12中英文雙語外文文獻翻譯成品對高斯定理在靜電場中的應(yīng)用及理解

1、<p>　　外文標題：Understanding and application of Gauss Theorem in electrostatic field</p><p>　　外文作者：Wang Xiaolan ，Wang Feng, Lanzhigao,Chen Ruian </p><p>　　文獻出處：International Conference on Intel

2、ligence Science and Information Engineering，2011，386-388</p><p>　　英文1078單詞，4865字符，中文1502漢字。</p><p>　　此文檔是外文翻譯成品，無需調(diào)整復(fù)雜的格式哦！下載之后直接可用，方便快捷！只需二十多元。</p><p>　　原文：Understanding and applic

3、ation of Gauss Theorem in electrostatic field</p><p>　　Wang Xiaolan ，Wang Feng, Lanzhigao,Chen Ruian </p><p>　　Abstract—Correct understanding and mastering electrostatic field of Gauss theorem i

4、s the key part of learning electromagnetism; the field strength E in Gaussian theorem is the total field strength excited by all charges in the space on the closed surface, E=Ein+Eout. The sum of E refers to the algebra

5、sum of charges in the closed surface; When charge distribution has certain symmetry(Spherical symmetry, axial symmetry, plane symmetry ), we can use Gauss theorem to calculate the distrubution of t</p><p>　　

6、Keywords- Gauss theorem; Gaussian surface; Electric field strength; Symmetry</p><p>　　INTRODUCTION</p><p>　　Gauss theorem is both important and difficult part in electrostatic field, and also in

7、 entire Electromagnetic, so understanding and mastering the Gauss theorem is the key of</p><p>　　by the mathematical expression of Gauss theroem learning Electromagnetic.</p><p>　　CONTENT OF GAU

8、SS THEOREM</p><p><b>　　Content:</b></p><p>　　Electrostatic field through any closed surface in the flux of electric field strength equal to the surface surrounded by the algebraic su

9、m of all the charges divided by the dielectric constant of vacuum.</p><p><b>　　Formula:</b></p><p><b>　　Note:</b></p><p>　　1In the Gauss theorem, the closed

10、surface we choose is often called Gauss Surface.</p><p>　　2Applicable scope of Gauss theorem: Electrostatic field; changing electric field.</p><p>　　3Gauss theorem show that electrostatic is act

11、ive field one of the basic properties of eletrostatic field.</p><p>　　UNDERSTANDING OF GAUSS THEOREM</p><p>　　1.The total flux 8e through the closed surface S in electric field is only related w

12、ith charges in the closed surface, and has nothing to do with the charge outside the closed surface, also the distribution of charges in the closed surface.</p><p>　　2. Refers to algebraic sum charge i

13、n the closed surface, if qi is positive charge, the Flux is positive, and negative versa.</p><p>　　3.The field strength E in Gaussian theorem is the total field strength excited by all charges in the space

14、 on the closed surface.</p><p><b>　　Thinking:</b></p><p><b>　　If</b></p><p>　　by the mathematical expression of Gauss theroem,</p><p><b>　

15、　Then:</b></p><p>　　E must equals to 0.</p><p><b>　　Discuss:</b></p><p>　　The power line began with closed surface within positive Charge equals to the power line

16、end with closed surface within negative charge,then the piercing the closed surface of power line and into the closed surface are equal, that is, the total flux is zero through all the closed surface.</p><p>

17、;　　But that doesn’t means the electric field strength E on the closed surface is 0,</p><p><b>　　FOR:</b></p><p>　　1. The field strength on Gaussian surface is the vector sum of field

18、 strength generated by both inside and outside the Gaussian surface, therefore within the Gaussian surface</p><p>　　E=0 can not be completely sure.</p><p>　　2.As in the type E and dS is the scal

19、ar product between</p><p>　　vector and therefore the direction of the two problems exist,if</p><p>　　and dS the side perpendicular to it, there are</p><p><b>　　If</b><

20、;/p><p>　　So it can not be </p><p>　　USE GAUSS THEOREM APPLIED FOR ELETRIC FIELD STRENGTH</p><p>　　When the charge distribution has some symmetry, the Gauss theorem can be applied to f

21、ind the electric field distribution. The following discussion of two cases:</p><p>　　1.Charge distribution is spherical symmetry: that is equidistant to the center of the sphere equal to the surface charge d

22、ensity</p><p>　　2. Electric field distribution: E along the radial direction. E is equal at any concentric spherical surface.</p><p><b>　　Eg1.</b></p><p>　　Find the fiel

23、d strength in and out the uniformly charged sephere with radius R and charge +q.</p><p>　　A:For the charge distribution is spherical symmetry, the field strength it Emerges has spherical symmetry, that is, a

24、s the center of the sphere r points on the sphere equal electric field strength and direction along the radius vector direction, outward.</p><p>　　For the concentric and the Gaussian surface of radius r Obta

