# Chapter 23

## Homework

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### **Problem 1**

A vertical electric field of magnitude $2.00 × 10^4 N/C$ exists above the Earth’s surface on a day when a [thunderstorm](https://dictionary.cambridge.org/zht/%E8%A9%9E%E5%85%B8/%E8%8B%B1%E8%AA%9E-%E6%BC%A2%E8%AA%9E-%E7%B9%81%E9%AB%94/thunderstorm)(雷雨) is brewing. A car with a rectangular size of 6.00 m by $3.00 m$ is traveling along a roadway sloping downward at 10.0°. Determine the electric flux through the bottom of the car.

**Question Translation:**

在雷暴即將來臨的日子，地表上方存在一個強度為 $2.00 \times 10^4 N/C$ 的垂直電場。一輛長 6.00 m、寬 3.00 m 的汽車沿著 $10.0^\circ$ 的下坡道路行駛。求汽車底部的電通量。

**Solution:**
電通量由公式計算：
$\Phi = E A \cos \theta$
其中：

- $E = 2.00 \times 10^4$ N/C(電場強度)
- $A = 6.00 \times 3.00 = 18.0 m^2$（底部面積）
- $\theta = 10.0^\circ$（傾斜角）

$\Phi = (2.00 \times 10^4) \times (18.0) \times \cos 10.0^\circ$

$\Phi \approx 3.54 \times 10^5 \, \text{N} \cdot \text{m}^2 / \text{C}$

**Answer:** $3.54 \times 10^5$ N·m²/C

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### **Problem 2**

A cone with base radius R and height h is located on a horizontal table. A horizontal uniform field E penetrates the cone, as shown in Figure P24.2. Determine the electric flux that enters the left-hand side of the cone.


**Question Translation:**
一個底半徑為 $R$、高為 $h$ 的圓錐放置於水平桌面上。一個水平均勻電場 $E$ 穿過圓錐，求進入圓錐左側的電通量。

**Solution:**
由於電場是水平的，只考慮側面積三角形。由高斯定律，電通量通過的部分為：
$\Phi = EA$

垂直錐體的平面: $A = \frac{1}{2} \times 2 R h$

$\Phi = E \times R h$

**Answer:**
$\Phi = E R h$

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### **Problem 3**

Four closed surfaces, S1 through S4, together with the charges –2Q, Q, and –Q are sketched in Figure P24.3. (The colored lines are the intersections of the surfaces with the page.) Find the electric flux through each surface.


**Question Translation:**
四個封閉曲面 $S_1$ 到 $S_4$ 包圍了電荷 $-2Q$、$Q$ 和 $-Q$，求每個表面的電通量。

**Solution:**
根據高斯定律：
$\Phi = \frac{q}{\varepsilon_0}$
對每個表面計算封閉的淨電荷，然後應用公式。

**Answer:**

$\Phi_{s1} = \frac{-Q}{\varepsilon_0}$

$\Phi_{s2} = 0$

$\Phi_{s3} = \frac{-2Q}{\varepsilon_0}$

$\Phi_{s4} = 0$

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### **Problem 4**

A point charge of $12.0 μC$ is placed at the center of a spherical shell of radius 22.0 cm. What is the total electric flux through
(a) the surface of the shell and
(b) any [hemispherical](https://dictionary.cambridge.org/zht/%E8%A9%9E%E5%85%B8/%E8%8B%B1%E8%AA%9E-%E6%BC%A2%E8%AA%9E-%E7%B9%81%E9%AB%94/hemispherical)(半球形) surface of the shell?
(c) Do the results depend on the radius? Explain.

**Question Translation:**
一個 $12.0 μC$ 的點電荷被放置在半徑為 $22.0 cm$ 的球殼中心。
(a) 求球殼表面的總電通量
(b) 求球殼任意半球面的電通
(c) 結果是否取決於半徑？

**Solution:**
根據高斯定律，封閉曲面的電通量僅與內部總電荷有關：
$\Phi = \frac{q}{\varepsilon_0}$

(a)
$\Phi = \frac{12.0 \times 10^{-6}}{8.85 \times 10^{-12}}$
$\Phi \approx 1.36 \times 10^6 \, \text{N} \cdot \text{m}^2 / \text{C}$

(b) 半球通量為總通量的一半：
$\Phi_{\text{hemisphere}} = \frac{1}{2} \Phi \approx 6.78 \times 10^5 \, \text{N} \cdot \text{m}^2 / \text{C}$
(c) 由於高斯定律，通量與半徑無關。

**Answer:**
(a) $1.36 \times 10^6$ N·m²/C

(b) $6.78 \times 10^5$ N·m²/C

(c) 不取決於半徑。

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### **Problem 5**

A charge of $170 μC$ is at the center of a cube of edge $80.0 cm$.
(a) Find the total flux through each face of the cube.
(b) Find the flux through the whole surface of the cube.
(c) What If? Would your answers to parts (a) or (b) change if the charge were not at the center? Explain.

**Question Translation:**
一個 170 μC 的電荷位於邊長 80.0 cm 的立方體中心。
(a) 求每個面的總電通量
(b) 求整個立方體的總電通量
(c) 如果電荷不是位於中心，答案會改變嗎？解釋。

**Solution:**
由高斯定律，
$\Phi_{\text{total}} = \frac{q}{\varepsilon_0}$
立方體有 6 個面，因此每個面的通量：
$\Phi_{\text{face}} = \frac{q}{6 \varepsilon_0}$

$\Phi_{\text{face}} = \frac{170 \times 10^{-6}}{6 \times 8.85 \times 10^{-12}}$

$\Phi_{\text{face}} \approx 3.20 \times 10^6 \, \text{N} \cdot \text{m}^2 / \text{C}$
**整個立方體的通量**：
$\Phi_{\text{total}} = 1.92 \times 10^7 \, \text{N} \cdot \text{m}^2 / \text{C}$

**Answer:**
(a) $3.20 \times 10^6$ N·m²/C
(b) $1.92 \times 10^7$ N·m²/C
(c) 總通量不變，但每個面不同

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### **Problem 6**

The charge per unit length on a long, straight filament is –90.0 μC/m. Find the electric field (a) 10.0 cm, (b) 20.0 cm, and (c) 100 cm from the filament, where distances are measured perpendicular to the length of the filament.

**Question Translation:**
一條長直細線上的線電荷密度為 $-90.0$ μC/m。
求距離細線 (a) 10.0 cm, (b) 20.0 cm, (c) 100 cm 的電場強度。

**Solution:**
根據高斯定律對無限長直導體：
$E = \frac{\lambda}{2 \pi \varepsilon_0 r}$
(a) $E = \frac{-90.0 \times 10^{-6}}{2 \pi \times 8.85 \times 10^{-12} \times 0.10}$
(b) $E = \frac{-90.0 \times 10^{-6}}{2 \pi \times 8.85 \times 10^{-12} \times 0.2}$
(c) $E = \frac{-90.0 \times 10^{-6}}{2 \pi \times 8.85 \times 10^{-12} \times 1}$
**Answer:**
(a) $1.62 \times 10^7 N/C$
(b) $8.09 \times 10^6 N/C$
(c) $1.62 \times 10^6 N/C$

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