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标准答案及评分标准一．Choice(20分，每题4分)1. a2. b3. c4. d5. a二．Blanks （20分）6. 0.5 （1分） 3 (1分) 1/2 (1分)π/6（2分） 3π or 9.42 (3pts)7. 1.26 （3分）8. 2.26°(3分)9. 3.56×10-28 (3分)10. 1.14eV (3分)三．Questions(10分)11. (5pts)The relativity principle: The laws of physics must be the same in all inertial reference frames. 一切物理规律在惯性系中相同。
(2分)The constancy of the speed of light: The speed of light in vacuum has the same value c in all inertial frames, regardless of the velocity of the observer or the velocity of the source emitting the light. 真空中的光速在任何惯性系中都是c ，与光源或观察者的运动无关。
（3分）12. (5pts)The maximum kinetic energy of photoelectrons is independent of light intensity. (2分)No electrons are emitted if the incident light frequency falls below some cutoff frequency fc , whose value is characteristic of the material being illuminated, regardless of the light intensity.（3分）四. Problems （50分）13(10pts) (1) The average transitional kinetic energy 21101.623-⨯==kT K t J ………………… 3pts (2) the rms speed s m MRT v rms /4803≈= ………………… 3pts (3) the internal energy RT i E 2= ………………… 1pts For O 2, i = 5, ………………… 1pts the internal energy 60912≈=RT i E J ………………… 2pts 14(10pts)(1) 1→2: Isothermal expansion )/ln(12V V nRT Q H H = ………………… 2pts 3→4: Isothermal compression )/ln(34V V nRT Q L L = ………………… 2pts (2) 2→3: Adiabatic expansion 1312--=γγV T V T L H ………………… 1pts4 1: Adiabatic compression. 1411--=γγV T V T L H ………………… 1pts Therfore, 4312V V V V = ………………… 1pts HL H L T T Q Q = ………………… 1pts The efficiency of Carnot engine is HL H L T T Q Q -=-=11η ………………… 2pts 15 (17pts)(1) The light path difference isθδs i nd = ………………… 2pts For bright fringesλθδm d ==s i n………………… 2pts The separation between the 2nd-order bright fringe and the 3rd-order bright fringe is423105.523tan tan -⨯==-≈-=∆d D d D d D D D x λλλθθm ………………… 3pts (2) The light path difference is()L n d 1sin -+=θδ ……………………………………… 3ptsFor bright fringes()λθδm L n d =-+=1s i n …………………………… 1pts At the central, point θ=0 ……………………………………… 1ptsThe central point on the viewing screen is now occupied by the seventh bright slid fringe (m =7) ()λδ710sin =-+=L n d ……………………………………… 1pts 61064.6)1/(7-⨯=-=n L λm ……………………………………… 2pts(3) No ……………………………………… 2pts16 (13pts)(1) The period measured by the observers on the spaceship is the propertime. ……………1pts126.020==km T πs ……………4pts (2) The mass and the spaceship is moving at a very high speed relative to the Earth. The effect of special theory of relativity must be taken into account. ………2ptsThe period measured by the observers on the Earth is not the propertime. ……………2ptsAccording to the special theory of relativity,209.0/1220=-=c v T T s ……………4pts。
