第七章 静电场中的导体和电介质
习题选解
7-1 如图所示，在一不带电的金属球旁有一点电荷q +，金属球半径为R ，已知q +与金属球心间距离为r 。试求：(1)金属球上感应电荷在球心处产生的电场强度E
及此时球心处的电势V ；(2)若将金属球接地，球上的净电荷为多少？
题7-1图
解：(1)由于导体内部的电场强度为零，金属球上感应的电荷在球心处产生的电场强度E 与点电荷q +在球心处产生的电场强度E
'大小相等，方向相反。
2
04r
q E E πε=
'= E
的方向由O 指向q +
点电荷q +在球心处的电势为
r
q V q 04πε=
金属球表面感应电荷在球心的电势为R V ，由于球表面感应电荷量总和为零，
??==
=
s
s
R dq
R
R dq V 041400
πεπε
故球心电势为q V 和R V 的代数和
r
q V V V R q 04πε=
+=
(2)若将金属球接地，金属球是一个等势体，球心的电势0=V 。设球上净电荷为
q '。球面上的电荷在球心处的电势为
??'=
=
=
s
s
R R
q dq R
R
dq
V 000
4414πεπεπε
点电荷q +在球心的电势为 r
q V q 04πε=
由电势叠加原理 0=+=q R V V V
q
R V V -=
r
q R q 0044πεπε-='
q
r
R q -
='
7-2 如图所示，把一块原来不带电的金属 板B 移近一块已带有正电荷Q +的金属板A ， 平行放置。设两板面积都是S ，板间距是d ， 忽略边缘效应。求：
(!)B 板不接地时，两板间的电势差；
Q
σ12σ34
题7-2图
(2)B 板接地时，两板间电势差。
解：(1)如图，设A 、B 两金属板各表面的面电荷密度分别为1σ、2σ、3σ、4σ。由静电平衡条件可知
???????=-++=---02222022220403020
10
4030201
εσεσεσεσεσεσεσεσ
解得 ??
?-==3
24
1σσσ
σ
又 4σ+3σ=0 Q S S =+21σσ 故 1σ=2σ=4σ=S
Q 2
3σ=S
Q 2-
两板间为匀强电场，电场强度 0
12εσ=
E +
22εσ—
32εσ—
42εσ=
S
Q 02ε
两板间的电势差 S
Qd Ed U 02ε=
=
(2)若B 板接地，则有
??
?
?
?=-===S Q 32410σσσσ
两板间的电场强度 =
E 0
22εσ—
32εσ=
S
Q
0ε
两板间的电势差 S
Qd
Ed U 0ε=
=
7-3 B A 、为靠得很近的两块平行的大金属平板，板的面积为S ，板间距离为d ，使B A 、板带电分别为A q 、B q ，且A q ＞B q 。求： (1)A 板内侧的带电量： (2)两板间的电势差。
解：(1)如图，设A 、B 两板各表面的 电荷面密度分别为1σ、2σ、3σ、4σ。 由题意 ??
?=+=+B
A q S S q S S 4321σσσσ ①
又由静电平衡条件（参考题7-2）得
???-==32
4
1σσσσ ②
题7-3图
，由①、②解得 ??
???
-=
-=+==S q q S q q B
A B A 223241
σσσσ 故Ａ板内侧的带电量 2
22B
A q q S q -==σ
(2)两板间为匀强电场，电场强度
12εσ=
E +
22εσ—
32εσ—
42εσ=
S
q q B A 02ε-
两板间电势差
=
=Ed U d S
q q B A 02ε-
7-4 如图所示，证明：对于两个无限大的平行平面带电导体板来说： (1)相向的两面（图中2和3）上，电荷的面密度总是大小相等而符号相反； (2)相背的两面（图中1和4）上，电荷的面密度总是大小相等而符号相同。
题7-4图
证明：(1)设两导体板各面的电荷面密度分别为1σ、2σ、3σ、4σ。取一个侧面垂直于带电导体板，端面分别在A 、B 板内的封闭圆柱形高斯面。由高斯定理：
?∑=
?s
q d 0
εS E
32
)(εσσ
S
S E S E d B A ?+=
?+?+??
内侧
内侧S E
由于侧侧S E
d ⊥ 故
=侧
侧0S E
d
又 0==内内B A E E
32
)(εσσ
S
?+=0 32σσ-=
(2)在导体B 内任取一点P ，P 的场强P E
=0，以向右为正向 =P E 1E +2E +3E +4E =0
12εσ+
22εσ+
32εσ—
42εσ
代入 32σσ-= 故 41σσ=
7-5 两平行金属板分别带等量的正负电荷，两板的电势差为120伏，两板的面积都是26.3cm ，两极相距mm 6.1，略去边缘效应，求两板间的电场强度和各板上所
带的电量。
解：两板间为均匀电场，电场强度
1
4
105.7-??==
m
V d
U E 又 E =
2εσ
+
2εσ
=
εσ
故 2701064.6--??==m C E εσ 各板所带电量 C S Q 101038.2-?==σ
7-6 如图所示，半径为1R 的导体球带有电荷q ，球外有一个内半径为2R 的同心导体球壳，壳上有电荷Q 。（1）求球与壳的电势差12U ；（2）用导线把球和壳联接在一起后，其电势为多少？
解：（1）导体球与球壳之间的电 场强度为 2
04r
q E πε=
题7-6图
球与壳的电势差 ?
