23. Electrostatic Energy & Capacitors靜電能和電容器1. Electrostatic Energy 靜電能2. Capacitors 電容器3. Using Capacitors 使用電容器4. Energy in the Electric Field 電場中的能

The lifesaving jolt of a defibrillator requires a large amount of energy delivered in a short time.除纖顫器救命的電擊需要在短時間內送出大批能量。Where does that energy come from?這些能量從何而來？

Capacitor 電容器

23.1. Electrostatic Energy 靜電能Electrostatic Energy = work done to assemble the charge configuration of a system.靜電能 = 裝組系統的電荷配置所需做的功。

Reference 參考 ( 0 energy 能 ): when all component charges are widely separated.當所有的組分電荷都相隔甚遠時。

Bringing q1 in place takes no work.把 q1 歸位不需做功 2 2 1 2W q V rBringing in q2 takes把 q2 歸位需功

1 1 0 0W q V

12qq ka

3 3 1 3 2 3W q V V r rBringing in q3 takes把 q3 歸位需功1 2

3q qq k ka a

Total electrostatic energy總靜電能 1 2 2U W W W 1 2 2 3 3 1k q q q q q qa

23.2. Capacitors 電容器Capacitor: pair of conductors carrying equal but opposite charges.電容器：帶著等量異性電荷的一對導體。

Usage: store electrical energy用途：存儲電能Parallel-Plate Capacitor 平行板電容器 :2 conducting plates of area A separated by a small distance d .兩塊相隔短距離 d ，面積為 A 的導體板。

Plates are initially neutral. 開始時兩塊板都是中性。They’re charged by connecting to a battery. 接上電瓶後起電。 Charge transfer plates are equal but oppositely charged.

電荷轉移 兩塊板的電荷量等而性異。Large A, small d E 0 outside. 大 A ，小 d 器外的 E 0 。

0

ˆinside

E zFar from the edges離邊遠處0

ˆQA

z ˆinsideV d E z0

Q dA

Capacitance 電容量0

QV dA

Parallel-plate capacitor:平行板電容器： 0 AQ Vd

C V

0 ACd

C = Q / V = capacitance 電容量Parallel-plate capacitor 平行板電容器See Probs 參閱習題 41 & 42

CCV

farad F

Practical capacitor ~ F ( 106 F) or pF ( 1012 F ) 實用電容器 0

C dA

/F m 1dV dQ

C

Charging / Discharging充 / 放電

dQCdV

Energy Stored in Capacitors 存在電容器內的電能When potential difference between capacitor plates is V, 電容器兩板之間的電位差為 V 時，work required to move charge dQ from to + plate is 把電荷 dQ 從 移至 + 板需作功

plate

plate

dW dQ d

E r V C dV

Work required to charge the capacitor from 0 to V is把電容器從 0 充電至 V 需作功0

VW C V dV 21

2C V = U = energy stored in capacitor存在電容器內的電能

Note 注意 :

In a “charged” capacitor, Q is the charge on the + plate.在一個”帶電”的電容器內， Q 是 + 板上的電荷。The total charge of the capacitor is always zero. 電容器的總電荷永遠為零。

plate platedQ V V E dr < 0

2

2QC

12QV

Example 23.1. Parallel-Plate Capacitor 平行板電容器A capacitor consists of two circular metal plates of radius R = 12 cm, separated by d = 5.0 mm.

一個電容器由兩塊圓形金屬板組成；板的半徑為 R = 12 cm ，相距 d = 5.0 mm 。Find 求(a) Its capacitance 其電容量，(b) the charge on the plates, and 板上電荷，和(c) the stored energy when the capacitor is connected to a 12-V battery. 電容器在接上一個 12-V 電瓶後所存的能量。

0ACd

100.8 10 F

22

9 3

12 1014 9 10 / 5.0 10

m

Vm C m

80 pF

(a)

(b)

(c)

Q CV 80 12pF V 960 pC

212

U CV 21 80 122

pF V 5760 pJ

5.76 nJ

CFV

23.3. Using Capacitors 使用電容器Computer memories: billions of 25 fF capacitors.電腦記憶體：數十億個 25 fF 電容器。Rectifiers 整流器： mF

Fuel-cells 燃料電池 : 102 F

220-mF electrolytic capacitor電解電容器

1 F

43 pF to 2.2 mF

Practical Capacitors 實用電容器

Inexpensive capacitors 廉價電容器 :