习题11-1 P 点相对于原点的位矢26=-+p r i j m , P 点到Q 点的位移42∆=-r i j m, 求Q 点相对于原点的位矢并画图.解：设Q 点相对与原点的位矢为Q r ，则：24=+∆=+Q p r r r i j1-2一质点作直线运动，它的运动方程是2ct bt x -=, b , c 是常数. (1) 求此质点的速度和加速度函数；(2) 作出x t -，t υ-和a t -图解：这是一个一维的问题.速度 (2)dx ct b dtυ==-+， 加速度 2d a c dtυ==-. 图略.1-3物体按照29.4t x =的规律运动，x 的单位为米，t 的单位为秒. (1) 计算下列各时间段内的平均速度：1s 到1.1s,1s 到1.01s,1s 到1.001s; (2) 求1s 末的瞬时速度；(3) 解释上述结果解：这也是一个一维的问题.(1) 平均速度 x tυ∆=∆. 1s 到1.1s 内： 224.9 1.1 4.911.11x t υ∆⨯-⨯==∆-＝10.29 （m/s ）, 1s 到1.01s 内：224.9 1.01 4.911.011x t υ∆⨯-⨯==∆-＝9.849（m/s ）, 1s 到1.001s 内：224.9 1.001 4.911.0011x t υ∆⨯-⨯==∆-＝9.8049（m/s ）. (2) 速度 9.8dx t dtυ==. 1-4一质点以110m s -⋅的恒定速率向东运动. 当它刚到达距出发点为d 的一点时，立即以120m s -⋅的恒定速率返回原处. 问: 质点在全过程中的平均速度和平均速率为多少？解：取出发点为原点，向东为x 轴正方向. 从原点到x ＝d 处，作匀速直线运动，时间 11s t υ∆∆=＝d/10.从x ＝d 处返回原点作匀速直线运动，时间22st υ∆∆=＝d/20 (全过程中，平均速率 12s d d t t t υ∆+===∆∆+∆13.3 （m/s ） 返回原处时，位移x ∆＝0，平均速度x tυ∆=∆＝0. 1－5 矿井里的升降机由井底从静止开始匀加速上升，经过3s 速度达到13m s -⋅，然后以这个速度匀速上升6s ，最后减速上升经过3s 后到达井口时刚好停止. (1) 求矿井深度；(2) 作出x t -，t υ-和a t -图.解：（1）以井底为原点，向上为x 轴正向.在0—3s 内，升降机作匀加速直线运动:210112x t a t υ∆=+ (1) 2210112a x υυ=+∆. (2)其中00υ=. 由（1）、（2）两式得：1x ∆＝4.5(m).在3—9s 内，升降机以1υ＝3m/s 作匀加速直线运动,21x t υ∆=＝18（m/s ） (3)在9—12s 内，升降机作匀减速直线运动231212x t a t υ∆=- (4) 2221232a x υυ=-∆, (5) 其中20υ=. 由（4）和（5）两式得3x ∆＝4.5(m)矿井深度 123H x x x =∆+∆+∆＝4.5＋18＋4.5＝27(m).1-6湖中有一小船，岸上有人用一根跨过定滑轮的绳子拉船靠岸。
Chapter 23 Exercises23-3 a) For surface S 1, the electric flux is ()()221CL j L k D j C i B S E -=-⋅-+-=⋅=Φ .For surface S 2, the electric flux is ()()222DL k L k D j C i B S E -=⋅-+-=⋅=Φ .For surface S 3, the electric flux is ()()223CL j L k D j C i B S E =⋅-+-=⋅=Φ .For surface S 4, the electric flux is ()()224DL k L k D j C i B S E =-⋅-+-=⋅=Φ.For surface S 5, the electric flux is ()()225BL i L k D j C i B S E -=⋅-+-=⋅=Φ.For surface S 6, the electric flux is ()()226BL i L k D j C i B S E =-⋅-+-=⋅=Φb) The total electric flux is 0654321=Φ+Φ+Φ+Φ+Φ+Φ=Φ.23-6 a) Because the S1 encloses the charge q1, the electric flux through the S1 isb) Because the S2 encloses the charge q2, the electric flux through the S2 isc) Because the S3 encloses the charge q1and q2, the electric flux through the S3 isd) Because the S4 encloses the charge q1 and q3, the electric flux through the S4 ise) Because the S5 encloses the charge q1 ,q2 and q3, the electric flux through the S5 is23-8 a) When the spherical surface ’s radius is 0.500m, the charges q1 and q2 aren ’t closed by this surface, so the electric flux is zero. b) When the spherical surface ’s radius is 1.50m, the second charge q2 is closed by this surface, so the electric flux is)/(1045.01085.81000.426126011C m N q ∙⨯=⨯⨯==Φ---ε)/(1088.01085.81080.726126022C m N q ∙⨯-=⨯⨯-==Φ---ε)/(1043.01085.81080.71000.42612660213C m N q q ∙⨯-=⨯⨯-⨯=+=Φ----ε)/(1072.01085.81040.21000.42612660314C m N q q ∙⨯=⨯⨯+⨯=+=Φ----ε)/(1016.01085.81040.21080.71000.4261266603215C m N q q q ∙⨯-=⨯⨯+⨯-⨯=++=Φ-----εc) When the spherical surface ’s radius is 2.50m, these two charges enclose by the surface, so the electric flux is23-24 a) When r <a, the