?
=
=
2
1
2
1
2
124R R R R r
dr q Edr U πε
)11(
42
1
R R q
-
=
πε
（2）用导线把球与球壳连接在一起后，导体球和导体球壳的电荷重新分布。静电平衡时，球与球壳为等势体，1221V V V ==。所有电荷（Q q +）均匀分布在球壳外表面。球壳外电场强度为
2
04r
q Q E πε+=
'
球与球壳的电势 ?
?
∞
∞
+=
+=
=
2
2
2
02
44'R R R q Q r
dr q Q Edr V πεπε
7-7 如图所示，同轴传输线由圆柱形长直导体和套在它外面的同轴导体管构成。设圆柱体的电势为1V ，半径为1R ；圆管的电势为2V ，内半径为2R ，求它们之间
离轴线为r 处(1R ＜r ＜2R )的电势。
题7-7图
解：设圆柱体表面沿轴线单位长度所带电量为λ，在距轴线为r 的任意一点
P
的场强为
r
E 02πελ
=
1R ＜r ＜2R
P
点与圆柱体的电势差
?
?
=
=
=
-r
R r
R P R r r
dr Edr V V 1
1
1
1ln
22πε
λ
πε
λ
①
圆管与圆柱体的电势差
?
?
=
=
=
-2
1
2
1
1
20
21ln
22R R R R R R r
dr Edr V V πε
λπε
λ ②
由①、②两式消去λ，得P 点电势
)
l n ()l n ()
(1
21211R R R r V V V V P --=
7-8 实验表明：在靠近地面处有相当强的电场，E
垂直于地面向下，大小约为
1
100-?m
V 。试求：
（1）地面的面电荷密度；
（2）地面的每平方米所受的库仑力。
解：设地球带电荷q 。由高斯定理，地球表面电场204e
R
q E πε=
E R q e 204πε=
电荷q 均匀分布于地球表面，则地面面电荷密度
2
10
022
010
85.844--??===
=
m
C E R
E R S q e e εππεσ
地面每平方米受库仑力 N E F 81085.8-?==σ
7-9 如图所示，一平行板电容器两极板 间充满了电容率为ε的均匀介质，已知极板上 的面电荷密度分别为0σ和0σ-。略去边缘效 应。求电介质中的电场强度E 、极化强度P
、
题7-9图
电位移D
，介质表面的极化电荷面密度σ'。
解：对于平行板电容器，两板间的电场强度为
0n E
ε
σ=
其中0n
为沿极板法线方向的单位矢量，方向从0σ极板指向–0σ极板。两极板电介质中的电位移为
00n E D
σε==
极化强度 0
000
00000)1(n n
n E D P
σε
εσε
εσε-
=-
=-=
由于极化电荷都在介质的上下两表面，故极化电荷体密度0='ρ。两极板间介质中的电场E
为板上自由电荷产生的电场0E
和介质表面束缚电荷产生的电场
E '
的叠加。设介质表面极化电荷面密度为σ'。
000εσ=
E 0
εσ'=
'E E E E '-==
00εσ
)(1
00
0σσεε
σ'-= 00)1(σε
εσ-='
对于靠近带正电荷极板的介质表面，极化电荷面密度为—σ'。靠近带负电荷极板的介质表面，极化电荷面密度为σ'。
7-10 如图所示，平行板电容器两极板相距为d ，接到电压为V 伏的电源上，在其间插入厚为x 、相对电容率为r ε的玻璃平板。略去边缘效应，求空隙中和玻璃中的电场强度。
题7-10图
解：设电容器极板上电荷面密度为0σ，则两极板间空气间隙中的场强为
00εσ=
E
玻璃平板中的场强为 r
E εεσε
σ000=
=
'
E E r '=ε0 ① 两极板间的电位差 x E x d E U '+-=)(0 ② 由①、②两式可得 x
x d U E r r +-=
)(0εε
x
x d U
E r +-=
')(ε
7-11 如图所示，一平行板电容器两极板 面积都是20.2m ，相距为mm 0.5，两板加
V
1000电压后，取去电源，再在其间充满两
层均匀介质，一层厚mm 0.2，0.51=r ε；另 一层厚mm 0.3，0.22=r ε，。略去边缘效应， 试求：
题7-11图
（1）各介质中电极化强度P
的大小；
（2）当电容器靠近介质2的极板接地（即电势为零）时，另一极板（正极板）的电势是多少？两介质接触面上的电势是多少？
解：（1）充介质之前电容器电容为
d
S
C 0ε=
两板加1000V 电压，板上带电荷量
U d
S
CU Q 0ε=
=
电容器极板的电荷面密度为
2
6
010
77.1--??==
=
m
C d
U
S Q εσ
去掉电源，板内充两层介质，各层介质中的场强和电位移分别为
1
01r E εεσ=
σ=1D
2
02r E εεσ=
σ=2D
各介质层中的电极化强度P 为
11011r E D P εσσε-
=-==261042.1--??m C
2
2022r E D P εσσε-
=-==271085.8--??m C
（2）带负电的极板2接地，极板1的电势为 2
22
011
02211108.3?=+
=
+=d d d E d E V r r εεσεεσV
两介质接触面上的电势 V d E V 222100.3?=='
7-12 在相对电容率为1r ε、半径为R 的均匀电介质球的中心有一点电荷q ，介质球外的空间充满相对电容率为2r ε的均匀电介质。求距q 为r (r ＜R )处的场强及电势（选无穷远处为电势零点）。
解：介质球中心的点电荷q 产生的电场具有球对称性。由高斯定理，介质球内外的场强分别为
2
1014r
q E r επε=
（r
＜R ）
22024r
q E r επε=
（r ＞R ）
选无穷远处为电势零点，距q 为r （r ? ? ∞ + - = + = R r r R r R q R r q dr E dr E V 2010214)11 ( 4επεεπε 7-13 解：设电介质的相对介电常量为r ε，则在插板前电容器两极板间的电势差为 02 .0εσεσ=?= d U 式中σ为电容器极板的面电荷密度，插入平板材料后电容器两极板间电势差为 ()t t d U r εεσεσ00 '+ -'= ' 式中t 为平板材料的厚度，t=1.5cm ，即： 015 .005.000 ?'+ ?'= 'r U εεσεσ 由于在插入过程中电容器的电荷保持不变，所以σσ'=，在由题意可知 02.06.0015.0005.000??=??? ? ? ? += 'εσ εεσr U 解得 14.2=r ε 7-14 有一面积为S 、间距为d 的平行板电容器。 (1)在板间平行于极板面插入厚度为3 d ，面积也为S 的相对电容率为r ε的均匀电 介质板，计算其电容； (2)若插入的是同样尺寸的导体板，求其电容； (3)上下平移介质板或导体板对电容有无影响？ 解：设电容器极板所带电荷面密度为σ (1)两极间电势差 332211d E d E d E V V B A ++=- = 30 2010 d d d r εσεεσεσ+ + 20310 )(d d d r εεσεσ+ += 由 3 31d d d d - =+， 3 2d d = r B A d d V V εεσεσ00 332+ = - 电容 d S d S V V Q C r r r r B A )12(3) 12( 300 +=+=-= εεεεεεσσ V V B A 题7-14图 (2)若插入导体板则02=E 3311d E d E V V B A +=-= 30 10 d d εσεσ+ = 310 32)3 ()(εσεσεσd d d d d = -= + 电容 d S d S V V Q C B A 233200 εεσσ== -= (3)上下移动介质板或导体板对电容无影响。 7-15 如图所示，一无限大平行板电容器，设B A 、两板相距cm 0.5，板上各带电荷26103.3--??=m C σ，A 板带正电，B 板带负电并接地（地的电势为零），求： (1)在两板之间距A 板cm 0.1处P 点的电势； (2)A 板的电势。 解：(1)平板电容器两板间场强 εσ= E P 点电势 V r d V r d E V B P 4 1049.1)()(?