📚 AS Physics Paper 4 Exam Report: Key Formula Derivations | AS物理试卷4考试报告：核心公式推导
AS Physics Paper 4 often includes questions that require you to derive key formulas from first principles. Examiner reports consistently highlight that students lose marks by memorising results without understanding the logical steps or the assumptions behind them. This article breaks down the most commonly tested derivations in mechanics, fields, and electricity, using the same approach that earns full marks in exams. Each section pairs a clear English explanation with its Chinese equivalent so you can cross-reference terminology and deepen your understanding.
AS物理试卷4经常包含要求从基本原理推导关键公式的题目。考官报告一再指出，学生因只会机械记忆而不理解推导逻辑或前提假设而失分。本文拆解了力学、场和电路中最常考的推导，采用与考试满分答案相同的思路。每一节都将清晰的英文讲解与中文对应内容配对，方便你对照术语并加深理解。
1. Centripetal Acceleration: a = v²/r | 向心加速度推导：a = v²/r
Consider an object moving at constant speed v in a circle of radius r. Although the speed is constant, the velocity changes direction continuously, so there is an acceleration towards the centre. In a small time interval Δt, the object moves from point A to point B through an angle Δθ. The velocity vectors at A and B have the same magnitude v but different directions. The change in velocity Δv is the vector from the tip of vA to the tip of vB.
考虑一个物体以恒定速率v在半径为r的圆周上运动。虽然速率不变，但速度方向不断改变，因此存在指向圆心的加速度。在很小的时间间隔Δt内，物体从A点运动到B点，转过角度Δθ。A点和B点的速度矢量大小均为v，但方向不同。速度的变化量Δv是从vA箭头指向vB箭头的矢量。
For small Δθ, the angle between vA and vB is Δθ, and the triangle formed by the two velocity vectors is similar to the triangle formed by the two radii OA and OB. From similarity, the ratio of the change in velocity to the speed equals the ratio of the arc length AB to the radius: |Δv| / v = arc AB / r. Since arc AB ≈ vΔt, we have |Δv| / v = vΔt / r, so |Δv| = v²Δt / r. The magnitude of the centripetal acceleration is a = |Δv| / Δt = v²/r.
对于很小的Δθ，vA和vB之间的夹角为Δθ，由这两个速度矢量构成的三角形与由两条半径OA和OB构成的三角形相似。根据相似性，速度变化量的大小与速率之比等于弧长AB与半径之比：|Δv| / v = 弧长AB / r。由于弧长AB ≈ vΔt，可得|Δv| / v = vΔt / r，因此|Δv| = v²Δt / r。向心加速度的大小为 a = |Δv| / Δt = v²/r。
a = v² / r
Examiners report that candidates often fail to state the vector nature of the quantities and lose marks for confusing arc length with chord length. Emphasise that the derivation assumes constant speed and a very small time interval.
考官报告指出，考生常常没有明确矢量的性质，并因混淆弧长与弦长而失分。要强调该推导假设速率恒定，且时间间隔非常小。
2. Conservation of Momentum from Newton’s Laws | 从牛顿定律推导动量守恒
Consider two objects, A and B, that interact with no external forces. According to Newton’s third law, the force FA on B exerted by A on B is equal in magnitude and opposite in direction to the force FB on A exerted by B on A. Therefore, FA on B = – FB on A.
考虑两个不受外力的物体A和B。根据牛顿第三定律，A对B的作用力FA on B与B对A的作用力FB on A大小相等、方向相反。因此，FA on B = – FB on A。
Newton’s second law states that the net force on an object equals the rate of change of its momentum: F = Δp / Δt. Applying this to each object during a time interval Δt, we get FB on A = ΔpA / Δt and FA on B = ΔpB / Δt. Substituting these into the third-law relation gives ΔpA / Δt = – ΔpB / Δt, which simplifies to ΔpA + ΔpB = 0.
牛顿第二定律指出，物体所受的合力等于其动量变化率：F = Δp / Δt。将该定律应用于时间间隔Δt内的每个物体，可得FB on A = ΔpA / Δt 和 FA on B = ΔpB / Δt。将这两个式子代入第三定律关系式中，得到ΔpA / Δt = – ΔpB / Δt，化简为ΔpA + ΔpB = 0。
Hence, the total change in momentum of the system is zero, meaning the total momentum before the interaction equals the total momentum after the interaction. This is the principle of conservation of linear momentum: in a closed system, total momentum remains constant.
因此，系统总动量变化为零，即相互作用前的总动量等于相互作用后的总动量。这就是动量守恒定律：在一个封闭系统中，总动量保持不变。
Total momentum before = Total momentum after
Many exam reports highlight that students forget to explicitly state the absence of external forces as a necessary condition. Always begin your derivation by specifying ‘in a closed system with no external resultant force’.