Gaussian surface doesn’t enclose any electric charges, so the magnitude of electric field is zero.When a <r <b, the Gaussian surface encloses the charge q, so the magnitude of electric field is When b <r <c, because the hollow sphere is a conductor, the electric field is zero inside it. When r>c, the Gaussian surface encloses the charge q, so the magnitude of electric field ise)b)c m N q 2612602/1068.01085.81000.6⨯-=⨯⨯-==Φ--εc m N q q 261266012/10025.01085.81000.61000.4⨯-=⨯⨯-⨯=+=Φ---ε204r qπε=Φ204r q πε=Φc) The negative charge –q is on the inner surface of the hollow sphere. d) The positive charge +q is on the outer surface of the hollow sphere.23-27 a) when r <a, the surface isn ’t closed any charge, so the electric field is E = 0 b) When a <r <b, the surface doesn ’t enclose any charge, so the electric field is E = 0 c) When b <r <c, the surface does enclose the charge +2q, so the electric field isd) When c <r <d, the surface doesn ’t enclose any net charge, so the electric flux is E = 0. e) When r > d, the surface does enclose the charge +2q and +4q, so the electric flux isThe total charge on the inner surface of the small shell is zero. The total charge on the outer surface of the small shell is +2q. The total charge on the inner surface of the large shell is -2q. The total charge on the outer surface of the large shell is +6q.23-28. a) when r <a, the surface doesn ’t closed any charge, so the electric field is E = 0. b) When a <r <b, the surface doesn ’t enclose any charge, so the electric field is E = 0. c) When b <r <c, the surface does enclose the charge +2q, so the electric field isd) When c <r <d, the surface doesn ’t enclose any net charge, so the electric flux is E = 0. e) When r > d, the surface does enclose the charge +2q and -2q, so the electric flux isO a b c d2042r q E πε=202046442r q r q q E πεπε=+=2042r q E πε=042220=-=rq q E πεThe total charge on the inner surface of the small shell is zero. The total charge on the outer surface of the small shell is +2q. The total charge on the inner surface of the large shell is -2q. The total charge on the outer surface of the large shell is zero.23-29. a) when r <a, the surface doesn ’t closed any charge, so the electric field is E = 0. b) When a <r <b, the surface doesn ’t enclose any charge, so the electric field is E = 0. c) When b <r <c, the surface does enclose the charge +2q, so the electric field ise) When r > d, the surface does enclose the charge +2q and -4q, so the electric flux isThe total charge on the inner surface of the small shell is zero. The total charge on the outer surface of the small shell is +2q. The total charge on the inner surface of the large shell is -2q. The total charge on the outer surface of the large shell is -2q.23-33 a) Between the cylinders, using the Gauss ’s law,the electric field isb) At any point outside the outer cylinder, the electric field isc)O a b c2c2042r q E πε=202042442r q r q q E πεπε-=-=⎰=⋅sq s d E 0ε 02ελπl l r E =⋅⋅r E 02πελ=r E 02πελ=。