25、ined by the Gauss theorem:</p><p>　　When r>R, Gaussian surface surrounding the charge q:</p><p>　　When r< R, no charge within the Gaussian surface:</p><p>　　Electric field dis

26、tribution of a uniformly charged sphere:</p><p>　　The charge distribution has axis symmetry: the charge density equal on the surface equidistant to the axis.</p><p>　　Electric field distribution

27、:</p><p>　　E along the vertical axis in the radial direction;</p><p>　　Any concentric cylindrical surface is equal to the points of E.</p><p><b>　　Eg2</b></p><

28、;p>　　Find infinite uniform electric field strength inside and outside the cylinder. Let cylindrical radius R, along with a charge per unit length of axis is + l</p><p>　　A: The field distribution should a

29、lso be column symmetrcial, the radicial direction, for a charged cylinder coaxial with the cylindrical Gaussian surface height l, radius</p><p>　　r. Obtianed by Gaussian theorem:</p><p>　　CONCLU

30、SION</p><p>　　Summary of the use of Gauss theorem seeking the conditions skills and procedures of field strength.</p><p>　　Condition: field strength has special symmetry: sephrical symmetry, axi

31、al Symmetry,plane symmetry.</p><p>　　If can integrated, we can use Gauss theorem no matter whether there is symmetry of the charge or the Electric field.)</p><p>　　Skills: choose the suitable

32、Gaussian surface, the</p><p>　　integral sign of the electric field strength E can form scalar</p><p>　　raised from the integral sign in.</p><p>　　Procedures:</p><p>　　(

33、1).Analysis the symmetry of field strength</p><p>　　(2).Choose suitable Gaussian surface</p><p>　　(3).Calculate the flux through the Gaussian surface </p><p>　　(4).Applied the Gauss

34、 theorem for electric field strength.</p><p>　　REFERENCES</p><p>　　[1] Ma Wenwei, physical [M].Beijing: Higher education press.2008.</p><p>　　[2] Chen Yincong, new century physics,

35、[M].Shanghai, East China normal University press, 2006.</p><p>　　[3] Shi Chuanzhu, Discussion of Gauss theorem, [J] .Qujing normal university. Press, 2002, 3.</p><p>　　[4] "Findings and Rec

36、ommendations to Enhance Reliability From the Summer of 1999", Final Report, Power Outage Study Team, U. S. Department of Energy, March 2000, p S-2</p><p>　　[5] IntelliTEAM Overview of Operations, "

37、 EnergyLine Systems, Alameda, CA. www.energyline.com</p><p><b>　　譯文：</b></p><p>　　對高斯定理在靜電場中的應(yīng)用及理解</p><p>　　Wang Xiaolan ，Wang Feng, Lanzhigao,Chen Ruian </p><

38、;p>　　摘要 - 在學(xué)習(xí)電磁學(xué)時，其重要的一環(huán)是要正確認識和掌握高斯定理的靜電場; 高斯定理中電場強度是指封閉曲面空間中所有電荷的總場強度，E = E內(nèi) + E外。 E的總值是指封閉曲面中電荷的總和; 當(dāng)電荷分布具有一定的對稱性（球?qū)ΨQ性、軸對稱性、平面對稱性）時，我們就可以用高斯定理來計算電場的分布。</p><p>　　關(guān)鍵詞 - 高斯定理; 高斯曲面; 電場強度; 對稱性</p>&l

39、t;p><b>　　引言</b></p><p>　　高斯定理不僅是靜電場的重要組成部分，而且也是整個電磁場的重要組成部分，因此通過高斯定理來學(xué)習(xí)電磁學(xué)的數(shù)學(xué)表達式對理解和掌握高斯定理非常關(guān)鍵。</p><p><b>　　高斯定理的內(nèi)容</b></p><p><b>　　內(nèi)容：</b><

40、;/p><p>　　靜電場通過任何封閉曲面的電場強度的通量等于由所有曲面的代數(shù)和，它是由所有曲面電荷除以真空介電常數(shù)得出。</p><p><b>　　其表達式是：</b></p><p><b>　　備注:</b></p><p>　　1在高斯定理中，我們選擇的閉合曲面通常稱為高斯曲面。</p&

41、gt;<p>　　2高斯定理的適用范圍：靜電場; 改變電場。</p><p>　　3高斯定理表明，靜電是激發(fā)的電場，是靜電場的基本性質(zhì)之一。</p><p><b>　　高斯定理的理解</b></p><p>　　1.電場中通過封閉曲面S的總流量Фe僅與封閉曲面中的電荷有關(guān)，與封閉曲面外的電荷以及封閉曲面中電荷的分布無關(guān)。<