许多考试报告强调，学生忘记明确指出“无外力”这一必要条件。推导时务必先声明“在一个无合外力的封闭系统中”。
3. Electric Field of a Point Charge: E = kQ / r² | 点电荷的电场：E = kQ / r²
Start from Coulomb’s law: the magnitude of the electrostatic force between two point charges Q and q separated by distance r is F = kQq / r², where k = 1/(4πε0). The electric field strength E at a point is defined as the force per unit positive charge at that point: E = F / q.
从库仑定律出发：两个点电荷Q和q相距r时，静电力大小为F = kQq / r²，其中k = 1/(4πε0)。电场强度E定义为该点处单位正电荷所受的力：E = F / q。
If a test charge q is placed at a distance r from the source charge Q, the force on it is F = kQq / r². Dividing this force by q gives the field strength due to Q at that location: E = F / q = (kQq / r²) / q = kQ / r². The direction of the field is radially outward from a positive Q and radially inward toward a negative Q.
如果把检验电荷q放在距离源电荷Q为r的位置，它所受到的力为F = kQq / r²。将力除以q就得到Q在该位置产生的电场强度：E = F / q = (kQq / r²) / q = kQ / r²。电场方向对于正Q沿径向向外，对于负Q沿径向向内。
E = kQ / r²
Be careful to distinguish between the source charge Q and the test charge q. A common exam error is to leave q in the final expression or to forget that E is a vector that depends on the sign of Q.
注意区分源电荷Q和检验电荷q。考试中常见的错误是在最终表达式中留下q，或是忘记E是矢量，其方向取决于Q的符号。
4. Uniform Electric Field and Potential: V = Ed | 匀强电场与电势：V = Ed
A uniform electric field exists between two parallel plates connected to a battery. The field strength E is constant in magnitude and direction. The potential difference V between the plates is defined as the work done per unit positive charge in moving a charge from one plate to the other against the field.
连接电池的两块平行板之间存在匀强电场。场强E的大小和方向恒定。两极板间的电势差V定义为将单位正电荷从一块极板逆着电场方向移动到另一块极板时所做的功。
Work done W on a charge q moving a distance d parallel to the field is W = Fd, where F = qE. Therefore, W = qEd. Since potential difference V = W / q, we get V = (qEd) / q = Ed. This relation holds strictly when the displacement is parallel to the field lines and the field is uniform.
对电荷q施加的力为F = qE，当它平行于电场移动距离d时，做功W = Fd，因此W = qEd。因为电势差V = W / q，可得V = (qEd) / q = Ed。这一关系在位移平行于电场线且电场均匀时严格成立。
E = V / d
The derivation shows that electric field strength can also be expressed in volts per metre (V m-1), which is equivalent to newtons per coulomb (N C-1). Many students fail to convert between these units correctly.
该推导表明电场强度也可表示为伏特每米（V m-1），等同于牛顿每库仑（N C-1）。许多学生未能正确换算这两种单位。
5. Force on a Current-Carrying Conductor: F = BIL sinθ | 载流导线所受的力：F = BIL sinθ
Place a straight conductor of length L carrying a current I in a uniform magnetic field of flux density B. The current consists of moving charges. A single charge q moving with drift velocity v experiences a magnetic force Fq = Bqv sinθ, where θ is the angle between v and B.
将长L、通有电流I的直导线置于磁通密度为B的匀强磁场中。电流由大量运动电荷构成。单个电荷q以漂移速度v运动，受到的磁力为Fq = Bqv sinθ，其中θ为v与B的夹角。
The total force on all charges in the conductor is the sum of the forces on each charge. The number of charges in the conductor is nAL, where n is the number density of charge carriers and A is the cross-sectional area. Current is given by I = nqvA. The total force is therefore F = (nAL) × (Bqv sinθ) = B (nqvA) L sinθ = BIL sinθ.
导线中所有电荷所受的总磁力等于各个电荷受力之和。导线中的电荷总数为nAL，其中n为载流子数密度，A为横截面积。电流由I = nqvA给出。总力因此为F = (nAL) × (Bqv sinθ) = B (nqvA) L sinθ = BIL sinθ。
F = BIL sinθ
The angle θ is taken between the direction of the current (which follows the drift velocity of positive charges) and the magnetic field. If the conductor is perpendicular to the field, sinθ = 1 and F = BIL. This microscopic derivation is a favourite in Paper 4, as it links several fundamental quantities.