Thin plastic sandwiched between aluminum foils & rolled into cylinder.薄塑料夾在兩鋁片間再捲成圓筒。Electrolytic capacitors 電解電容器 (large capacitance 大電容量 ):

Insulating layer developed by electrolysis. 絕緣層以電解生成。

Capacitors in IC circuits IC 電路中的電容器 (small capacitance 小電容量 ):

Alternating conductive & insulating layers. 導體層和絕緣層交替。

Dielectrics 介電體Dielectrics: insulators containing molecular dipoles but no free charges.介電體：含有分子雙極但沒有自由電荷的絕緣體

Dielectric layer lowers V between capacitor plates by factor 1/ ( > 1).電容器兩板間的介電質層把 V 降低 1/ ( > 1) 倍。

0Ad

QCV

= dielectric constant 介電常數

Molecular dipoles aligned by E0 .分子雙極依 E0 排向

Dipole fields oppose E0. 雙極場與 E0 反向。Net field reduced to E = E0 / . 淨場減至 E = E0 / 。Hence V = V0 / . 故 V = V0 / 。Q is unchanged, so C = C0 . Q 不變，所以 C = C0 。

: 2 ~ 10 mostly

Working voltage V = Max safe potential < Ebkd d工作電壓 V = 最大安全電位 < Ebkd d

崩潰電場介電常數介電材料表 23.1 常見介電體的性質

氧化鋁

Example 23.2. Which Capacitor? 那個電容器？A 100-F capacitor has a working voltage of 20 V, 一個 100-F 電容器的運作電壓為 20

V ，while a 1.0-F capacitor is rated at 300 V. 另 一個 1.0-F 電容器可承受 300 V 。Which can store more charge? More energy? 那一個可以儲存較多電荷？較多能量？

100 FQ CV 6100 10 20F V 32 10 C 2 mC

61 1 10 300FQ F V

43 10 C 0.3mC

2100

12FU CV 1 2 20

2mC V

12QV 20 mJ

11 0.3 3002FU mC V 45mJ

GOT IT? 23.1.

You need to replace a capacitor with one that can store more energy.你需要把一個電容器換成一個存能較多的。Which will give you greater energy increase:下面那一個會給你較多的能量：(a) a capacitor with twice the capacitance and same working voltage as the old one,一個電容量是舊的二倍，運作電壓卻和它一樣的電容器，

or 或(b) a capacitor with the same capacitance and twice the working voltage?一個電容量和舊的一樣，運作電壓卻是它二倍的電容器？

Connecting Capacitors 連接電容器Two ways to connect 2 electronic components: parallel & series兩個方法連接兩個電子元件：並聯和串聯

Parallel : Same V for both components 並聯： 兩元件的 V 一樣Q CV 1 2Q Q 1 2C V C V

1 2C C C

Series : Same I (Q) for both components 串聯： 兩元件的 I (Q) 一樣QVC

1 2V V 1 2

Q QC C

1 2

1 1 1C C C

1 2C C or C

1 2C C or C

Conceptual Example 23.1. Parallel & Series Capacitors 並聯與串聯電容Using parallel-plate capacitors, explain why capacitance should increase with capcitors in parallel an decrease with capacitors in series.用平板電容器來闡述為何並聯會增大電容量，串聯則減少電容量。What happens to the working voltage in each case? 工作電壓又會怎樣？

Parallel-plate capacitor :平板電容器 :0 ACd

in parallel 並聯A = A1 + A 2 C increases 增大

in series 串聯d = d1 + d 2 C decreases 減少Vworking

= min(Vw1 ,Vw2 )

Vworking < Vw1 +Vw2

Making the Connection 連起來You’ve got two 10-F capacitors rated at 15 V. 你有兩個 15 V 級的 10-F 電容量。What are the capacitances & working voltages of their parallel & series combinations?它們並聯和串聯後的電容量和工作電壓為何 ?

Parallel :並聯 :2 10 20C F F

15workingV V

Series :串聯 :1 10 52

C F F

2 15 30workingV V V 1 212

V V V

1 2V V V

GOT IT? 23.2.

You have 2 identical capacitors with capacitance C.你有兩個一樣的電容器，電容量都是 C 。How would you connect them to get equivalent capacitances你會如何把它們連接來達到以下的電容量？(a) 2 C, and 和(b) ½ C ?

