A2 Physics: Common Mistake Questions Explained | A2 物理：易错题精讲

📚 A2 Physics: Common Mistake Questions Explained | A2 物理：易错题精讲

A2 Physics requires not only mathematical fluency but also deep conceptual clarity. Over the years, examiners have noted that certain mistakes appear again and again, often because students rely on rote memory or intuitive but incorrect pictures. In this article we unpack eight classic trouble spots, explain why the wrong ideas are so tempting, and show how to think about them correctly. Each section is paired with the kind of question you might meet in a real exam.

A2 物理不仅要求熟练的数学运算，更需要清晰的概念理解。多年来，考官注意到一些错误反复出现，往往是因为学生依赖死记硬背或凭直观但错误的图像作答。本文拆解八个经典易错点，分析错误想法的诱因，并给出正确思路。每一节都配合真实考试中可能遇到的题型。

1. Mistaking Centripetal Force as a Separate Force | 误将向心力当作单独的力

A very common free-body diagram error is to draw an arrow labelled ‘centripetal force’ pointing towards the centre, as if it were a new force like gravity or tension. In reality, centripetal force is the name we give to the resultant force directed towards the centre of circular motion. It is provided by one or more of the real forces: tension, friction, the normal reaction, or a component of gravity. Treating it as an extra force leads to double-counting or wrong equations.

一个非常常见的受力分析错误，是画上一个指向圆心的箭头并标注“向心力”，就好像它是像重力或绳拉力一样的新力。实际上，向心力是指向圆周运动圆心的合力，由一个或多个真实力提供：拉力、摩擦力、支持力或重力的分量。把它当成额外的力会导致重复计算或方程错误。

Consider a car of mass m passing over a convex bridge at speed v. The forces are the weight mg downwards and the normal reaction N upwards. At the highest point, the acceleration is towards the centre of curvature (downwards). The net force towards the centre is mg − N, so we write:

考虑一辆质量为 m 的汽车以速率 v 驶过凸形桥。受力有向下的重力 mg 和向上的支持力 N。在最高点，加速度指向曲率中心（向下）。指向圆心的合力为 mg − N，因此有：

mg − N = mv² / r

This gives N = mg − mv²/r. Students often incorrectly state that N = mg + mv²/r or that the centripetal force is an extra push downward. The correct picture is that the resultant of the real forces provides the centripetal requirement. As speed increases, N drops; when v = √(gr), the car loses contact with the road because N becomes zero.

由此得出 N = mg − mv²/r。学生常错误地写成 N = mg + mv²/r，或认为向心力是一个额外的向下推力。正确的理解是真实力的合力提供向心力需求。随着速度增加，N 减小；当 v = √(gr) 时，N 变为零，汽车飞离路面。

2. Electric Field Strength vs Electric Potential | 电场强度与电势混淆

It is tempting to assume that where the electric field is strong, the electric potential must also be high, and that on an equipotential surface the field strength is uniform. Both ideas are false. Electric field E is the negative potential gradient: E = − dV/dr. In a uniform field, the potential changes linearly with distance; near a point charge, E and V follow different distance dependences (E ∝ 1/r², V ∝ 1/r).

很容易想当然地认为电场强的地方电势一定高，等势面上电场强度处处相等。这两种想法都是错误的。电场强度 E 等于电势梯度的负值：E = − dV/dr。在匀强电场中，电势随距离线性变化；在点电荷附近，E 与 V 遵循不同的距离关系（E ∝ 1/r²，V ∝ 1/r）。

Consider two charged parallel plates 0.10 m apart with a p.d. of 500 V. The field strength is E = V/d = 5000 V m⁻¹, constant everywhere between the plates. The potential decreases uniformly from 500 V to zero. If you place a positive test charge at the midpoint, it experiences a force but the potential is 250 V — not zero and not 500 V. Many students incorrectly believe that the midpoint has either no field or half the field strength.

考虑两带电平行板相距 0.10 m，电势差 500 V。电场强度 E = V/d = 5000 V m⁻¹，在板间处处恒定。电势从 500 V 均匀降至零。如果在正中间放一个正检验电荷，它受到电场力，但该点电势为 250 V——既不是零也不是 500 V。许多学生错误地认为中点没有电场，或电场强度减半。

3. Capacitor Time Constant vs Half-Life | 电容器时间常数与半衰期

When a capacitor discharges through a resistor, the voltage decays exponentially: V = V₀ e⁻ᵗ/ᴿᶜ. The time constant τ = RC is often misinterpreted as the time for the voltage to drop to half its initial value. In reality, τ is the time for the voltage to fall to V₀/e ≈ 37% of its original value. The half-life t₁/₂ = RC ln 2 ≈ 0.693 RC.

