圆柱电容公式
圆柱形电容器是一种在少量区域内存储大量电流的设备。它由一个同心的空心球形圆柱体围绕着一个空心或实心圆柱导体组成。电动机、谷物碾磨机、电动榨汁机和其他电器都使用圆柱形电容器。电容器旨在将电场线限制在有限区域内。结果，即使电场非常强，电容器的两个导体之间的电位差也非常小。尽管可以通过多种方式对电容器进行分组以获得必要的电容，但最常见的配置是串联和并联。本文介绍了圆柱电容的计算公式。
圆柱形电容器
Capacitance per unit length is the capacitance for a cylindrical geometry.
每个电容器的电位差都是独一无二的。有许多电路，其中电容器必须以正确的顺序排列以获得所需的电容。串联电容器和并联电容器是两种最普遍的配置。法拉是电容 (F) 的标准单位。对于电荷的存储，通常使用圆柱形电容器。
圆柱电容公式
圆柱形电容器的电容可以使用以下公式计算：
C = 2πε0 × (L / ln(b/a))
Where,
- C = Capacitance of Cylinder,
- ε0 = Permittivity of free space,
- a = Inner radius of cylinder,
- b = Outer radius of cylinder,
- L = Length of cylinder.
示例问题
问题一：为什么有这么多圆柱形电容器？
回答：
Two conducting plates are separated by an insulating gap in the general form of the capacitor. Metal foils, metalized polymer films, electrolytes, and other materials can all be used to create the insulating and conductive components. All of the items mentioned above can be built in a flat rectangular shape. Some capacitors aren’t spherical in shape. Stacking layers of insulating and conducting material alternately form them. Multilayer ceramic chip capacitors, for instance, are a good example.
问题2：圆柱形电容器的组成部分是什么？给几个用途。
回答：
A cylindrical capacitor consists of a cylindrical conductor encircled by a hollow spherical cylinder. Electric motors, grain mills, electric juicers, and other electrical instruments are among the most common uses for cylindrical capacitors.
问题 3：计算一个用纸包装的圆柱形电容器的电容，该电容器由内半径和外半径分别为 2 厘米和 5 厘米的圆柱体组成。
回答：
We substitute the permittivity, which equals 3.85 for paper, in the formula for the Capacitance of Cylindrical Capacitor, resulting in:
Given : ε0 = 8.85 × 10-12, L = 3.85, b = 5 cm, a = 2 cm.
Since,
C = 2πε0 × (L/ln(b/a))
= 2 × 3.14 × 8.85 × 10-12 × (3.85 / ln(5/2))
= 538.22 × 10-12 F
问题 4：圆柱形电容器由两个内径为 6 厘米、外径为 12 厘米的环组成。根据 16 cm 的长度计算电容器的电容。
回答：
Given : a = 6 cm, b = 12 cm, L = 16 cm, ε0 = 8.85 × 10-12
Since,
C = 2πε0 × (L/ln(b/a))
= 2 × 3.14 × 8.85 × 10-12 × (16 × 10-2 / ln(12/6))
= 2.954 × 10-11 F
问题5：一个8厘米长的圆柱形电容器由两个内半径为3厘米，外半径为6厘米的同心环组成。计算电容器的电容。
回答：
Given : a = 3 cm, b = 6 cm, L = 8 cm, ε0 = 8.85 × 10-12
Since,
C = 2πε0 × (L/ln(b/a))
C = 2 × 3.14 × 8.85 × 10-12 × (8 × 10-2 / ln(6/3))
= 1.477 × 10-11 F
问题 6：圆柱形电容器的长度为 9 厘米。它由两个内半径和外半径分别为 2 厘米和 7 厘米的同心环组成。计算电容器的电容。
回答：
Given : a = 2 cm, b = 7 cm, L = 9 cm, ε0 = 8.85 × 10-12
Since,
C = 2πε0 × (L/ln(b/a))
C = 2 × 3.14 × 8.85 × 10-12 × (9 × 10-2 / ln(7/2))
= 9.19488 × 10-12 F