=-= +-=εσ 题7-15图 (2)A 板的电势 V d V Ed V B A 4 1086.1?== +=εσ 7-16 面积是0.2平方米的两平行导体板放在空气中相距0.5毫米，两板电势差为 1000 伏特，略去边缘效应。试求： (1)电容C ； (2)各板上的电量Q 、电荷的面密度σ和板间电场强度E 的值。 解：(1)平板电容器电容 F F d S C με3 9 010 54.310 54.3--?=?== (2)各板上的电量 C CU Q AB 6 10 54.3-?== 板上电荷的面密度 2 6 1077.1--??== m C S Q σ 板间电场强度E 的值 1 50 100.2-??== C N E εσ 7-17 如图所示，电容器由三片面积 都是20.6cm 的锡箔构成，相邻两箔间距离 都是mm 10.0，外边箔片联在一起成为一 题7-17图 极，中间箔片作为另一极， (1)求电容C ； (2)若在这电容器上加V 220电压，问三箔上电荷的面密度各是多少？ 解：(1)三片锡箔组成的电容器，其电容相当两个电容器的并联。 pF F d S d S C C C 2 10 00211006.110 06.1?=?=+ = +=-εε (2)总电量 C CU Q 81034.2-?== 对于中间一片锡箔，总电量Q 均匀分布在箔的两面，故锡箔面电荷密度 2 5 1095.12--??== m C S Q σ 7-18 如图所示，同心球电容器内外半径分别为1R 和2R ，两球间充满相对电容率为r ε的均匀介质，内球带电量Q ，试求： (1)电容器内外各处电场强度E 和两球的电势差U ； (2)电介质中电极化强度P 和极化电荷面密度σ'； (3)电容U 。 解：(1)内球所带电荷Q 在外球壳内外两表面 感应出电荷－Q 和＋Q ，两球间及球外电场 题7-18图 具有球对称性，由高斯定理 02 014r r Q E r επε= （1R ＜r ＜2R ） 02 024r r Q E πε= （r ＞2R ） 两球的电势差 ? ? - = = =2 1 2 1 )11( 442 1 001R R r r R R R R Q r dr Q dr E U επεεπε (2)电介质中的极化强度 1010)(E E D P εεε-=-== 02 4)1(r r Q r r πεε- 极化电荷分布在靠近内外球表面的球面上，极化电荷面密度分别为 2 1 114)1(cos R Q P r r πεεθσ-= ='（靠近内球表面 πθ=1 ） 2 2 22 4)1(cos R Q P r r πεεθσ-=='（靠近内球表面02=θ） (3)由两球的电势差= U ) 11(42 1 0R R Q r -επε，电容器电容为 1 22 102 1 04) 11( 41R R R R R R Q U Q C r r -=- = = επεεπε 7-19 两个电容器1C 和2C ，分别标明为V pF C 5002001：，V pF C 9003002：。把它们串联后，加上V 1000电压，是否被击穿？ 解： 两电容串联后的等效电容 pF C C C C C 1202 121=+= 电容所带电量 CU Q ==C 7102.1-? 1C 上的电压 V C Q U 6001 1== 2C 上的电压 V C Q U 4002 2== 1U 大于1C 的耐压值，1C 先被击穿。接着V 1000电压加在2C 上，2C 也被击穿。 7-20 解：并前总能量()C Q C Q C Q W 252222 2 2 = + = 并后C C 2=总 Q Q 3=总 C Q C Q 23U == 总 总 总能量C Q C Q C CU W 49232 122 2 2 = ?? ? ???=? =' 并后总能量变化C Q W W 42 = -' 7-21 一电容率为0ε的无限大均匀介质中有一个半径为R 的导体球，带电荷Q 。求电场的能量。 解：导体球的电荷均匀分布在外表面，球内不存在电场，电场只存在于球体外，其空间分布为 01=E 0＜r ＜R 2 024r Q E πε= r ＞R 此时，电场的能量为 ? ? ? ∞ ∞ ∞ = = = = R R R R Q r dr Q dr r r Q dV E W 02 2 2 2 22 0022 0884)4( 2 12 1πεπε ππεεε 7-22 一空气球形电容器内外球壳的半径分别为1R 和2R ，分别带有等量异号电荷，电势差为U 。试求：（1）电势能； （2）电场的能量。 解：（1）介质为空气的球形电容器的电容为 1 22104R R R R C -= πε 电势能为 = = = 2 2 2 121AB CU C Q W 1 22 2104R R U R R -πε （2）球形电容器介质层中的电场强度为 2 04r Q E πε= 其中 AB CU Q ==1 22104R R U R R -ε 故 2 041r E πε= 1 22104R R U R R -πε=2 1221)(r R R U R R - 由于内球壳中场强为零，外球壳外场强也为零，故电场能量储存在内外球壳之间。在两球壳间取体积元dr r dV 24π=，其电场能量 dr r E dr r E dV E dW 2 202202 0242 12 1πεπεε== = 全部电场中的能量 ?? ? -=?? ????-==2 1 2 1 2 122 222102 212212 0)(2)(2R R R R r dr R R U R R dr r R R U R R r dW W πεπε 1 22 2102 1 2 122 222 102)11( ) (2R R U R R R R R R U R R -= - -= πεπε 7-23 解：（1）这个由导体球和同心导体球壳的系统将空间分为4个部分，分别为导体球内（0＜r ≤R ＝；球与壳之间（R 1＜r ＜R 2＝；球壳之间（R 2≤r ≤R 3）及球壳之外（r>R 3）。由于球体与球壳均为导体，所以导体球内与球壳之间的电场强度为零。由高斯定理易得此时全空间场强的表达式为 2 014r Q πε （R 1＜r ＜R 2） =E 2 0214r Q Q πε+ （r>R 3） 系统储藏电能为 ? ?? ∞ ??? ? ??++??? ? ??= = 3 2 1 22 2021022 201044214421R v R R e e dr r r Q Q dr r r Q dW W ππεεππεε ()302 2121 2 118118R Q Q R R Q πεπε ++???? ??-= (2) 如果用导线将球与壳连在一起，此时电荷只分布在球壳外表面上。由高斯定 理知：球壳外表面以内电场强度为0，球壳外表面以外电场强度表达式为 2 0214r Q Q E πε+= （r>R 3） 系统储藏电能为 ? ? ∞ ' ??? ? ??+== '3 22 202104421R v e e dr r r Q Q dW W ππεε() 302 218R Q Q πε+= 7-24 如图所示，两平行导体板面积为S 、间距为d ，在它们中间平行地插入一层厚为t 、电容率ε的电介质，求下 列两种情况下，插入介质后能量改变的值。 （1）维持两极板电荷Q 不变时插入介质； （2）维持两极板电压U 不变时插入介质。 解：未插入介质极前，平板电容器的电容为 题7-24图 d S C 0ε= 插入介质后，平板电容器的电容变为 t t d S C 00)(εεεε+-= ' （1）维持两极板电荷Q 不变时插入介质板，插入前后电场的能量分别为 C Q W 22 = ， C Q W ' = '22 能量改变的值 ?? ??? ?