电容器通过电阻放电时，电压按指数衰减：V = V₀ e⁻ᵗ/ᴿᶜ。时间常数 τ = RC 常被误解为电压降至初始值一半所需的时间。实际上，τ 是电压降至初始值的 V₀/e ≈ 37% 所用的时间。半衰期 t₁/₂ = RC ln 2 ≈ 0.693 RC。

If a 100 μF capacitor discharges through a 20 kΩ resistor, τ = 2.0 s. Students often say the voltage halves in 2.0 s, which would give V = ½V₀. However, after 2.0 s the voltage is V₀/e ≈ 0.368V₀. The correct half-life is 2.0 × ln2 ≈ 1.39 s. These errors become critical when analysing timing circuits or interpreting oscilloscope traces.

如果一个 100 μF 的电容器通过 20 kΩ 的电阻放电，τ = 2.0 s。学生常说电压在 2.0 s 内减半，即 V = ½V₀。然而，2.0 s 后电压为 V₀/e ≈ 0.368V₀。正确的半衰期是 2.0 × ln2 ≈ 1.39 s。在分析定时电路或解读示波器波形时，这类错误影响很大。

4. Radioactive Decay: Activity and the Number of Nuclei | 放射性衰变：活度与核数目

Many candidates confuse the activity A with the number of undecayed nuclei N, or forget that A = λN directly links the two. Another common slip is to use the half-life T₁/₂ incorrectly in the decay equation. The decay constant λ is given by λ = ln 2 / T₁/₂, and the number of nuclei after time t is N = N₀ e⁻λt. The activity follows the same exponential law: A = A₀ e⁻λt.

很多考生混淆活度 A 与未衰变核的数目 N，或者忘记 A = λN 直接将两者联系起来。另一个常见失误是在衰变方程中错误地使用半衰期 T₁/₂。衰变常数 λ = ln 2 / T₁/₂，经时间 t 后核数目 N = N₀ e⁻λt。活度遵循相同的指数规律：A = A₀ e⁻λt。

Suppose a source has an initial activity of 800 Bq and a half-life of 10 minutes. After 20 minutes, the activity is not 400 Bq — one half-life gives 400 Bq, and a second gives 200 Bq. Using the formula, λ = ln2 / 10 min⁻¹, t = 20 min, so A = 800 e⁻λ×²⁰ = 800 × (1/2)² = 200 Bq. Students who misapply the time constant often divide incorrectly. Remember that the units of λ must be consistent with time (s⁻¹, min⁻¹, etc.).

假设一个放射源初始活度为 800 Bq，半衰期为 10 分钟。20 分钟后，活度不是 400 Bq——经过一个半衰期降为 400 Bq，再经第二个半衰期降为 200 Bq。用公式计算，λ = ln2 / 10 min⁻¹，t = 20 min，得 A = 800 e⁻λ×²⁰ = 800 × (1/2)² = 200 Bq。错误运用时间常数的学生往往会算错。注意 λ 的单位必须与时间一致（s⁻¹、min⁻¹ 等）。

5. Standing Waves: Node and Antinode Separations | 驻波：波节与波腹距离

A standing wave on a string fixed at both ends produces nodes at the fixed ends. The distance between adjacent nodes (or adjacent antinodes) is half a wavelength, λ/2. A common mistake is to label the entire length of a fundamental mode as one wavelength, when in reality the string length L = λ/2 for the fundamental. For the nth harmonic, L = n(λ/2).

两端固定的弦上形成的驻波，在固定端为波节。相邻波节（或相邻波腹）之间的距离为半个波长，即 λ/2。常见错误是把基频模式下弦的全长标为一个波长，而实际上弦长 L = λ/2。对于第 n 次谐波，L = n(λ/2)。

When a string vibrates in its second harmonic (first overtone), it has three nodes and two antinodes. The length L equals one full wavelength, because L = 2 × (λ/2) = λ. If the frequency is 100 Hz and the string is 0.80 m long, then λ = 0.80 m and the wave speed v = fλ = 80 m s⁻¹. Students often mistakenly take λ = 0.40 m, halving the wavelength and producing a wave speed of only 40 m s⁻¹. Drawing a clear diagram with node–node distances marked helps avoid this slip.

当弦以二次谐波（第一泛音）振动时，共有三个波节和两个波腹。弦长 L 等于一个完整的波长，因为 L = 2 × (λ/2) = λ。若频率为 100 Hz，弦长 0.80 m，则 λ = 0.80 m，波速 v = fλ = 80 m s⁻¹。学生常误认为 λ = 0.40 m，波长减半，算出的波速只有 40 m s⁻¹。画一个清晰的图标出波节间距，可避免此类失误。

6. Photoelectric Effect: Stopping Potential Graph | 光电效应：遏制电压图

The photoelectric experiment yields a straight-line graph of stopping potential V_s against frequency f. Einstein’s equation gives e V_s = h f − φ, where φ is the work function. Rearranging: V_s = (h/e) f − φ/e. Many students incorrectly state that the gradient of this graph is Planck’s constant h. The gradient is actually h/e, because V_s is plotted, not the maximum kinetic energy K_max directly.

