# AP物理2 电容

> AP 物理 2 · 第3单元：电场力、电场与电势
> 来源: https://www.owlsprep.com/zh/study/ap-physics-2-u3-capacitance/

本模块讲解AP物理2的核心电容概念，包括定义、平行板电容、串并联组合、电介质和储能。包含例题、常见误区和符合考试要求的解题策略。

**先修:** [电势和电势差](https://www.owlsprep.com/zh/study/ap-physics-2-electric-potential/); 匀强电场与电势的关系; [电场高斯定律](https://www.owlsprep.com/zh/study/ap-physics-2-gauss-law/)

## 学习目标

- 定义电容并识别其固有属性
- 计算有、无电介质时平行板电容器的电容
- 求解串联和并联电容器组合的等效电容
- 计算不同题目条件下电容器储存的能量
- 解决AP风格的带电电容器插入电介质相关问题

## 什么是电容？

电容描述了一对相互分开的导体储存分离电荷，进而储存电势能的能力。它占AP物理2考试总分的2-3%，在选择题（MCQ）和简答题（FRQ）中都会考查，常结合电场/电势概念或电路问题出题。

**电容** — 一个导体上储存的电荷量大小$Q$与两个导体之间电势差大小$V$的比值：$C = \frac{Q}{V}$。国际单位制单位是法拉（F），其中$1\ \text{F} = 1\ \text{C/V}$。电容是固有属性，仅取决于导体的几何形状和极板间的材料，与储存的电荷量或施加的电势差无关。

*记法:* $C$

*例:* 实际电容器的电容值大多在$10^{-12}\ \text{F}$（皮法，pF）到$10^{-6}\ \text{F}$（微法，μF）之间。

> **tip**
>
> 对于固定电容器，改变$Q$或$V$不会改变$C$，只会改变另一个量。只有几何形状或极板间材料改变时，$C$才会改变。

## 平行板电容

AP物理2考试中最常考的电容器结构是平行板电容器：两块相同的平行导电极板，间距$d$均匀，极板间为真空或空气。

**推导:** 推导真空填充平行板电容器的电容

*起点:* 电场高斯定律，以及匀强电场与电势差的关系

1. 两块极板分别带电荷$+Q$和$-Q$，总极板面积为$A$。面电荷密度为$\sigma = \frac{Q}{A}$。
2. 根据高斯定律，极板间的电场为：
3. $$E = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}$$
4. 对于匀强电场，极板间电势差为$V = E d$。代入$E$：
5. $$V = \frac{Q d}{\epsilon_0 A}$$
6. Rearrange using definition $C = \frac{Q}{V}$ to get the final capacitance:

*结论:* 平行板电容器的电容与极板面积$A$成正比，与极板间距$d$成反比。

$$C = \frac{\epsilon_0 A}{d}$$

**例题:** 一个空气填充的平行板电容器，极板面积为$2.0 \times 10^{-3}$ m²，极板间距为0.10 mm。电容器连接到12 V电池完成充电。计算(a)电容和(b)正极板储存的总电荷。

1. Convert all values to SI units: plate separation $d = 0.10$ mm = $1.0 \times 10^{-4}$ m, $\epsilon_0 = 8.85 \times 10^{-12}$ C²/N·m².
2. Substitute into the parallel-plate capacitance formula:

   $$C = \frac{\epsilon_0 A}{d} = \frac{(8.85 \times 10^{-12})(2.0 \times 10^{-3})}{1.0 \times 10^{-4}} = 1.77 \times 10^{-10} \text{ F} = 177 \text{ pF}$$
3. Use the definition of capacitance to solve for charge:

   $$Q = C V = (1.77 \times 10^{-10} \text{ F})(12 \text{ V}) = 2.12 \times 10^{-9} \text{ C} = 2.1 \text{ nC}$$

> **考试提示:** 代入电容公式前一定要先将单位转换为国际单位制；极板间距通常以毫米或微米为单位给出，忘记转换会导致答案偏差3或6个数量级，这是选择题中常见的陷阱。