-+-= -'=?S d S t t d Q W W W 0002)(21 εεεεε S tQ S d t t d Q εεεεεεεεε02 0002 2)())(( 21-- =-+-= （2）维持两极板电压U 不变时插入介质板，插入前后电场的能量分别为 2 21CU W = 2 2 1U C W '= ' 能量改变的值 ))(( 210002 d S t t d S U W W W εεεεε- +-= -'=? [] [] t d d StU t t d d St St Sd S d U )(2)()(2 102 0002 00002 εεεεεεεεεεεεεεε---= +--+-= 7-25 一平行板电容器，板的面积为S 、极板间距离为d ,把它充电到两极板电势差为U 时去掉电源，然后把两极板拉开到距离为d 2。略去边缘效应，试求： （1）分开两极板所需的功；（2）两极板的电势差；（3）电容器所储存的能量。 解：当极板间距为d 时，电容器电容为 d S C 01ε= 当极板间距为2d 时，电容器电容为2 2102C d S C = = ε （1）由于在拉开极板前电池已撤去，所以板上电荷量不变， d SU U C Q 01ε= = 分开两极板所需的功为电容器储存能量的增加量 d SU d SU S d S d C Q C Q W W A 2))( 2( 21 21212 02 0001 2 2 2 12εεεε= - = - = -= （2）两极板的电势差 U S d d SU C Q V V B A 22002 == = -εε （3）两极板拉开后电容器所储存的能量 d SU U d S U C W AB 2 02 02242212 1εε= ?= = 7-26 一平行板电容器极板的面积为S ，极板间距离为d ，在它中间有一块厚度为t 、相对电容率r ε的介质板，把两极板充电到电势差为U 。略去边缘效应。 题7-26图 （1）断开电源，把这个介质板抽出，问抽出时要做多少功？ （2）如果在不断开电源的情况下抽出，则要做多少功？ 解：介质板未抽出前，电容器的电容为 t t d S C r r +-= )(01εεε 介质板抽出后，电容器的电容 d S C 02ε= （1）断开电源，把介质板抽出，极板电量Q 不变 t t d SU U C Q r r AB +-= =)(01εεε 抽出介质板需要作的功等于电容器电能的增量 2 0001 2 2 2 12)()(212121?? ? ? ??+-??????+--=- = -=t t d SU S t t d S d C Q C Q W W A r r r r εεεεεεε = [] [] 2 202 2 2 2 2 00)1(2)1()()1(21t d StU t t d U S S t r r r r r r r r ---= +--εεεεεεεεεεε （2）不断开电源抽出介质板，两极板电势差不变，抽出介质板要做的功等于电容器电能的增量 2122122 12 1AB AB U C U C W W A - =-= []d t d U St U t t d S U d S r r r r r )1(2)1()(21 212 02 02 0----= +-- = εεεεεεεε 翻译在最后 Conversation One M: Guess what? The worst food I've ever had was in France. W: Really? That's odd. I thought the French were all good cooks. M:Yes. That's right. I suppose it's really like anywhere else, though. You know, some places are good. Some bad. But it's really all our own fault. W: What do you mean? M: Well, it was the first time I'd been to France. This was years ago when I was at school. I went there with my parents' friends, from my father's school. They'd hired a coach to take them to Switzerland. W: A school trip? M: Right. Most of them had never been abroad before. We'd crossed the English Channel at night, and we set off through France, and breakfast time arrived, and the coach driver had arranged for us to stop at this little café. Th ere we all were, tired and hungry, and then we made the great discovery. W: What was that? M: Bacon and eggs. W: Fantastic! The real English breakfast. M: Yes. Anyway, we didn't know any better— so we had it, and ugh...! W: What was it like? Disgusting? M: Oh, it was incredible! They just got a bowl and put some fat in it. And then they put some bacon in the fat, broke an egg over the top and put the whole lot in the oven for about ten minutes. W: In the oven! You're joking. You can't cook bacon and eggs in the oven! M:Well. They must have done it that way. It was hot, but it wasn't cooked. There was just this egg floating about in gallons of fat and raw bacon. W: Did you actually eat it? M: No! Nobody did. They all wanted to turn round and go home. You know, back to teabags and fish and chips. You can't blame them really. Anyway, the next night we were all given another foreign speciality. W: What was that? M: Snails. That really finished them off. Lovely holiday that was! Questions 1 to 4 are based on the conversation you have just heard. Question 1. What did the woman think of the French? Question 2. Who did the man travel with on his first trip to Switzerland? Question 3. What does the man say about the breakfast at the little French café? Part III Listening Comprehension (35 minutes) Section A Directions: In this section, you will hear 8 short conversations and 2 long conversations. At the end of each