光电效应实验得出遏制电压 V_s 对频率 f 的直线图。爱因斯坦方程给出 e V_s = h f − φ，其中 φ 为功函数。整理得：V_s = (h/e) f − φ/e。许多学生错误地认为该图斜率为普朗克常数 h。因为纵坐标是 V_s 而非最大动能 K_max，斜率实际上是 h/e。

If the graph has a gradient of 4.14 × 10⁻¹⁵ V s, the value of h/e is 4.14 × 10⁻¹⁵ V s. Multiplying by the elementary charge e ≈ 1.60 × 10⁻¹⁹ C gives h = 6.63 × 10⁻³⁴ J s. Another mistake is to think that the intercept on the f-axis is φ/h — indeed the threshold frequency f₀ = φ/h. But the intercept on the V_s‑axis is −φ/e. When answering questions, be prepared to convert between V_s and K_max using e. Drawing the axes clearly avoids confusion.

若图线斜率为 4.14 × 10⁻¹⁵ V·s，则 h/e = 4.14 × 10⁻¹⁵ V·s。乘以基本电荷 e ≈ 1.60 × 10⁻¹⁹ C，得到 h = 6.63 × 10⁻³⁴ J·s。另一个错误是以为 f 轴截距为 φ/h——实际上阈值频率 f₀ = φ/h 没错，但 V_s 轴截距是 −φ/e。答题时需能利用 e 在 V_s 与 K_max 间进行转换。清楚标注坐标轴可避免混淆。

7. Magnetic Flux Linkage and Faraday’s Law | 磁链与法拉第定律

Magnetic flux φ = BA cosθ refers to the flux through a single turn; flux linkage is Nφ, where N is the number of turns on a coil. Faraday’s law states that the induced e.m.f. is equal to the rate of change of flux linkage: ε = − d(Nφ)/dt. A persistent error is to use φ alone in Faraday’s law for a coil with multiple turns, forgetting to multiply by N. This mistake can cost several marks in calculation questions.

磁通量 φ = BA cosθ 指穿过单匝的磁通量；磁链为 Nφ，其中 N 是线圈匝数。法拉第定律指出感生电动势等于磁链的变化率：ε = − d(Nφ)/dt。一个常见错误是对多匝线圈使用 φ 代入法拉第定律，忘记乘以 N。这个失误在计算题里可能导致大量失分。

Consider a solenoid with 200 turns, area 5.0 × 10⁻⁴ m², placed in a magnetic field that changes from 0.20 T to zero in 0.050 s. The flux φ through one turn is initially 1.0 × 10⁻⁴ Wb. The change in flux linkage is N × Δφ = 200 × (−1.0 × 10⁻⁴) = −0.020 Wb. The magnitude of the induced e.m.f. is |ε| = 0.020 / 0.050 = 0.40 V. If the N is forgotten, students obtain an e.m.f. of only 0.0020 V. The direction is given by Lenz’s law.

考虑一个 200 匝的螺线管，截面积 5.0 × 10⁻⁴ m²，置于磁场中，磁感应强度在 0.050 s 内从 0.20 T 均匀减至零。单匝磁通量初始为 1.0 × 10⁻⁴ Wb。磁链变化量为 N × Δφ = 200 × (−1.0 × 10⁻⁴) = −0.020 Wb。感生电动势大小为 |ε| = 0.020 / 0.050 = 0.40 V。若遗漏 N，学生会得到仅 0.0020 V 的电动势。方向由楞次定律决定。

8. Binding Energy per Nucleon and Nuclear Stability | 核结合能每核子与核稳定性

The graph of binding energy per nucleon against nucleon number A shows a peak around iron-56 (⁵⁶Fe). Nuclei in this region are the most stable, not the heaviest ones like uranium. A higher binding energy per nucleon means more energy has been released when the nucleus was assembled, making it harder to pull apart. Many students incorrectly assume that because uranium releases energy in fission, it must have the highest binding energy per nucleon.

结合能每核子对核子数 A 的曲线在铁-56（⁵⁶Fe）附近达到峰值。该区域的原子核最稳定，而非铀等最重的核。每核子结合能越高，意味着核子结合成原子核时释放的能量越多，越难将它拆散。许多学生错误地认为，铀在裂变时释放能量，因此它必定具有最高的每核子结合能。

The curve explains why both fission and fusion can release energy. Very heavy nuclei split into fragments with higher binding energy per nucleon, releasing energy (fission). Very light nuclei fuse to form nuclei with higher binding energy per nucleon, also releasing energy (fusion). Iron sits at the top: it cannot release energy by either process. A typical exam question might show a mass defect calculation and ask for the binding energy per nucleon; students often forget to divide by the number of nucleons. Always check your denominator.

该曲线解释了为什么裂变和聚变都能释放能量。很重的核分裂成每核子结合能更大的碎块，释放能量（裂变）；很轻的核聚变成每核子结合能更大的核，也释放能量（聚变）。铁位于曲线顶峰：无论裂变还是聚变，它都不能释放能量。考试常见题型给出质量亏损，要求计算每核子结合能；学生常忘记除以总核子数。务必检查分母。

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A2 Physics: Common Mistake Questions Explained | A2 物理：易错题精讲