⼤学物理双语（下）试题A卷2009─2010年第⼆学期《⼤学物理》双语试卷( A 卷)注意：1、本试卷共 3 页； 2、考试时间: 120 分钟 3、姓名、学号必须写在指定地⽅ 4、可以携带计算器Ⅰ. Filling the Blanks(每⼩题 2 分，共 20 分)1．⼀空⽓平⾏板容器，两板相距为d ，与⼀电池连接时两板之间相互作⽤⼒的⼤⼩为F ，在与电池保持连接的情况下，将两板距离拉开到3d ，则两板之间的相互作⽤⼒的⼤⼩是2. There is a point charge of electric quantity Q at the center of a cube, the electric flux through one surface of cube is3．如图1，⼀根⽆限长直导线通有电流I ，在P 点处被弯成了⼀个半径为2R 的圆，且P 点处⽆交叉和接触，则圆⼼O 处的磁感强度⼤⼩为_______________，⽅向为______________4. As shown in the figure 2, in the vacuum let the metal sphere with radius R be grounded, place a point charge q with a distance r (r>R) away from the center O of the sphere, the total induced charge on the surface of the metal sphere is5. 如图3所⽰，AOC 为⼀折成形的⾦属导线（AO =OC = L ），位于xoy 平⾯上. 磁感应强度为B 的匀强磁场垂直于xoy 平⾯. 当AOC 以速度v 沿x 轴正向运动时，导线上A 、C 两点间的电势差U AC = ，当以速度v 沿y 轴正向运动时A 、C两点中点电势⾼.6．⼀空⽓平⾏板电容器，接电源充电后电容器中储存的能量为W 0，在保持电源接通的条件下，在两极间充满相对电容率为r ε的各向同性均匀电介质，则该电容器中储存的能量W 为________________7． The period of a pendulum(单摆) is measured to be 3.0s in the reference frame of the pendulum. The period when measured by an observer moving at a speed of 0.95c relative to the pendulum is8．把⼀个静⽌质量为0m 的粒⼦，由静⽌加速到0.6v c =(c 为真空中的光速)需做功为Ⅱ．Choose the Correct Answer(每⼩题 3 分，共 30 分)1. 如图4所⽰，在真空中半径分别为R 和2R 的两个同⼼球⾯，其上分别均匀地带有电量+2q 和-2q ，今将⼀电量为+Q 的带电粒⼦从内球⾯处由静⽌释放，则该粒⼦到达外球⾯时的动能为：（）(A)04Qq Rπε.(C) 08Qq Rπε. (D)038QqRπε.2．In following statements which one is correct ( ) (A) The electric potential is definitely zero at the point where the electric field2qFigure 4Figure 3is zero.(B) The electric potential is non-zero at the point where the electric field is not zero.(C) The electric field is definitely zero at the point where the electric potential is zero.(D) The electric field is definitely zero in this area where the electric potential is constant3．有两个长直密绕螺线管，长度及线圈匝数均相同，半径分别为r 1和r 2．管内充满均匀介质，其磁导率分别为µ1和µ2．设r 1∶r 2=1∶2，µ1∶µ2=4∶1，当将两只螺线管串联在电路中通电稳定后，其⾃感系数之⽐L 1∶L 2与磁能之⽐Wm 1∶Wm 2分别为： ( )(B) L 1∶L 2=1∶2，Wm 1∶Wm 2 =1∶1 (C) L 1∶L 2=1∶2，Wm 1∶Wm 2 =1∶2 (D) L 1∶L 2=2∶1，Wm 1∶Wm 2=2∶14. One circular wire loop carries a clockwise current I in a uniform magnetic field B directed to the right as in the figure5. If B increases, what will happen to the net magnetic force and the torque on the current loop,respectively? ( ) (A) increase, increase (B) both keep the same (C) keep the same, decrease (D) keep the same, increase 5. 尺⼨相同的铁环与铜环所包围的⾯积中，通以相同变化率的磁通量，则环中： ( ) (A) 感应电动势相同，感应电流相同 (B) 感应电动势不同，感应电流不同 (C) 感应电动势相同，感应电流不同 (D) 感应电动势不同，感应电流相同Figure 56. The segment of wire in Figure 6 carries a current of I, where the radius of the circular arc is R, the magnitudeof the magnetic field at the origin O is ( )(A) R I 80µ (B) R I 40µ (C) RI πµ80 (D) R Iπµ207. 