conversation, one or more questions will be asked about what was said. Both the conversation and the questions will be spoken only once. After each question there will be a pause. During the pause, you must read the four choices marked A), B), C) and D), and decide which is the best answer. Then mark the corresponding letter on Answer Sheet 2 with a single line through the centre. Now let’s begin with the eight short conversations: 11. M: I’m asked to pick up the guest speaker Bob Russel at the airport this afternoon, do you know what he looks like? W: Well, he’s in his sixties, he stands out, he’s bald, tall and thin and has a beard. Q: What do we conclude from the woman’s remarks about Bob Russel? 12. M: I am considering dropping my dancing class. I am not making any progress. W: If I were you, I stick with it. It’s definit ely worth time and effort. Q: What does the man suggest the woman do? 13. W: You see I still have this pain in my back, this medicine the doctor gave me was supposed to make me feel better by now. M: Maybe you should’ve taken it three times a day as you w ere told. Q: What do we learn from the conversation? 14. M: Frankly, when I sat the back of the classroom, I can’t see the words on the board clearly. W: Well, you’ve been wearing those same glasses as long as I’ve known you. Why not get a new pair? It wo uldn’t cost you too much. Q: What does the woman imply about the man’s glasses? 15. W: How come the floor is so wet? I almost slipped, what happened? M: Oh, sorry! The phone rang the moment I got into the shower, anyway, I’ll wipe it up right now. 2002年12月听力原文 Section A: 1. M: How well are you prepared for your presentation? Your turn comes on next Wednesday. W: I spend a whole week searching on the net. But it came up with nothing valuable. Q: What did the women say about her presentation? 2. W: Good morning, Jack. Late again! What’s the excuse this time? M: I’m awfully sorry I must have turned the alarm off and gone back to sleep again. Q: What do we learn from the conversation? 3. W: Excuse me, but could you tell me where I can change American Dollars into British Pounds? M: There is a bank around the corner, but I’m afraid it’s already past it’s closing time. Why don’t you try the one near the railway station? Q: What does the man mean? 4. M: Could I speak to Dr. Chen? She told me to call her today. W: She’s not available right now. Would you like to try around three? Q: What does the woman tell the man to do? 5. W: Oh dear, I’m afraid I’ll fail again in the national test. It’s the third time I took it. M: Don’t be too upset. I have the same fate. Let’s try a fourth time. Q: What does the man mean? 6. W: Professor Smith, I really need the credits to graduate this summer. M: Here of this school, the credits are earned, not given. Q: What do we learn from the conversation? 7. M: How did you go to Canada, Jane? Did you fly? W: I was planning to, because it’s such a long trip by bus or by train, but Fred decided to drive and invited me to join him. It took us two days and one night. Q: What can we infer from the conversation? 8. M: How do you like the way I’ve arranged the furniture in my living room? W: Fine, but I think the walls could do with a few paintings. Q: What does the woman suggest the man do? 9. W: I don’t imagine you have any interest in attending my lecture on drawing, do you? M: Oh, yes, I do. Not that you remind me of it. Q: What do we learn about the man from the conversation? 