势差 V 加速后的速度：

Q V = ½ m v²

which gives:

由此可得：

v = √(2 Q V / m)

This formula appears frequently in CIE A-Level exam questions, especially in the context of electron guns and particle accelerators.

该公式在 CIE A-Level 考试中频繁出现，尤其是在电子枪和粒子加速器相关题目中。

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7. Comparing Electric Force and Gravitational Force | 电场力与万有引力的比较

Both electric and gravitational forces are field forces that act at a distance. However, there are important differences. The electric force can be either attractive or repulsive, because charges can be positive or negative, whereas the gravitational force is always attractive between masses.

电场力与万有引力都属于场力，可以在一定距离内发生作用。但两者存在重要区别：电场力可以是引力也可以是斥力，因为电荷有正负之分；而万有引力在质量之间总表现为吸引力。

Property | 性质	Electric Force | 电场力	Gravitational Force | 万有引力
Acts on | 作用对象	Charges | 电荷	Masses | 质量
Attraction or repulsion | 吸引或排斥	Both | 两者都有	Attraction only | 只有吸引
Relative strength | 相对强度	Much stronger | 强得多	Much weaker | 弱得多
Depends on medium | 是否依赖介质	Yes, permittivity | 依赖介电常数	No | 不依赖

In many A-Level problems involving charged particles, gravitational effects are negligibly small compared with electric effects, so gravity can be ignored unless the question explicitly asks about it.

在许多涉及带电粒子的 A-Level 题目中，与电场力相比，重力作用往往可以忽略不计。除非题目明确要求考虑重力，否则一般只分析电场力。

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8. Electric Force on Multiple Charges | 多个电荷所受的电场力

If several charges are present, the total electric force on a given charge is the vector sum of the individual forces due to each other charge. This is known as the principle of superposition.

当存在多个电荷时，某一个电荷所受的总电场力等于其他各电荷单独作用时对该电荷产生的电场力的矢量之和。这就是电场力的叠加原理。

For each pair of charges, the magnitude of the force is given by Coulomb’s law:

对于任意一对电荷，其相互作用力的大小由库仑定律给出：

F = k Q₁ Q₂ / r²

where k is the Coulomb constant, Q₁ and Q₂ are the magnitudes of the charges, and r is the distance between them. The direction of each force must be drawn separately and the vector resultant found using components or the parallelogram law.

其中 k 为库仑常数，Q₁ 和 Q₂ 是两个电荷的电荷量大小，r 是它们之间的距离。求解合力时，必须先分别画出每个分力的方向，再通过正交分解或平行四边形法则求矢量和。

In field notation, the total electric field at a point is the vector sum of the fields due to individual charges, and the force on a charge placed at that point is then F = Q E_total.

在场的概念中，某一点的总电场等于各个电荷在该点产生的电场的矢量和，放置在该点的电荷所受电场力则为 F = Q E_total。

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9. Worked Example | 典型例题分析

A proton is accelerated from rest through a potential difference of 500 V. Calculate the final speed of the proton. The mass of a proton is 1.67 × 10⁻²⁷ kg and its charge is 1.60 × 10⁻¹⁹ C.

一个质子从静止开始经过 500 V 的电势差加速。已知质子质量为 1.67 × 10⁻²⁷ kg，电荷量为 1.60 × 10⁻¹⁹ C，求质子的末速度。

Using energy conservation:

由能量守恒：

Q V = ½ m v²

Substituting the values:

代入数值：

v = √(2 Q V / m) = √(2 × 1.60 × 10⁻¹⁹ × 500 / 1.67 × 10⁻²⁷)

Calculating this gives:

计算得：

v = 3.10 × 10⁵ m s⁻¹

Notice that we did not need to know the electric field strength or the distance travelled, because the potential difference alone determines the energy gained. This shows the advantage of using energy methods in electric field problems.

注意，我们并不需要知道电场强度或质子移动的距离，因为电势差单独决定了质子获得的能量。这体现了在电场问题中使用能量方法的优势。

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10. Common Mistakes and Exam Tips | 常见错误与考试技巧

One common mistake is confusing electric field direction with electric force direction. Always check the sign of the charge before determining the direction of the force.

常见错误之一是混淆电场方向与电场力方向。在确定受力方向之前，务必先判断电荷的正负。

Another common mistake is forgetting that E = V / d only applies to uniform electric fields. For radial or non-uniform fields, this equation is not valid.

另一个常见错误是忘记 E = V / d 仅适用于匀强电场。对于径向电场或非匀强电场，该公式不适用。

In exam calculations, make sure to use consistent units. Distances in metres, charges in coulombs, and masses in kilograms will give results in SI units.

考试计算中要确保单位统一。距离用米、电荷用库仑、质量用千克，这样得到的结果才是国际单位制下的数值。

• Draw a clear diagram showing the direction of the electric field and the charge sign.
• 画一个清晰的示意图，标出电场方向和电荷符号。
• Decide whether the motion is parallel or perpendicular to the field.
• 判断粒子的运动方向是与电场平行还是垂直。
• Use energy conservation when the question involves potential difference and speed.
• 当题目涉及电势差和速度时，优先考虑能量守恒。
• Use Newton’s second law when the question asks about acceleration or time.
• 当题目涉及加速度或时间时，使用牛顿第二定律。

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