1、22. Electric Potential電位1. Electric Potential Difference電位差2. Calculating Potential Difference計(jì)算電位差3. Potential Difference & the Electric Field電位差和電場(chǎng)4. Charged Conductors帶電導(dǎo)體This parasailer landed on a 138,000-volt power line.這名滑翔傘手落在一條138,000伏的電力線上 Why didnt he get electrocuted?為甚麼他沒(méi)有被電死？He tou

2、ches only 1 line theres no potential differences & hence no energy transfer involved.他祇碰到一條線 ：沒(méi)有電位差，所以沒(méi)有涉及能量轉(zhuǎn)移。22.1. Electric Potential Difference電位差Conservative force 守恆力:ABBAUUUABW BAd FrElectric potential difference electric potential energy difference per unit charge電位差 每單位電荷的電位能差BAd ErABABU

3、VqBVif reference potential VA = 0.若參考電位 VA = 0. V = J/C = Volt 伏特 = V 伏For a uniform field 均勻場(chǎng):ABABV E rBA Err( path independent ) 與路徑無(wú)關(guān)rABEE points at direction of most rapidly decreasing V.E 指向 V 遞減最快的方向。Table 22.1.Force & Field, Potential Energy & Electric Potential 表22.1.力和場(chǎng)，位能和電位Quantit

4、y 量量Symbol / Equation符號(hào)符號(hào) / 公式公式Units單位單位BABAUd FrUVqBABAVd ErForce 力Electric field 電場(chǎng)Potential energy difference位能差Electric potential difference電位差FNE = F / qN/C or V/mJJ/C or VU FV EPotential Difference is Path Independent電位差與路徑無(wú)關(guān)Potential difference VAB depends only on positions of A & B.電位差

5、VAB 祇與 A 和 B 的位置有關(guān)。Calculating along any paths (1, 2, or 3) gives VAB = E r.沿著任何一條路徑 (1, 2, 或 3) 來(lái)算都得到 VAB = E r 。GOT IT? 22.1What would happen to VAB in the figure if以下情況下，圖中的 VAB 會(huì)怎樣？(a) E were doubled;E 加倍；(b) r were doubled; r 加倍；(c) the points were moved so the path lay at right angles to E;移動(dòng)點(diǎn)

6、使路徑與 E 正交；(d) the positions of A & B are interchanged.A 與 B 的位置互換doubles 加倍doubles 加倍becomes 變成 0reverses sign變符號(hào)The Volt & the Electronvolt伏特和電子伏特 V = J/C = Volt = VUqVE.g., for a 12V battery, 12J of work is done on every 1C charge that moves from its negative to its positive terminals.例， 1

7、2V 電瓶：每次把 1C 電荷從負(fù)極移至正極，就要作功 12J 。Voltage = potential difference when no B(t) is present.電壓 = 無(wú) B(t) 時(shí)的電位差Electronvolt (eV) = energy gained by a particle carrying 1 elementary charge when it moves through a potential difference of 1 volt. 電子伏特 (eV) = 帶有 1 基本電荷的粒子通過(guò) 1 伏電位差後所增加的能量。1 elementary charge 基

8、本電荷 = 1.61019 C = e1 eV = 1.61019 JTable 22.2. Typical Potential Differences表22.2. 典型電位差Between human arm & leq due to1 mV hearts electrical activity因心臟的電性活動(dòng)而產(chǎn)生在手與腿之間的Across biological cell membrane80 mV生物細(xì)胞膜的兩邊Between terminals of flashlight battery1.5 V手電筒電池兩極之間Car battery 汽車電瓶12 V Electric ou

9、tlet (depends on country)100-240 V電源插座(因國(guó)別而異)Between long-distance electric 365 kVtransmission line & ground長(zhǎng)程輸電線與地之間Between base of thunderstorm cloud & ground100 MV雷雨雲(yún)層底與地之間INTERNATIONAL VOLTAGE國(guó)際電壓Afghanistan 220 VAustralia 240 VBahamas 120 VBrazil 120/220 VCanada 120 VChina 220 VFinland

10、230 VGuam 110 VHongKong 220 VJapan 100 VMexico 127 VSpain 230 VUnited Kingdom 230 VUnited States 120 VVietnam 127/220 VGOT IT? 22.2(a) A proton ( charge e ), 一粒質(zhì)子 ( 電荷為 e ) ，(b) an particle ( charge 2e ), and一粒 粒子( 電荷為 2e ) ，和(c) a singly ionized O atom一粒 O 離子each moves through a 10-V potential diff