10. M: You are my campaign manager. What do you think we should do to win the election? I’m convinced I’m the best candidate for the chairman of the Student Union. W: We won’t be able to win unless you get the majority votes from the women students. Q: What is the man doing? Section B Passage One There are some serious problems in the Biramichi river. The local chamber of commerce, which represents the industry in the area, hired me as a consultant to do a one year study on fishing in the Biramichi river and write a rep ort for them. This is my report: One of the major problems in the Biramichi river is that the level of oxygen in the water is too low. Several chemicals have displaced the oxygen. This chemical pollution has two sources: the 1．拓扑排序的结果不是唯一的，试写出下图任意2个不同的拓扑序列。 2．写出求以下AOE网的关键路径的过程。要求：给出每一个事件和每一个活动的最早开始时间和最晚开始时间。 【解析】解题关键是弄清拓扑排序的步骤 （1）在AOV网中，选一个没有前驱的结点且输出；（2）删除该顶点和以它为尾的弧；（3）重复上述步骤直至全部顶点均输出或不再有无前驱的顶点。 【答案】（1）0132465 （2）0123465 【解析】求关键路径首先求关键活动，关键活动ai的求解过程如下 （1）求事件的最早发生时间ve(j), 最晚发生时间vl(j); （2）最早发生时间从ve(0)开始按拓扑排序向前递推到ve(6), 最晚发生时间从vl(6)按逆拓扑排序向后递推到vl(0); （3）计算e(i),l(i)：设ai由弧 关键路径为：a0->a4->a6->a9 7.1选择题 1．对于一个具有n个顶点和e条边的有向图，在用邻接表表示图时，拓扑排序算法时间复杂度为（B） A）O(n) B）O(n+e) C）O(n*n) D）O(n*n*n) 2．设无向图的顶点个数为n，则该图最多有（B）条边。 A）n-1 B）n(n-1)/2 C）n(n+1)/2 D）n2 3．连通分量指的是（B） A）无向图中的极小连通子图 B）无向图中的极大连通子图 C）有向图中的极小连通子图 D）有向图中的极大连通子图 4．n个结点的完全有向图含有边的数目（D） A）n*n B）n（n+1） C）n/2 D）n*（n-1） 5．关键路径是（A） A）AOE网中从源点到汇点的最长路径 B）AOE网中从源点到汇点的最短路径 C）AOV网中从源点到汇点的最长路径 D）AOV网中从源点到汇点的最短路径 6．有向图中一个顶点的度是该顶点的（C） A）入度B）出度C）入度与出度之和D）（入度+出度）/2 7．有e条边的无向图，若用邻接表存储，表中有（B）边结点。 A） e B）2e C）e-1 D）2(e-1) 8．实现图的广度优先搜索算法需使用的辅助数据结构为（B） 巨微英语 2016 年 12 月英语六级听力原文及参考答案听力稿原文 section A Conversation 1 气候变化和全球经济发展 W： Professor Henderson could you give us a brief overview of what you do ，where you work and your main area of research ？ M： Well the Center for Climate Research where I work links the science of climate change to issues around economics and policy。 Some of our research is to do with the likely impacts of climate change and all of the associated risks 。 W：And how strong is th e evidence that clima te change is happening that it ‘ s really something we need to be worried about 。 M： Well most of the science of climate change particularly that to do with global warming is simply fact 。 But other aspects of the science are less certain or at least more disputed 。And so we ‘ re really talking about risk what the economics tells us is that it ’s probably cheaper to avoid climate change to avoid the risk than it has to deal with the likely consequences 。 W： So what are we doing ？ What can we do about it ？ M：Well I would argue that we need to develop the scienc e specifically to understa nd the likely impacts of climate change in different contexts 。 六级听力试题及答案（红色字体为答案） Part III Section A 11. A)He is quite easy to recognize B)he is an outstanding speaker C)he looks like a movie star D)he looks young for his age 12. A)consult her dancing teacher B)take a more interesting class C)continue her dancing class D)improve her dancing skills 13. A)the man did not believe what the woman said B)the man accompanied the woman to the hospital C)the woman may be suffering from repetitive strain injury D)the woman may not followed the doctor’s instructions 14. A)they are not in style any more B)they have cost him far too much C)they no longer suit his eyesight D)they should be cleaned regularly 15. A)he spilled his drink onto the