半径分别为2R 和R 的两个⾦属球，相距很远. ⽤⼀根长导线将两球连接，并使它们带电.在忽略导线影响的情况下，两球表⾯的电荷⾯密度之⽐为： ( )(A) 1/2 (B) 4 (C) 1/4 (D) 28. Which of the following statements is NOT true: ( ) (A) No two electric field lines can cross each other(B) The electric field vector is tangent to the electric field line at each point. (C) Magnetic field lines are always closed curves(D) The magnetic fields can be produced by current, so magnetic fields have sources9. 在某地发⽣两件事，静⽌位于该地的甲测得时间间隔为3s ，若相对甲以4c/5(c 表⽰真空中光速)的速率作匀速直线运动的⼄测得时间间隔为：(A) 2.4s (B) 4s (C) 3.6s (D) 5s ( )10. A spaceship is measured to be 120.0 m long and 20.0 m in diameter while at rest relative to an observer. If this spaceship now flies by the observer with a speed of 0.99c, the diameter does the observer measure is ( ) (A) 17m (B) 19m (C) 15m (D) 20mⅢ．如图7所⽰，两个同⼼的均匀带电球⾯，内球⾯半径为R 1，带电量Q 1,外球⾯半径为R 2，带电量为Q 2.设⽆穷远处为电势零点，求（1）空间各处电场强度的分布；（2）在内球⾯内，距中⼼为r 处的P 点的电势。
Chapter 24 exercises:24-3 a) When the speed of q 2 is 22.0m/s, the kinetic energy of q 2 is21121mv K =; the potential energy of system is 102114r q q U πε=.According to the conservation of energy , when the two spheres are 0.400m apart, the energy is22221mvK =, 202124r q q U πε=. 2211U K U K +=+. We get the speed of q 2 iss m v /5.122=.b) When the q 2 closes to the q 1, the speed of q 2 will slowly until its zero. So the system energy is113U K U +=;10212130214214r q q mv r q q πεπε+=. The distance between the two charges ism r 323.03=.24-7 a) When q 3 is placed at x = +10.0cm, the potential energy of system isJr q q r q q r q q U U U U 7129912991299303120321021*******.310.01085.814.341000.21000.410.01085.814.341000.2)1000.3(20.01085.814.34)1000.3(1000.4444----------⨯-=⨯⨯⨯⨯⨯⨯⨯+⨯⨯⨯⨯⨯⨯⨯-+⨯⨯⨯⨯⨯-⨯⨯=++=++=πεπεπεb) When the potential energy of the system is zero, the q 3 is placed 0444303120321021321=++=++=r q q r q q r q q U U U U πεπεπε3312321210r q q r q q r q q ++=; 2.023=+r rm r 0743.03=.24-9 If zero net work is required to place the three charges at the corners of the triangle, the third charge must be0444000='+'+rq q rq q rqq πεπεπε, 2q q -='.24-25 a) The electron will move with an acceleration to the center of the ring. After through the center, the electron will leave it until the speed is zero. Then it will return to the center and through it to move other side. So the electron will do oscillation.b) According to the conservation of energy, we get K U =,2002144mvReq req =-πεπε.2311291922129191011.92115.01085.814.341024106.130.015.01085.814.341024106.1v⨯⨯⨯=⨯⨯⨯⨯⨯⨯⨯--+⨯⨯⨯⨯⨯⨯⨯--------s m v /1067.17⨯=.24-26 We can suppose the zero potential energy at point p. The distance of point p is R. So we get the initial energy of proton is 2121ln2mv Rr q K U +=+πελ. The final energy of proton isR r q U '=ln202πελ. According to the conservation of energy, we know 22121mv U U =-.221ln2mv r r q ='πελ. The final distance between the line