Capacitance — 题型分析

Question Type 1: Definition of Capacitance

如何识别

题目包含 "Define capacitance" 或 "State what is meant by the capacitance of a capacitor"。

标准解题方法

解题要点

• Capacitance = charge per unit potential difference
• 需明确是 "charge on one plate" 和 "potential difference across the plates"

评分标准

评分模式

• M1: charge per unit potential (difference)
• A1: charge on one plate and p.d. across the plates

完整原题

Example 1 — 9702/41/O/N/20 Q6(a)(i) (2 marks):

Define the capacitance of a parallel plate capacitor.

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MS:

• M1: charge per unit potential (difference)
• A1: charge on one plate and potential difference across the plates

Example 2 — 9702/41/M/J/21 Q7(a) (2 marks):

State what is meant by the capacitance of a parallel plate capacitor.

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MS:

• M1: charge per unit potential (difference)
• A1: (charge stored) on one plate / p.d. across the capacitor

常见陷阱

常见陷阱

• 只说 "charge per unit potential difference" 可能不够——需要提 "on one plate"
• 不是 "charge on both plates"——正负电荷分别在两个板上

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Question Type 2: Combined Capacitance Calculations

如何识别

题目给出多个电容，要求计算总电容。

标准解题方法

解题步骤

1. 串联：$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \dots$
2. 并联：$C = C_1 + C_2 + \dots$
3. 注意：电容串并联公式与电阻相反！
4. 复杂电路：先找纯并联或纯串联的部分，逐步化简

评分标准

评分模式

• C1: 写出正确的公式/计算部分组合
• A1: 最终答案（含单位）

完整原题

Example 1 — 9702/41/O/N/20 Q6(b) (2 marks):

A student has four capacitors, each 24 $\mu$F. They are connected as shown — three in series (two parallel branches with two capacitors) between X and Y. Calculate the combined capacitance.

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MS:

• C1: series combination: $1/C = 1/24 + 1/24 + 1/24$ → $C = 8$ $\mu$F
• A1: total $C = 8 + 24 = 32$ $\mu$F

Example 2 — 9702/41/M/J/21 Q7(b)(ii) (2 marks):

A capacitor of capacitance C is connected into the circuit shown. For V = 150 V and f = 60 Hz, the average current is 4.8 $\mu$A. Calculate the capacitance in pF.

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MS:

• C1: $I = fCV$
• C1: $C = I/(fV) = 4.8 \times 10^{-6} / (60 \times 150)$
• A1: $= 5.3 \times 10^{-10}$ F $= 530$ pF

常见陷阱

常见陷阱

• 电容串联公式与电阻相反：串联 $\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$，并联 $C = C_1 + C_2$
• 两个相同电容串联：$C = C_1/2$，不是 $C_1/4$
• 单位：1 $\mu$F $= 10^{-6}$ F, 1 pF $= 10^{-12}$ F

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Question Type 3: Energy Stored in a Capacitor

如何识别

题目要求计算电容器储存的能量，或使用 $W = \frac{1}{2}QV$。

标准解题方法

解题步骤

1. 从 $V$-$Q$ 图，面积 = $\frac{1}{2} QV$ = 储能
2. 公式：$W = \frac{1}{2} QV = \frac{1}{2} CV^2 = \frac{1}{2} Q^2 / C$
3. 选择最方便的公式（取决于已知量）

常见陷阱

常见陷阱

• 能量不是 $QV$，而是 $\frac{1}{2} QV$（因为充电过程中 $V$ 从 0 增加到最终值）
• $W = \frac{1}{2} CV^2$ 中的 $V$ 是最终电压，不是变化量

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Question Type 4: RC Discharge — Graphs and Calculations

如何识别

题目给出 RC 放电电路，要求计算时间常数、电流、电荷等，或 sketch 放电曲线。

标准解题方法

解题步骤

1. 初始值（$t = 0$）：
   • $Q_0 = CV_0$
   • $I_0 = V_0 / R$
2. 时间常数：$\tau = RC$（单位：s）
3. 放电公式：
   • $Q = Q_0 e^{-t/RC}$
   • $V = V_0 e^{-t/RC}$
   • $I = I_0 e^{-t/RC}$
4. $t = RC$ 时：$x = x_0 e^{-1} \approx 0.37 x_0$
5. 半衰期：$t_{1/2} = RC \ln 2 \approx 0.693 RC$

评分标准

评分模式

• C1: 写出 $Q = CV$ 或 $I = V/R$ 求初始值
• C1: $\tau = RC$ 或指数公式
• A1: 正确答案（含单位）
• 作图：B1 指数衰减形状 + B1 标注初始值或时间常数

完整原题

Example 1 — 9702/41/O/N/22 Q5 (11 marks):

A capacitor of 470 $\mu$F is connected to a 24 V battery. Switch S is moved to Y so the capacitor discharges through wire P (5.6 k$\Omega$). (a)(i) Calculate the charge on the capacitor at t = 0. (a)(ii) Calculate the current in wire P at t = 0. (a)(iii) Calculate the time constant of the discharge circuit. (a)(iv) On Fig. 5.2, sketch the variation of I with t.

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MS:

• (a)(i) C1: $Q = CV$
• (a)(i) A1: $= 470 \times 10^{-6} \times 24 = 1.13 \times 10^{-2}$ C
• (a)(ii) A1: $I = V/R = 24 / 5600 = 4.3 \times 10^{-3}$ A
• (a)(iii) C1: $\tau = RC$
• (a)(iii) A1: $= 5600 \times 470 \times 10^{-6} = 2.6$ s
• (a)(iv) B1: exponential decay starting at $I_0$
• (a)(iv) B1: correct shape approaching zero asymptotically

Example 2 — 9702/41/M/J/23 Q5 (9 marks, part)

Fig. 5.2 shows the variation of $V_{IN}$ with t. Fig. 5.3 shows $V_{OUT}$ with t. Show that the time constant $\tau$ for the discharge of the capacitor through the resistor is 0.038 s. Calculate the capacitance C.

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MS:

• (b)(ii) C1: $V_{OUT} = V_0 e^{-t/RC}$
• (b)(ii) C1: $0.50 = 7.5 e^{-t/RC}$ or reads time from graph
• (b)(ii) A1: shows that $\tau = 0.038$ s
• (b)(iii) C1: $C = \tau / R = 0.038 / 14000$
• (b)(iii) A1: $= 2.7 \times 10^{-6}$ F (unit required)

常见陷阱

常见陷阱

• 放电是 $x = x_0 e^{-t/RC}$（指数衰减），不是 $x = x_0 (1 - e^{-t/RC})$（这是充电公式）
• 时间常数 $\tau = RC$，$R$ 和 $C$ 都用 SI 单位（$\Omega$ 和 F），结果才是秒
• 放电时电流方向与充电时相反
• $t = RC$ 时 $x / x_0 = e^{-1} \approx 0.37$，不是 $0.5$
• 半衰期 $t_{1/2} = RC \ln 2 \approx 0.693 RC$