## 电容器的组合

电容器在电路中几乎都是组合使用的。需要记住的关键点：电容器的组合规则与电阻的组合规则相反。并联电容器的所有极板电势差相同，而串联电容器的所有极板储存电荷量相同。

For capacitors in parallel, total charge stored is the sum of individual charges. Substituting $Q = C V$ and canceling the common $V$ gives:

$$C_{eq, parallel} = C_1 + C_2 + ... + C_n$$

For capacitors in series, total potential difference across the combination is the sum of individual potential differences. Substituting $V = Q/C$ and canceling the common $Q$ gives:

$$\frac{1}{C_{eq, series}} = \frac{1}{C_1} + \frac{1}{C_2} + ... + \frac{1}{C_n}$$

> **tip**
>
> Intuition check: adding capacitors in parallel increases effective plate area, so equivalent capacitance is larger than any individual. Adding capacitors in series increases effective plate separation, so equivalent capacitance is smaller than any individual.

**例题:** Three capacitors with capacitances 2 μF, 3 μF, and 6 μF are connected as follows: 2 μF and 3 μF are in parallel with each other, and this parallel combination is in series with the 6 μF capacitor. Find the total equivalent capacitance of the combination.

1. First simplify the innermost parallel combination, working outward from nested combinations:

   $$C_{parallel} = C_1 + C_2 = 2 \mu\text{F} + 3 \mu\text{F} = 5 \mu\text{F}$$
2. Now this 5 μF combination is in series with the 6 μF capacitor. Apply the series rule:

   $$\frac{1}{C_{eq}} = \frac{1}{C_{parallel}} + \frac{1}{C_3} = \frac{1}{5 \mu\text{F}} + \frac{1}{6 \mu\text{F}}$$
3. Calculate the sum of reciprocals:

   $$\frac{1}{C_{eq}} = \frac{6 + 5}{30 \mu\text{F}} = \frac{11}{30 \mu\text{F}}$$
4. Invert to get equivalent capacitance, check against intuition:

   $$C_{eq} = \frac{30}{11} \approx 2.7 \mu\text{F}$$

> **考试提示:** 记住电容组合规则与电阻组合规则相反，混淆规则是本知识点最常见的错误。

## 电介质与电容器储能

Most practical capacitors use an insulating material called a dielectric between their plates. Dielectrics increase capacitance by a dimensionless factor called the dielectric constant $\kappa$, where $\kappa > 1$ for all insulating materials. Dielectrics polarize in the electric field between plates, reducing the net electric field for a given stored charge, which increases capacitance per the definition $C = Q/V$.

When a dielectric fills the entire gap between plates of a parallel-plate capacitor, the capacitance becomes:

$$C = \kappa \frac{\epsilon_0 A}{d}$$

Work done to separate charge on a capacitor is stored as electric potential energy. There are three equivalent forms for stored energy:

$$U = \frac{1}{2} Q V = \frac{1}{2} C V^2 = \frac{Q^2}{2 C}$$

The energy is stored in the electric field between the plates, with energy density (energy per unit volume):

$$u = \frac{1}{2} \kappa \epsilon_0 E^2$$

**例题:** A parallel-plate capacitor with capacitance 10 μF is connected to a 9 V battery to charge it. After charging, the battery is disconnected, and a dielectric with $\kappa = 2.5$ is inserted between the plates, filling the entire gap. Find the new energy stored in the capacitor after insertion.

1. Calculate initial charge before insertion. Since the battery is disconnected, $Q$ remains constant:

   $$Q = C_i V_i = (10 \times 10^{-6} \text{ F})(9 \text{ V}) = 9 \times 10^{-5} \text{ C}$$
2. Inserting the dielectric increases capacitance by a factor of $\kappa$:

   $$C_f = \kappa C_i = 2.5 \times 10 \mu\text{F} = 25 \mu\text{F} = 25 \times 10^{-6} \text{ F}$$
3. Use the energy formula that depends on constant $Q$ to avoid errors:

   $$U = \frac{Q^2}{2 C_f}$$
4. Substitute values to get final energy:

   $$U = \frac{(9 \times 10^{-5})^2}{2 (25 \times 10^{-6})} = 1.62 \times 10^{-4} \text{ J} = 162 \mu\text{J}$$

**概念自测**

Test your understanding with this AP-style multiple choice question:

1. A parallel-plate air-filled capacitor is connected to a battery that maintains a constant potential difference $V$ across its plates. The separation between the plates is slowly doubled, while the plate area remains unchanged. Which of the following correctly describes the resulting change in capacitance $C$ and total stored charge $Q$?

