# Chapter 25

## Homework

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### **Problem 1**

Regarding the Earth and a cloud layer 800 m above the Earth as the “plates’’ of a capacitor, calculate the capacitance. Assume the cloud layer has an area of $1.00 km^2$ and that the air between the cloud and the ground is pure and dry. Assume charge builds up on the cloud and on the ground until a uniform electric field of $3.00 × 10^6 N/C$ throughout the space between them makes the air break down and conduct electricity as a lightning bolt. What is the maximum charge the cloud can hold?

**Question Translation:**

將地球與地面上方 $800 m$ 的雲層視為電容器的「電極板」，計算其電容值。
假設雲層的面積為 $1.00 km^2$，雲層與地面之間的空氣純淨且乾燥。假設電荷在雲層和地面上積聚，直到在其間產生 $3.00 × 10^6 N/C$ 的均勻電場，使空氣擊穿並導電形成閃電。
求雲層可承載的最大電荷量。

**Solution:**
電容量：
$\epsilon_{0}\frac{A}{d} = 11.06$ nF

最大電荷量:
$Q = CV = 11.06 \times 10^{-9} \times 3 \times 10^6 \times 800 = 26.54C$

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### **Problem 2**

An air-filled spherical capacitor is constructed with inner and outer shell radii of $7.00$ and $14.0 cm$ , respectively.
(a) Calculate the capacitance of the device.
(b) What potential difference between the spheres results in a charge of $4.00 μC$ on the capacitor?

**Question Translation:**
一個充滿空氣的球形電容器，內外殼半徑分別為 $7.00 cm$ 和 $14.0 cm$。
(a) 計算該電容器的電容值。
(b) 如果電容器儲存了 $4.00 μC$ 的電荷，球殼間的電勢差為多少？

**Solution:**
(a)
$C = 4\pi\epsilon_0 \frac{ab}{b-a} = 4\pi8.85\times10^{-12} \times\frac{14\times7\times10^{-6}}{7\times10^{-3}} = 15.6$ pF
(b)
$V = \frac{Q}{C} = \frac{4 \times10^{-6}}{15.6\times10^{-12}} = 256kV$
**Answer:**
(a) $C = 15.6$pF
(b) $V = 256kV$

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### **Problem 3**

Three capacitors are connected to a battery as shown in below figure. Their capacitances are $C1 = 3C$ , $C2 = C$ , and $C3 = 5C$ .
(a) What is the equivalent capacitance of this set of capacitors?
(b) State the ranking of the capacitors according to the charge they store, from largest to smallest.
(c) Rank the capacitors according to the potential differences across them, from largest to smallest.
(d) What If? If C3 is increased, what happens to the charge stored by each of the capacitors?

[解題影片](https://youtu.be/MkQCxFdIzfo)

**Question Translation:**
三個電容器如圖所示連接到電池。已知 C₁ = 3C，C₂ = C，C₃ = 5C。
(a) 求此電容器組合的等效電容值。
(b) 根據電荷量從大到小對這些電容器排序。
(c) 根據電容器兩端的電勢差從大到小排序。
(d) 如果 增大 C₃，這些電容器所儲存的電荷會如何變化？

**Solution:**
(a)
根據電容串並聯特性:
並聯電容相加 $C_2 + C_3 = 6$
最後串聯$\frac{6 \times 3}{3 + 6} = 2C$
(b)
$Q_1 > Q_3 > Q_2$
(c)
$V_1 > V_3 = V_2$
(d)
$Q_1$ increase
$Q_2$ decrease

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### **Problem 4**

Find the equivalent capacitance between points a and b for the group of capacitors connected as shown in below figure. Take $C1 = 5.00 μF$ , $C2 = 10.0 μF$ , and $C3 = 2.00 μF$ .

**Question Translation:**

求等效電容

**Solution:**

**Answer:**
$6.04\mu$F

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### **Problem 5**

(a) A $3.00μF$ capacitor is connected to a $12.0V$ battery. How much energy is stored in the capacitor? (b) If the capacitor had been connected to a $6.00V$ battery, how much energy would have been stored?

**Question Translation:**
(a) 一個 $3.00 μF$ 的電容器連接到 $12.0 V$ 電池，求電容器儲存的能量。
(b) 如果該電容器改接到 $6.00 V$ 電池，儲存的能量會是多少？

**Solution:**
$U = \frac{1}{2}CV^2$

**Answer:**

(a) $2.16\times10^{-4}J$
(b) $5.40\times 10^{-5}J$

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### **Problem 6**

A parallel-plate capacitor is charged and then disconnected from a battery. By what fraction does the stored energy change (increase or decrease) when the plate separation is doubled?

**Question Translation:**
當平行板距離加倍求能量變化
**Solution:**

$C = \epsilon_{0}\frac{A}{d}$
$U = \frac{Q^2}{2C}$
電容量減小能量變多

**Answer:**
increase

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### **Problem 7**

Determine (a) the capacitance and (b) the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of $1.75 cm^2$ and plate separation of $0.04mm$ .

**Question Translation:**
求以下 [Teflon](https://zh.wikipedia.org/zh-tw/%E8%81%9A%E5%9B%9B%E6%B0%9F%E4%B9%99%E7%83%AF)

填充的平行板電容器的相關參數：
(a) 電容值。
(b) 可施加的最大電勢差。
已知該電容器的板面積為 1.75 cm²，板間距為 $0.0400 mm$。

**Solution:**
(a)
$C = \kappa\epsilon_{r}\frac{A}{d} = 2.1\times8.85\times10^{-12}\times\frac{1.75\times10^{-4}}{0.04\times10^{-3}}=8.13pF$
(b)
Teflon最大電場 $6\times10^7V/m$
**Answer:**
(a) $8.13\times 10^{-11}F$
(b) $2.4kV$

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