高斯面

高斯面（Gaussian surface），又称高斯曲面，是三维空间一闭合曲面，用于运用高斯定理计算向量场的通量，例如重力场、电场和磁场。是任意形状的封闭曲面1=S = ∂V（3维V)流形边界），通过对其进行曲面积分运算，可以求出曲面中包含的场源总量，例如重力场中包含的物质总量和静电场场源中包含的总电荷量等等，也可以反过来从场源推算它产生的场。例如这里所举的最常见的情况，运用高斯曲面和高斯定理计算电场的时候，运用对称性选择恰当的高斯面，可以简化所研究的问题，使曲面积分更简单。如果高斯曲面上的每一点都能使该点垂直曲面的电场分量为常数，进行曲面积分的时候就能大大简化运算，因为常数可以从积分式中被提取出来。

## 常见的高斯曲面

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大多数使用高斯曲面计算都是从研究高斯电定律开始：

$\Phi_E =$ $\scriptstyle S$ $\mathbf E\;\cdot\mathrm{d}\mathbf A = \frac{Q_{\text{enc}}}{\varepsilon_0}.$

从而Qenc是被高斯曲面包围的电荷。

这是结合了高斯散度定理和库仑定律的高斯定律。

### 高斯球面

当找到由以下任何一种产生的电场或通量时，可使用高斯球面：

- a point charge
- a uniformly distributed spherical shell of charge
- any other charge distribution with spherical symmetry

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting 1=QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

With the same example, using a larger Gaussian surface outside the shell where r > R, Gauss's law will produce a non-zero electric field. This is determined as follows.

球面S通量为：

$\Phi_E =$ $\scriptstyle \partial S$ $\mathbf{E}\cdot d \mathbf{A} = \iint_c E dA\cos 0^\circ = E \iint_S dA$

The surface area of the sphere of radius r is $$\iint_S dA = 4 \pi r^2$$ which implies $$\Phi_E = E 4\pi r^2$$

By Gauss's law the flux is also $$\Phi_E =\frac{Q_A}{\varepsilon_0}$$ finally equating the expression for ΦE gives the magnitude of the E-field at position r:

$$E 4\pi r^2 = \frac{Q_A}{\varepsilon_0} \quad \Rightarrow \quad E=\frac{Q_A}{4\pi\varepsilon_0r^2}.$$

This non-trivial result shows that any spherical distribution of charge acts as a point charge when observed from the outside of the charge distribution; this is in fact a verification of Coulomb's law. And, as mentioned, any exterior charges do not count.

### 高斯圆柱面

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当找到由以下任何一种产生的电场或通量时，可使用高斯圆柱面：

- 一条无限长的均匀电荷线
- 一个无限均匀电荷平面
- 一个无限长的均匀电荷圆柱体

例如“无限线电荷附近的场”如下所示。

Consider a point P at a distance r from an infinite line charge having charge density (charge per unit length) λ. Imagine a closed surface in the form of cylinder whose axis of rotation is the line charge. If h is the length of the cylinder, then the charge enclosed in the cylinder is $$q = \lambda h ,$$ where q is the charge enclosed in the Gaussian surface. There are three surfaces a, b and c as shown in the figure. The differential vector area is dA, on each surface a, b and c.

The flux passing consists of the three contributions:

$\Phi_E =$ $\scriptstyle A$ $\mathbf{E} \cdot d\mathbf{A} = \iint_a \mathbf{E} \cdot d\mathbf{A} + \iint_b\mathbf{E} \cdot d\mathbf{A} + \iint_c\mathbf{E} \cdot d\mathbf{A}$

For surfaces a and b, E and dA will be perpendicular.

For surface c, E and dA will be parallel, as shown in the figure.

$$\begin{align} \Phi_E & = \iint_a E dA\cos 90^\circ + \iint_b E d A \cos 90^\circ + \iint_c E d A\cos 0^\circ \\ & = E \iint_c dA \end{align}$$

The surface area of the cylinder is $$\iint_c dA = 2 \pi r h$$ which implies $$\Phi_E = E 2 \pi r h.$$

By Gauss's law $$\Phi_E = \frac{q}{\varepsilon_0}$$ equating for ΦE yields $$E 2 \pi rh = \frac{\lambda h}{\varepsilon_0} \quad \Rightarrow \quad E = \frac{\lambda}{2 \pi\varepsilon_0 r}$$

### 高斯盒（pillbox）

This surface is most often used to determine the electric field due to an infinite sheet of charge with uniform charge density, or a slab of charge with some finite thickness. The pillbox has a cylindrical shape, and can be thought of as consisting of three components: the disk at one end of the cylinder with area πR2, the disk at the other end with equal area, and the side of the cylinder. The sum of the electric flux through each component of the surface is proportional to the enclosed charge of the pillbox, as dictated by Gauss's Law. Because the field close to the sheet can be approximated as constant, the pillbox is oriented in a way so that the field lines penetrate the disks at the ends of the field at a perpendicular angle and the side of the cylinder are parallel to the field lines.

## 进一步阅读

- Electromagnetism (2nd Edition), I.S. Grant, W.R. Phillips, Manchester Physics, John Wiley & Sons, 2008,