floor B)he has just finished wiping the floor C)he was caught in a shower on his way home D)he rushed out of the bath to answer the phone 16. A)fixing some furniture B)repairing the toy train C)reading the instructions D)assembling the bookcase 17. A)urge Jenny to spend more time on study B)help Jenny to prepare for the coming exams C)act towards Jenny in a more sensible way D)send Jenny to a volleyball training center 18. A)The building of the dam needs a large budget B)the proposed site is near the residential area C)the local people fel insecure about the dam D)the dam poses a threat to the local environment Question19 to21 are based on the conversation you have just heard 19 A. It saw the end of its booming years worldwide 第七章图 一、选择题 1．图中有关路径的定义是（ A ） A．由顶点和相邻顶点序偶构成的边所形成的序列 B．由不同顶点所形成的序列 C．由不同边所形成的序列 D．上述定义都不是 2．设无向图的顶点个数为n，则该图最多有（B ）条边。 A．n-1 B．n(n-1)/2 C． n(n+1)/2 D．0 E．n2 3．一个n个顶点的连通无向图，其边的个数至少为（ A ）。 A．n-1 B．n C．n+1 D．nlogn； 4．要连通具有n个顶点的有向图，至少需要（ B ）条边。】 A．n-l B．n C．n+l D．2n 5．n个结点的完全有向图含有边的数目（ D ）。 A．n*n Ｂ．n（n＋１） C．n／2 D．n*（n－l） 6．一个有n个结点的图，最少有（ B ）个连通分量，最多有（ D ）个连通分量。 A．0 B．1 C．n-1 D．n 7．在一个无向图中，所有顶点的度数之和等于所有边数（ B ）倍，在一个有向图中，所有顶点的入度之和等于所有顶点出度之和的（ C ）倍。 A．1/2 B．2 C．1 D．4 8．用有向无环图描述表达式(A+B)*（（A+B）/A），至少需要顶点的数目为( A)。 A．5 B．6 C．8 D．9 9．用DFS遍历一个无环有向图，并在DFS算法退栈返回时打印相应的顶点，则输出的顶点序列是(A )。 A．逆拓扑有序 B．拓扑有序 C．无序的 10．下面结构中最适于表示稀疏无向图的是（ C ），适于表示稀疏有向图的是（ BDE ）。 A．邻接矩阵 B．逆邻接表 C．邻接多重表 D．十字链表 E．邻接表 11．下列哪一种图的邻接矩阵是对称矩阵？（ B ） A．有向图 B．无向图 C．AOV网 D．AOE网 12．从邻接阵矩可以看出，该图共有（B）个顶点；如果是有向图该图共有（B）条弧；如果是无向图，则共有（D）条边。 ①．A．9 B．3 C．6 D．1 E．以上答案均不正确 ②．A．5 B．4 C．3 D．2 E．以上答案均不正确 ③．A．5 B．4 C．3 D．2 E．以上答案均不正确 14．用相邻矩阵A表示图，判定任意两个顶点Vi和Vj之间是否有长度为m的路径相连，则只要检查（ C ）的第i行第j列的元素是否为零即可。 A．mA B．A C．A m D．Am-1 15.下列说法不正确的是（ C ）。 A．图的遍历是从给定的源点出发每一个顶点仅被访问一次 C．图的深度遍历不适用于有向图 B．遍历的基本算法有两种：深度遍历和广度遍历 D．图的深度遍历是一个递归过程 16．无向图G=(V,E),其中：V={a,b,c,d,e,f},E={(a,b),(a,e),(a,c),(b,e),(c,f),(f,d),(e,d)}，对该图进行深度优先遍历，得到的顶点序列正确的是（ D ）。 A．a,b,e,c,d,f B．a,c,f,e,b,d C．a,e,b,c,f,d D．a,e,d,f,c,b 17.设图如右所示，在下面的5个序列中，符合深度优先遍历的序列有多少？（ D ） a e b d f c a c f d e b a e d f c b a e f d c b a e f d b c A．5个 B．4个 C．3个 D．2个 18.下图中给出由7个顶点组成的无向图。从顶点1出发，对它进行深度优先遍历得到的序列是(C)，而进行广度优先遍历得到的顶点序列是（C）。 ①．A．1354267 B．1347652 C．1534276 D．1247653 E．以上答案均不正确 ②．A．1534267 B．1726453 C．l354276 D．1247653 E．以上答案均不正确 19．下面哪一方法可以判断出一个有向图是否有环（AB）： A．深度优先遍历 B.拓扑排序 C.求最短路径 D.求关键路径 20.在图采用邻接表存储时，求最小生成树的 Prim算法的时间复杂度为( B )。 A. O(n) B. O(n+e) C. O(n2) D. O(n3) 21.下面是求连通网的最小生成树的prim算法：集合VT，ET分别放顶点和边，初始为（ C），下面步骤重复n-1次: a：（ A）；b：（ B）；最后：（A）。 （1）．A．VT，ET为空 B．VT为所有顶点，ET为空 C．VT为网中任意一点，ET为空 D．VT为空，ET为网中所有边 （2）．A.选i属于VT，j不属于VT，且（i，j）上的权最小 B．选i属于VT，j不属于VT，且（i，j）上的权最大 2007年6月大学英语四级考试听力原文 Part ⅢListening Comprehension （35 minutes） Section A Directions: In his section, you will hear 8 short conversations and 2 long conversations. At the end of each conversation, one or more questions will be asked about what was said. Both the conversation and the questions will be spoken only once. After each question there will a pause. During the pause, you must read the four choices marked A ,B ,C and D, and decide which is the best answer. Then mark the corresponding letter on Answer Sheet 2 with a single line through the centre. 