高斯面

高斯面（Gaussian surface），又稱高斯曲面，是三維空間一閉合曲面，用於運用高斯定理計算向量場的通量，例如重力場、電場和磁場。是任意形狀的封閉曲面1=S = ∂V（3維V)流形邊界），通過對其進行曲面積分運算，可以求出曲面中包含的場源總量，例如重力場中包含的物質總量和靜電場場源中包含的總電荷量等等，也可以反過來從場源推算它產生的場。例如這裡所舉的最常見的情況，運用高斯曲面和高斯定理計算電場的時候，運用對稱性選擇恰當的高斯面，可以簡化所研究的問題，使曲面積分更簡單。如果高斯曲面上的每一點都能使該點垂直曲面的電場分量為常數，進行曲面積分的時候就能大大簡化運算，因為常數可以從積分式中被提取出來。

## 常見的高斯曲面

大多數使用高斯曲面計算都是從研究高斯電定律開始：

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

從而Qenc是被高斯曲面包圍的電荷。

這是結合了高斯散度定理和庫侖定律的高斯定律。

### 高斯球面

當找到由以下任何一種產生的電場或通量時，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通量為：

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

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.

### 高斯圓柱面

當找到由以下任何一種產生的電場或通量時，可使用高斯圓柱面：

- 一條無限長的均勻電荷線
- 一個無限均勻電荷平面
- 一個無限長的均勻電荷圓柱體

例如「無限線電荷附近的場」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.

## 進一步閱讀

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