42、/p><p>　　2. 指封閉曲面中的電荷的總數(shù)，當(dāng)qi是正電荷，則電磁流量為正，反之亦然。</p><p>　　高斯定理中的電磁場強E是封閉曲面空間中所有電荷激發(fā)的總電磁場強度。</p><p><b>　　思考一下:</b></p><p><b>　　當(dāng)</b></p><p

43、>　　高斯定理的數(shù)學(xué)表達式為,</p><p><b>　　那么:</b></p><p><b>　　E 必須等于 0.</b></p><p><b>　　探討:</b></p><p>　　正電荷內(nèi)封閉曲面的電力線初始端等于負電荷內(nèi)封閉曲面的電力線末端，然后穿過電力

44、線的封閉曲面并進入封閉曲面，也就是通過所有封閉曲面的總通量為零。</p><p>　　但這并不表示著封閉曲面上的電場強度E為0，</p><p><b>　　對于：</b></p><p>　　1.高斯曲面上的場強是高斯面內(nèi)外產(chǎn)生的場強的向量和，因此在高斯曲面內(nèi)</p><p>　　E=0 不能完全確定。</p&g

45、t;<p>　　E和dS是向量之間的數(shù)積，因此著兩個問題存在方向的問題，</p><p><b>　　當(dāng)</b></p><p>　　dS 是垂直它的面, 那么就有：</p><p><b>　　當(dāng)</b></p><p><b>　　則不會出現(xiàn)：</b><

46、;/p><p>　　高斯定理在電場強度中的應(yīng)用</p><p>　　當(dāng)電荷分布具有某種對稱性時，可以應(yīng)用高斯定理來找到電場分布。 以下討論兩種情況：</p><p>　　1.電荷分布是球形對稱：即等距的球體中心等于曲面電荷密度</p><p>　　2.電場分布：E沿半徑方向。 E在任何同心球面上都是相等的。</p><p>

47、;<b>　　例1</b></p><p>　　求出半徑為R和電荷+ q的均勻帶電球內(nèi)外的場強。</p><p>　　答：電荷分布是球?qū)ΨQ的，其出現(xiàn)的場強具有球?qū)ΨQ性，即球體r的中心指向球體的電場強度和沿半徑向量方向的場外強度相等。</p><p>　　圖一 帶電球結(jié)構(gòu)</p><p>　　半徑為r的同心和高斯曲面，

48、由高斯定理可以得到：</p><p>　　當(dāng) r>R, 高斯曲面環(huán)繞的電荷 q:</p><p>　　當(dāng) r< R, 高斯曲面內(nèi)無電荷:</p><p>　　圖二 均勻帶電球的電場分布</p><p><b>　　圖三 電場分布</b></p><p>　　電荷分布具有軸對稱性：

49、電荷密度與曲面到軸等距。</p><p><b>　　電場分布：</b></p><p>　　E沿半徑方向的垂直軸線;</p><p>　　任何同心圓柱面都等于E點。</p><p><b>　　例2</b></p><p>　　在圓柱體內(nèi)外發(fā)現(xiàn)無限均勻分布的電場強度。 假設(shè)

50、圓柱形半徑R以及每單位長度軸的電荷為+ 1</p><p>　　答：場分布也應(yīng)該是沿著半徑柱狀對稱，同軸帶電圓柱體圓柱形高斯曲面高度l，半徑r可用高斯定理求出：</p><p><b>　　圖四 圓柱圖</b></p><p><b>　　圖五 圓柱圖 </b></p><p><b&

51、gt;　　圖六 電場分布</b></p><p><b>　　結(jié)論</b></p><p>　　利用高斯定理總結(jié)求解場強的條件技巧和步驟。</p><p>　　條件：場強具有特殊的對稱性：球?qū)ΨQ性，軸對稱性，平面對稱性。</p><p>　　如果能對 進行整合, 則無論電荷還是電場是否具有對稱性，我們都可

52、以使用高斯定理。）</p><p>　　技巧：選擇合適的高斯面，電場強度E的積分符號可以從積分符號中形成標量。</p><p><b>　　步驟：</b></p><p>　?。?）分析場強的對稱性</p><p>　　（2）選擇合適的高斯曲面</p><p>　?。?）計算通過高斯面的通量<

53、;/p><p>　?。?）在電場強度中應(yīng)用高斯定理</p><p><b>　　參考文獻</b></p><p>　　[1] Ma Wenwei, physical [M].Beijing: Higher education press.2008.</p><p>　　[2] Chen Yincong, new centur

54、y physics, [M].Shanghai, East China normal University press, 2006.</p><p>　　[3] Shi Chuanzhu, Discussion of Gauss theorem, [J] .Qujing normal university. Press, 2002, 3.</p><p>　　[4] "Findi

55、ngs and Recommendations to Enhance Reliability From the Summer of 1999", Final Report, Power Outage Study Team, U. S. Department of Energy, March 2000, p S-2</p><p>　　[5] IntelliTEAM Overview of Operati