Which combination would have the higher working voltage?那一個組合有較高的運作電壓？

parallel 並聯series 串聯

Example 23.2. Connecting Capacitors 連接電容器Find the equivalent capacitance of the combinations shown in the Figure.求圖中所示組合的相當電容量。If the maximum voltage to be applied between points A and B is 100 V,如果加到 A 和 B 之間的最高電壓是 100 V ，what should be the working voltage of C1 ? C1 的運作電壓為何？

23 2 3C C C 3.0 1.0F F 4.0 F

123 1 23

1 1 1C C C

1 1

12.0 4.0F F

123 3.0C F

13.0 F

11

1

QVC

1 123Q Q 123 ABC V

3.0 100F V

130012.0

CVF

25V ( min. working voltage ) 最小運作電壓

300 C

Bursts of Power 爆發性電力

San Francisco’s BART train 三藩市的灣區捷運 :

KE of deceleration stored as EE in ultracapacitor.減速時的 KE 以 EE 存於超級電容器中。Stored EE is used to accelerate train. 儲存的 EE 再用來加速。

Capacitors deliver higher energy much more quickly than batteries.電容器送出的能量可以比電瓶的更大和更快。

Flash light 閃光燈 :

Battery charges capacitor,which then discharges to give flash.電容器經電池充電後放電以產生閃光。

Other examples 其他例子 :

Defibrillator, controlled nuclear fusion, amusement park rides, hybrid cars, …除纖顫器，受控的核子融合，遊樂場的設施，混合動力汽車，…

23.4. Energy in the Electric Field 電場中的能Charging a capacitor rearranges charges energy stored in E

電容器充電時把電荷重組 能量存到 E 中Energy density = energy per unit volume 能量密度 = 單位體積內的能量

Parallel-plate capacitor :平行板電容器：

2

2QUC

2

02

QAd

EUuA d

Energy density :能量密度：2

202QA

2

02

20

12

E0

E

is universal是通用的20

12Eu E 3/Eu J m

2

012 U dVE

Example 23.4. A Thunderstorm 雷雨Typical electric fields in thunderstorms average around 105 V/m.雷雨中的電場一般平均有 105 V/m 。Consider a cylindrical thundercloud with height 10 km and diameter 20 km,

茲有一圓筒型雷雨雲塊，其高為 10 km ，其直徑為 20 km ，and assume a uniform electric field of 1105 V/m. 假定其均勻電場為 1105 V/m 。Find the electric energy contained in this storm.求此風暴所藏電能。

111.39 10 J

20

12Eu E

EU u V 2 25 3 39 2 2

1 1 10 / 10 10 10 102 4 9 10 /

V m m mNm C

139 GJ

~ 1400 gallons of gasoline.加侖汽油

Example 23.5. A Shrinking Sphere 一個縮小的球A sphere of radius R1 carries charge Q distributed uniformly over its surface.一半徑為 R1 的球帶有一均勻分佈於其表面的電荷 Q 。How much work does it take to compress the sphere to a smaller radius R2 ?把球壓至較小半徑 R2 時所需作的功為何？

2ˆ

Qkr

E r2

012 U dVE

1

2

22

1 12

R

Rk Q dr

r 1

2

22

2

1 48

R

R

Qk r drk r

2

2 1

1 1 12k Q

R R

2 10U for R R Work need be done to shrink sphere把球縮小要作功

表面積為 4r2 …

… 厚度為 dr

2 1U U U 2 1

2 20

1 42 R R

r dr

E

Extra energy stored here新增的能量存於止

You’re at point P a distance a from a point charge +q. 你在 P 點與一點電荷 +q 相距 a 。

You then place a point charge q a distance a on the opposite side of P as shown.

之後你把一點電荷 q 放在 P 的另一邊，離它 a 處。What happens to 以下各項會有怎麼變化？(a) the electric field strength and 在 P 點的電場強度和(b) the electric energy density at P ? 電能密度？(c) Does the total electric energy U = ∫ uE dV of the entire field increase, decrease, or remain the same?

整個電場的總電能 U = ∫ uE dV 會增加，減少，還是不變？

GOT IT? 23.3.

doubles 加倍quadruples 四倍

decrease 減少

Negative work done to bring in –q.把 – q 加進來要作的功是負的。