θ角取电流方向（沿正电荷漂移速度方向）与磁场方向之间的夹角。若导线与磁场垂直，sinθ = 1，则F = BIL。这一微观推导是试卷4中的常见考点，因为它将多个基本量串联了起来。
6. Force on a Moving Charge: F = Bqv sinθ | 运动电荷所受的力：F = Bqv sinθ
Starting from the macroscopic formula F = BIL sinθ, we can derive the force on a single free charge. In a conductor, I = nqvA, and L can be expressed in terms of drift velocity v and time t: L = vt. Substituting these into F = B (nqvA) L sinθ gives the total force on all charges in a segment of length L.
从宏观公式F = BIL sinθ出发，可以推导出单个自由电荷所受的力。在导线中，I = nqvA，且L可通过漂移速度v和时间t表示为L = vt。将这些代入F = B (nqvA) L sinθ，得到长度为L的导线段内所有电荷所受的总力。
The number of charges in the segment is N = nAL = nA(vt). The total force is F = (BIL sinθ) = B (nqvA) (vt) sinθ = (nAvt) × (Bqv sinθ). Dividing the total force F by the number of charges N gives the force per charge: Fq = F / N = Bqv sinθ.
该段内的电荷总数为N = nAL = nA(vt)。总力为F = (BIL sinθ) = B (nqvA) (vt) sinθ = (nAvt) × (Bqv sinθ)。将总力F除以电荷总数N即可得到单个电荷所受的力：Fq = F / N = Bqv sinθ。
F = Bqv sinθ
Alternatively, this can be explained directly: the magnetic force on a charge is always perpendicular to both its velocity and the magnetic field, and its magnitude is proportional to q, v, B, and sinθ. Many examiner reports note that candidates struggle to identify θ correctly when the velocity is not perpendicular to the field.
也可以直接解释：电荷在磁场中所受的力始终垂直于其速度和磁场，大小与q、v、B以及sinθ成正比。许多考官报告指出，当速度不与磁场垂直时，考生往往难以正确确定θ角。
7. Faraday’s Law of Electromagnetic Induction: ε = – N ΔΦ / Δt | 法拉第电磁感应定律：ε = – N ΔΦ / Δt
Faraday’s law can be introduced through the concept of changing magnetic flux. Magnetic flux Φ through a coil of area A is defined as Φ = BA cosθ, where θ is the angle between the magnetic field B and the normal to the area. When the flux linking a coil changes, an electromotive force (emf) is induced.
法拉第定律可通过变化的磁通量概念引出。穿过面积为A的线圈的磁通量定义为Φ = BA cosθ，其中θ为磁场B与面积法线之间的夹角。当与线圈交链的磁通量发生变化时，就会感应出电动势（emf）。
Experiments show that the induced emf in a coil is directly proportional to the rate of change of flux linkage. For a coil of N turns, flux linkage is NΦ. The magnitude of the induced emf is therefore |ε| = N |ΔΦ| / Δt. To include the direction of the induced emf, Lenz’s law states that the induced current opposes the change in flux, giving the negative sign: ε = – N ΔΦ / Δt.
实验表明，线圈中感应出的电动势与磁通链的变化率成正比。对于N匝线圈，磁通链为NΦ。因此，感应电动势的大小为|ε| = N |ΔΦ| / Δt。为了包含感应电动势的方向，楞次定律指出感应电流总是阻碍磁通量的变化，从而引入负号：ε = – N ΔΦ / Δt。
ε = – N ΔΦ / Δt
The wording ‘rate of change of flux linkage’ is crucial in the derivation. A common pitfall in exams is to write simply ε = NΦ/t without the change symbol, or to omit the negative sign without referring to Lenz’s law. Always state Lenz’s law explicitly.
“磁通链变化率”这一表述在推导中至关重要。考试中常见的错误是写成ε = NΦ/t而没有表示变化的符号，或忽略负号且不提楞次定律。务必明确写出楞次定律。
8. Acceleration of a Charged Particle in an Electric Field | 电场中带电粒子的加速
When a particle of charge q accelerates from rest through a potential difference V, the work done by the electric field is W = qV. If there are no energy losses, this work is converted entirely into kinetic energy. The initial kinetic energy is zero, so the final kinetic energy is Ek = ½ mv².
当一个电荷量为q的粒子在电势差V中从静止加速时，电场做的功为W = qV。如果没有能量损失，这些功完全转化为动能。初动能为零，因此末动能为Ek = ½ mv²。
By energy conservation: qV = ½ mv². Solving for the final speed gives v = √(2qV / m). This derivation is valid for non-relativistic speeds and assumes the electric field is conservative. It is used extensively in problems involving electron guns and cathode ray tubes.
由能量守恒：qV = ½ mv²。解得末
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