and the proton is cm r 8.15='.24-57 a) Take V = 0, r = b, when r < a, the potential isab dr rr d E r d E r d E V babrarbaln2200πελπελ==⋅+⋅=⋅=⎰⎰⎰⎰.When a < r < b, the potential is rb dr rV brln2200πελπελ==⎰.When r > b, the potential is v = 0.b) Between the inner cylinder and outer cylinder, the electric field is rE 02πελ=. So thepotential difference is ⎰==baab ab dr rV ln2200πελπελ.c) Use Eq. (24-23), we get ()()r a b V r b b rdrdV r E ab 1/ln 220⋅=⎪⎭⎫⎝⎛-⨯⨯-=-=πελ.24-73 a) Use the small element on the rod xdq dV 04πε=, at point P , we getxa x aQ xa x xdxxdq dV V ax xax x+=+====⎰⎰⎰++ln4ln4440000πεπελπελπε.b) At point R, we get⎪⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛++=+===⎰⎰⎰2002201ln 444y a y a a Q yx dxrdqdV V aπεπελπε.c) In part a, when the x is much larger than a, we can think the charged rod like a point charge, so the potential is xQ V 04πε=. In part b, the potential is yQ V 04πε=.24-78 a) From the equation ()2223),,(z y x A z y x V +-=, we can get at any point the electricfield is ()k z j y i x A k z V j y V i x V E 262+--=⎪⎪⎭⎫⎝⎛∂∂+∂∂+∂∂-=. b) At point()()m z y x 250.0,0,0,,=, the potential is )(0625.01V A V ⨯=. At point()()0,0,0,,=z y x , the potential is 02=V . The work done by the electric force isW q V V =-)(21. 651050.10625.01000.6--⨯⨯⨯=⨯A ; 640=Ac) At the point ()()m z y x 250.0,0,0,,=, the electric field is k k E320250.02640-=⨯⨯-=.d) In every plane parallel to the xz-plane, the equation is changed()2223640,,z x yz y x V +=+.So the equipotential contours are circles.e) The radius of the equipotential contour corresponding to V =1280V and y = 2.00m ism r 74.3)00.236401280(212=⨯+=.。
1. S: 2kv dtdva -==2kv dxdvv dt dx dx dv -==k d x v dvxx vv -=⎰⎰)(ln00x x k v v--= )(00x x k e v v --= (answer)2. S: j t i t dt rd v )3cos 15()3sin 15(+-== jt i t dtv d a )3sin 45()3cos 45(-+-==()()j t i t j t i t v r)3cos 15()3sin 15()3sin 5()3cos 5(+-⋅+=⋅j j t t i i t t⋅⋅+⋅⋅-=)3c o s 3s i n 75()3sin 3cos 75( 0= (proved c)3. S: dtdv v m k m f a =-==dt mkv dv t t v v -=⎰⎰0)(0t mkv t v -=0)(ln t m ke v t v -=0)( (answer) D: t m k e v dtdxv -==0dt e v dx t m k tt x -⎰⎰=00)(0kmv x e kmv ekmv t x t m k t t mk 0max 00),1()(=-=-=--4. S: )()32(j y d i dx j i x r d f dw+⋅+=⋅=dy xdx dw w fi32+==⎰⎰dy xdx 323342⎰⎰--+== -6 J (answer)5. S: 23230.60.4)0.30.4(t t t t t dtddt d +-=+-==θω, t t t dtddt d 60.6)30.60.4(2+-=+-==ωα 0.40300.60.4)0(2=⨯+⨯-=ω (answer of a)0.28)0.4(30.40.60.4)0.4(2=⨯+⨯-=ω rad/s (answer of a ) 60.266)0.2(=⨯+-=α rad/s 2 (answer of b )t t 60.6)(+-=α is time varying not a constant (answer of c) 6. S: ω20031222ML L v m L mv +⋅= MLmv ML L mv 4343020==ω (answer a))c o s 1(2)31(21m a x 22θω-=LMg ML ]1631[cos 2221maxgLM v m -=-θ (answer b) 7. G: m =1.0g, M =0.50kg, L =0.60m, I rod =0.0602m kg ⋅,s rod /5.4=ωR:I sys , v 0S: I sys =I rod +(M+m)L2=0.060+(0.50+0.0010)×0.602= 0.24 2m kg ⋅(answer)the system ’s angular momentum about rotating axis is conservative in the collision.sysI L mv ω=0s m mL I v sys/108.160.00010.024.05.430⨯=⨯⨯==ω (answer )D: The bullet momentum 0v m p=(before impact), its angular momentumabout rotating axis can be expressed as L mv 0(a scalar) 8. S:γ==00.800x xt v c -∆==0811800.600 3.0010t t γ∆=∆=⨯⨯ 9. S: 202202)(mc E cp E E γγ==+=222c p m c m c m c =10. S: 0i n t =-=∆n e t n e t W Q E n e t n e t W Q = 1(3010)(4.0 1.0)2=-- J 30= (answer)11. S: from nRT PV =and K T A 300= we can get:KT K T C B 100300== (answer of a)Change of internal energy between A and B:0)(23int =-=∆A B T T k n E (answer of b)The net work of the cycle:))(100300()13(2121m N AC BC W ⋅-⋅-=⋅=J 200= (answer of c) From the first law : W E Q +∆=int we can derive:the net heat of the whole cycle is J W Q 200== (answer)12. S: 131)(320===⎰⎰∞F v Av dv Av dv v p F33FvA =(answer of a ) F F v a v g v Av dv vAv v F4341420===⎰13. G: T 1=T 2=T , m 1, p 1, v rms,1, m 2, p 2=2p 1, v avg,2 = 2v rms,1 R: m 1 / m 2 S: v avg,2 =1.602m kTv rms,1 = 1.731m kTv avg,2 = 2v rms,167.4)60.173.12(221=⨯=m m (answer) 14. S: dE int =dQ – dWd Q = dE int + dW = n C v dT+pdV VdVnR T dT nC dV T p T dT nC T dQ dS v v +=+==if i f v VV v T T V V nR T T nC V dVnR T dT nC ds S f i filnln +=+==∆⎰⎰⎰ 15. S: dA E q θεcos 0⎰=212100)0.60100(1085.8⨯-⨯⨯=- C 61054.3-⨯= 16. S: 2041)(r Qr E πε=(R < r <∞) dr rQ dr r E udV dU 2022208421πεπε=⋅== RQ r dr Q udV U R0220288πεπε===⎰⎰∞(answer) RQ r dr Q U r r Rεπεεπε02202*88==⎰∞(answer ) 18. S: in the shell of r – r + drdr r R r dV r dq 204)/1()(πρρ-==)34(31)/(4)(4303200r Rr dr R r r dq r q r-=-==⎰⎰πρπρfrom the shell theorems , within the spherical symmetry distribution )34(12)(41)(20020r Rr Rr r q r E -==ερπε (answer of b)R r r R Rdr dE 320)64(12*00=⇒=-=ερ 00200*max 9])32(3324[12)(ερερRR R R R r E E =-⨯== 19. S: j yV i x V V gradV y x E∂∂-∂∂-=-∇=-=),( )0.20.2(y x x VE x +-=∂∂-= x yV E y 0.2-=∂∂-= )/(480.2)0.20.2()0.2,0.2(m V j i j x i y x E--=-+-=20. S: Q in = - q , Q out = q (answer ) 1010241241)0(R qq V q πεπε==104)0(R qV in πε-=204)0(R q V o u t πε=)0()0()0()0(out in q V V V V ++= )11(4210R R q +=πε21. S: from the planar symmetry and superposition principle, Emust in normal direction of the plates and 1σ,2σ,3σ,4σ must be const. Fromcharge conservationA Q S =+)(21σσ ⇒ SQ A=+21σσ (1) B Q S =+)(43σσ ⇒ SQ B=+43σσ (2) Apply Gauss ’ law in the closed surface shown in Fig. 032=+σσ (3)within the metal, 0=p Ewhich leads to002222432104030201=-++⇒=-++σσσσεσεσεσεσFrom(1), (2), (3), (4) yield:⎪⎪⎩⎪⎪⎨⎧-=-=+==S Q Q SQ Q B AB A 223241σσσσ (answer of a) (6 points) 004030201122222εεσεσεσεσS Q Q E BA p -=--+= (1 point) 004030201222222εεσεσεσεσS Q Q E BA p +=+++=(1 point) (answer of b) d S Q Q d E d E V BA p AB 012ε-==⋅= (2 points) (answer of c)27.33()(32)18w F x dx x dx J ==+=⎰⎰；at x=3m212W mV =， 6/V m s =。
