AP物理2 电容

AP 物理 2· AP物理2 CED — 电场力、电场与电势· 14 分钟阅读

1. 什么是电容？★☆☆☆☆⏱ 2 min

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。

📘 定义

电容

$C$

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 $1 F = 1 C/V$ 。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。

例:

实际电容器的电容值大多在 $1 0^{- 12} F$ （皮法，pF）到 $1 0^{- 6} F$ （微法，μF）之间。

2. 平行板电容★★☆☆☆⏱ 4 min

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

目标:

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 $σ = \frac{Q}{A}$ 。
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:

结果:

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 $2.0 \times 1 0^{- 3}$ 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 = $1.0 \times 1 0^{- 4}$ m, $ϵ_{0} = 8.85 \times 1 0^{- 12}$ 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:
$Q = C V = (1.77 \times 10^{-10} \text{ F})(12 \text{ V}) = 2.12 \times 10^{-9} \text{ C} = 2.1 \text{ nC}$

Exam tip:

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。

3. 电容器的组合★★★☆☆⏱ 4 min

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}

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

Exam tip:

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。

4. 电介质与电容器储能★★★☆☆⏱ 4 min

Most practical capacitors use an insulating material called a dielectric between their plates. Dielectrics increase capacitance by a dimensionless factor called the dielectric constant $κ$ , where $κ > 1$ 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 $κ = 2.5$ 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 $κ$ :
$C_f = \kappa C_i = 2.5 \times 10 \mu\text{F} = 25 \mu\text{F} = 25 \times 10^{-6} \text{ F}$
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:
$U = \frac{(9 \times 10^{-5})^2}{2 (25 \times 10^{-6})} = 1.62 \times 10^{-4} \text{ J} = 162 \mu\text{J}$

✓ 快速检测

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

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
显示答案
1 —
For a parallel-plate capacitor, $C = \frac{ϵ _{0} A}{d}$ , so doubling $d$ halves $C$ . Since $V$ is constant (battery connected), $Q = C V$ is also halved.

Exam tip:

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。

5. 常见陷阱

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

6. 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)	$C = \frac{ϵ _{0} A}{d}$	A = plate area, d = plate separation
Parallel-Plate (with dielectric)	$C = \frac{κ ϵ _{0} A}{d}$	$κ > 1$ = dielectric constant, fills full gap
Parallel Equivalent	$C_{e q} = C_{1} + C_{2} + ... + C_{n}$	All capacitors share same potential difference V
Series Equivalent	$\frac{1}{C _{e q}} = \frac{1}{C _{1}} + ... + \frac{1}{C _{n}}$	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	$u = \frac{1}{2} κ ϵ_{0} E^{2}$ ,

真题中的出现

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