11、erence.每粒都通過(guò) 10-V電位差。Whats the work in eV done on each?對(duì)每粒所作的功有幾 eV？10 eV20 eV10 eVExample 22.1. X RaysIn an X-ray tube, a uniform electric field of 300 kN/C extends over a distance of 10 cm, from an electron source to a target; the field points from the target towards the source.在一X光管中，從電子源到標(biāo)靶之間的 1

12、0 cm 距離內(nèi)，有一 300 kN/C 均勻電場(chǎng)。Find the potential difference between source & target and the energy gained by an electron as it accelerates from source to target ( where its abrupt deceleration produces X-rays ).求電子源到標(biāo)靶間的電位差和電子在電子源到標(biāo)靶間加速時(shí)所增加的能量 (電子在碰上標(biāo)靶時(shí)因突然減速而產(chǎn)生 X光) 。Express the energy in both electr

13、onvolts & joules.能量同時(shí)以電子伏特和焦耳表示。ABVEr 300/0.10kN Cm30 kVABABUeV 30 keV 30ABABKUkeV 154.8 10J4.8 fJ電子源標(biāo)靶Example 22.2. Charged Sheet帶電片An isolated, infinite charged sheet carries a uniform surface charge density .一張孤立的無(wú)限寬帶電片上有一均勻面電荷密度。Find an expression for the potential difference from the sheet

14、to a point a perpendicular distance x from the sheet.求電片至與其垂直距離為 x 的一點(diǎn)間的電位差。00 xVE x 02x ECurved Paths & Nonuniform Fields彎曲路徑和非均勻場(chǎng)Staight path, uniform field:直路徑，均勻場(chǎng)：ABABV E rCurved path, nonuniform field:彎曲路徑，非均勻場(chǎng)：0limABiiiV rErBAd ErThe figure shows three straight paths AB of the same length,

15、 each in a different electric field.圖示三條在不同電場(chǎng)中的等長(zhǎng)直路徑 AB 。The field at A is the same in each.在各A點(diǎn)的場(chǎng)都是一樣。Rank the potential differences VAB.請(qǐng)為各電位差 VAB 排序。GOT IT? 22.3Largest VAB .最大Smallest VAB .最小22.2. Calculating Potential Difference 計(jì)算電位差Potential of a Point Charge點(diǎn)電荷的電位ABBAVVVBAd Er2BAk qdr rr2BAr

16、ABrk qVdrr ddr rrr rdr11BAk qrr For A,B on the same radialA,B 皆在同一徑線上For A,B not on the same radial, break the path into 2 parts,1st along the radial & then along the arc. 若 A, B 不在同一徑線上，應(yīng)把路徑拆成兩段： 前一段沿徑線走，後一段沿弧線走。Since, V = 0 along the arc, the above equation holds.因弧線上 V = 0 ，上列式子還是對(duì)的。The Zero

17、of Potential電位的零點(diǎn)Only potential differences have physical significance.祇有電位的差別才有物理意義。Simplified notation:簡(jiǎn)化的符號(hào)RAARAVVVVSome choices of zero potential 一些電位零點(diǎn)一些電位零點(diǎn)R = point of zero potential電位原(零)點(diǎn)VA = potential at A. A 的電位Power systems / Circuits電力系統(tǒng) / 電路Earth ( Ground ) 地球(地)Automobile electric sys

18、tems汽車的電力系統(tǒng)Cars body 車身Isolated charges孤立電荷Infinity無(wú)限遠(yuǎn)處GOT IT? 22.4You measure a potential difference of 50 V between two points a distance 10 cm apart in the field of a point charge. 在一個(gè)點(diǎn)電荷的場(chǎng)中，你量得相距 10 cm 的兩點(diǎn)之間的電位差為 50 V 。If you move closer to the charge and measure the potential difference over an

19、other 10-cm interval, will it be如果你往電荷靠近，再量另一相距 10 cm 的兩點(diǎn)之間的電位差。結(jié)果會(huì)(a) greater,比較大，(b) less, or比較小，還是(c) the same?一樣？Example 22.3. Science MuseumThe Hall of Electricity at the Boston Museum of Science contains a large Van de Graaff generator, a device that builds up charge on a metal sphere.波士頓科學(xué)博物館