   - A) $C$ doubles, $Q$ doubles
   - B) $C$ is halved, $Q$ is halved
   - C) $C$ doubles, $Q$ remains constant
   - D) $C$ is halved, $Q$ remains constant

   *答案:* B) $C$ is halved, $Q$ is halved

   *解析:* For a parallel-plate capacitor, $C = \frac{\epsilon_0 A}{d}$, so doubling $d$ halves $C$. Since $V$ is constant (battery connected), $Q = CV$ is also halved.

> **考试提示:** 插入/取出电介质前，一定要先检查电容器是否还连接在电池上（$V$恒定）还是已经断开（$Q$恒定）。这会改变各物理量的变化规律，以及你应该使用的公式。

## 常见错误

- **错误做法:** 电容器连接电池时，假设几何变化后$Q$保持不变来计算电容
  - 原因: 学生错误地认为无论电池是否连接Q都是恒定的，而当V保持恒定时Q实际上会变化
  - 正确做法: 解决任何几何变化或电介质变化的问题前，一定要先确定电容器是连接电池（V恒定）还是断开电池（Q恒定）
- **错误做法:** 对电容使用电阻的串并联规则（例如并联电容取倒数相加）
  - 原因: 学生混淆了相反的规则，分不清每种组合类型中哪个物理量恒定
  - 正确做法: 并联电容电势差V相同，因此直接加电容；串联电容电荷量Q相同，因此取电容的倒数相加
- **错误做法:** 计算平行板电容时，忘记将极板间距从毫米转换为米
  - 原因: 题目为了方便会用毫米表示小间距，学生因为数值很小就跳过单位转换
  - 正确做法: 每个平行板问题开始时，先写下所有给定值并完成单位转换，再代入公式
- **错误做法:** Using $U = 1/2 C V^2$ when Q is constant (battery disconnected) and concluding energy increases when a dielectric is inserted
  - 原因: Students pick the wrong energy formula without checking which quantity is constant
  - 正确做法: After identifying which quantity is constant, pick the energy formula that uses that constant quantity to avoid errors from changing variables
- **错误做法:** Assuming that changing the voltage applied to a capacitor changes its capacitance
  - 原因: Students confuse the definition $C = Q/V$ with a proportionality, thinking C depends on V or Q
  - 正确做法: Remember that C is intrinsic to geometry and dielectric, so changing V or Q only changes the other variable, not C
- **错误做法:** Calculating equivalent capacitance for mixed combinations by simplifying outer combinations first
  - 原因: Students don't map the circuit correctly and simplify the wrong combination first
  - 正确做法: Start simplifying from the innermost (most nested) combination, working outward toward the battery terminals

## 速查表

| 类别 | 公式 | 注释 |
| --- | --- | --- |
| Definition of Capacitance | $C = \frac{Q}{V}$ | C is intrinsic, independent of Q/V; 1 F = 1 C/V |
| Parallel-Plate (vacuum/air) | $C = \frac{\epsilon_0 A}{d}$ | A = plate area, d = plate separation |
| Parallel-Plate (with dielectric) | $C = \frac{\kappa \epsilon_0 A}{d}$ | $\kappa >1$ = dielectric constant, fills full gap |
| Parallel Equivalent | $C_{eq} = C_1 + C_2 + ... + C_n$ | All capacitors share same potential difference V |
| Series Equivalent | $\frac{1}{C_{eq}} = \frac{1}{C_1} + ... + \frac{1}{C_n}$ | All capacitors share same stored charge Q |
| Stored Energy | $U = \frac{1}{2} Q V = \frac{1}{2} C V^2 = \frac{Q^2}{2 C}$ | Use form matching your constant: Q if disconnected, V if connected |
| Electric Field Energy Density | $u = \frac{1}{2} \kappa \epsilon_0 E^2$, |  |

---

来自 [OwlsPrep](https://www.owlsprep.com) —— A-Level / IB / AP / IGCSE 免费学习指南，依据官方考纲编写。原页面：https://www.owlsprep.com/zh/study/ap-physics-2-u3-capacitance/