20、的電力廳用一部大型范得格拉夫起電機(jī)將一個(gè)金屬球起電。The sphere has radius R = 2.30 m and develops a charge Q = 640 C.球的半徑是 R = 2.30 m ，總電荷可達(dá) Q = 640 C 。Considering this to be a single isolate sphere, find把它當(dāng)成一個(gè)孤立的球，求(a) the potential at its surface 它表面上的電位，(b) the work needed to bring a proton from infinity to the spheres su

21、rface,把一質(zhì)子從無(wú)限遠(yuǎn)處帶到球面上所需的功，(c) the potential difference between the spheres surface & a point 2R from its center.球面與離球心 2R 處的電位差。 QV RkR69640 109.0 10/2.30CVm Cm(a)2.50 MV WeV R(b)2.50 MeV191.6 102.50CMV134.0 10J ,22RRVVRV R(c)2QQkkRR2QkR 1.25 MV 徑距離Example 22.4. High Voltage Power Line高壓電力線A long

22、, straight power-line wire has radius 1.0 cm & carries line charge density = 2.6 C/m.一條半徑為 1 cm 的長(zhǎng)而直的電力線上帶有線電荷密度 = 2.6 C/m 。Assuming no other charges are present,whats the potential difference between the wire & the ground, 22 m below?假定沒(méi)有其他電荷，電力線與在它下面 22 m 處的地面之間的電位差為何？BArABrVd Er0 2BArrdrr

23、 r r012BArrdrr 0ln2BArr 962229.0 10/2.6 10/ln0.01mV m CC mm 360 kV Finding Potential Differences Using Superposition以疊加求電位差Potential of a set of point charges:一組點(diǎn)電荷的電位 iiPiqV PkrrPotential of a set of charge sources:一組電荷的電位 iiV PV PExample 22.5. Dipole Potential雙極電位An electric dipole consists of poi

24、nt charges q a distance 2a apart.一個(gè)雙極由相距 2a 的點(diǎn)電荷 q 組成。Find the potential at an arbitrary point P, and approximate for the casewhere the distance to P is large compared with the charge separation.求任一點(diǎn) P 的電位，並取得到 P 的距離比電荷間距離大很多時(shí)的近似值。 12qqV Pkkrr1211kqrr22212cosrrar a22222cosrrar a22214cosrrr a2112rrrr

25、r a 212cosrra 212rrV Pk qr22cosqakr2cospkrp = 2qa = dipole moment 雙極距2121rrkqr r+q: hill 丘丘 q: hole 洞洞V = 0GOT IT? 22.5The figure show 3 paths from infinity to a point P on a dipoles perpendicular bisector.圖示從無(wú)限遠(yuǎn)處到一雙極的中垂線上一點(diǎn) P 的三條路徑。Compare the work done in moving a charge to P on each of the paths.

26、比較在這些路徑上把一個(gè)電荷移至 P 所需做的功。V is path independent V與路徑無(wú)關(guān) work on all 3 paths are the same. 三條路徑上要做的功都一樣。Work along path 2 is 0 since V = 0 on it.路徑 2 上 V = 0 ，所以要做的功為 0。Hence, W = 0 for all 3 paths.故三條路徑上都是 W = 0 。P123qqContinuous Charge Distributions連續(xù)電荷分佈Superposition:疊加VdVdqkrdVkr 3d rVkrrrrExample 2

27、2.6. Charged Ring 帶電環(huán)A total charge Q is distributed uniformly around a thin ring of radius a.一半徑為 a 的幼環(huán)上均勻地分佈了 Q 電荷。Find the potential on the rings axis.求環(huán)軸上的電位。 dqV xkr22kdqxa22kQxaSame r for all dq所有 dq 的 r 都一樣QkxaxExample 22.7. Charged Disk 帶電盤(pán)A charged disk of radius a carries a charge Q distributed uniformly over its surface.一半徑為 a 的盤(pán)上均勻地分佈了 Q 電荷。Find the potential at a point P on the disk axis, a distance x from the disk.求盤(pán)軸上離盤(pán) x 遠(yuǎn)的 P 點(diǎn) 的電位。 V xdV22kdqxr2222k Qxaxa22202akQr draxr222