11. W: Did you watch the 7 o’clock program on Cannel (频道) 2 yesterday evening? I was about to watch it when someone came to see me. M: Y eah, it reported some major breakthrough (突破) in cancer research. People over 40 would find the program worth watching. Q: What do we learn from the conversation about the TV program? C) 12. W: I won the 1st prize in the National Writing Contest and I got this camera as an award(奖品). M: It’s a good camera. Y ou can take it when you travel. I had no idea you were a marvelous writer. Q: What do we learn from the conversation? D) 13. M: I wish I hadn’t thrown away that reading list! W: I thought you might regret it. That’s why I picked it up from the waste paper basket and left it on the desk. Q: What do we learn from the conversation? B) 14. W: Are you still teaching at the junior high school? M: Not since June. My brother and I opened a restaurant as soon as he got out of the army. Q: What do we learn about the man from the conversation? A) 15. M: Hi, Susan! Have you finished reading the book Professor Johnson recommended? W: Oh, I haven’t read it through the way I read a novel. I just read a few chapters which interested me. Q: What does the woman mean? C) 16. M: Jane missed class again, didn’t she? I wonder why. W: Well, I knew she had been absent all week. So I called her this morning to see if she was sick. It turned out that her husband was badly injured in a car accident. Q: What does the woman say about Jane? D) 17. W: I’m sure the Smiths’ new house is somewhere on this street, but I don’t know exactly where it is. M: But I’m told it’s two blocks from their old home. Q: What do we learn from the conversation? 2007年6月 11. A) Surfing the net. B) Watching a talk show. C) Packing a birthday gift. D) Shopping at a jewelry store. 12. A) He enjoys finding fault with exams. B) He is sure of his success in the exam. C) He doesn’t know if he can do well in the exam. D) He used to get straight A’s in the exams he took. 13. A) The man is generous with his good comments on people. B) The woman is unsure if there will be peace in the world. C) The woman is doubtful about newspaper stories. D) The man is quite optimistic about human nature. 14. A) Study for some profession. B) Attend a medical school. C) Stay in business. D) Sell his shop. 15. A) More money. B) Fair treatment. C) A college education. D) Shorter work hours. 16. A) She was exhausted from her trip. B) She missed the comforts of home. C) She was impressed by Mexican food. D) She will not go to Mexico again. 17. A) Cheer herself up a bit. B) Find a more suitable job. C) Seek professional advice. D) Take a psychology course. 18. A) He dresses more formally now. 第二篇 数学物理方程 ——物理问题中的二阶线性偏微分方程及其解法 Abstracts:1、根据物理问题导出数理方程—偏微分方程； 2、给定数理方程的附加条件：初始条件、边界条件、物理条件 (自然条件，连接条件)，从而与数理方程一起构成定解问题； 3、方程齐次化； ４、数理方程的线性导致解的叠加。 一、数理方程的来源和分类（状态描述、变化规律） 1、来源 I ．质点力学：牛顿第二定律F mr = 连续体力学2222()(,)(,)0(()0;v 1()0(Euler eq.).u r t a u r t t v t v v p f t ρρρ?????-?=?????????+??=????-?+??=+=????? 弹性定律弦弹性体力学 杆 振动：波动方程);膜流体力学：质量守恒律：热力学物态方程： II.麦克斯韦方程 ;;00;().,,,D D E l B s E B B B H l j D s H j D E u B A u A σρτρσ??=???=?=????=????=???=?=+????=+??=-?=????????? ???????????d d d d d d d 满足波动方程。Lorenz 力公式力学方程；Maxwell eqs.+电导定律电报方程。 III. 热力学统计物理 220;0.T k T t D t ρρ??-?=??????-?=??? 热传导方程：扩 散方程：特别: 稳态（0t ρ?=?）:20ρ?= (Laplace equation). IV . 量子力学的薛定谔方程： 22.2u i u Vu t m ?=-?+? 二、数理方程的导出 推导泛定方程的原则性步骤： （1）定变量：找出表征物理过程的物理量作为未知数（特征量），并确定影响未知函数的自变量。 （2）立假设：抓主要因素，舍弃次要因素，将问题“理想化” ---“无理取闹”（物理趣乐）。 （3）取局部：从对象中找出微小的局部（微元），相对于此局部一切高阶无穷小均可忽 略---线性化。 （4）找作用：根据已知物理规律或定律，找出局部和邻近部分的作用关系。 （5）列方程：根据物理规律在局部上的表现，联系局部作用列出微分方程。 Chapter 7 一维波动方程的傅里叶解 第一节 一维波动方程-弦振动方程的建立 7.1.1 弦横振动方程的建立 （一根张紧的柔软弦的微小振动问题） （1）定变量：取弦的平衡位置为x 轴。表征振动的物理量为各点的横向位移),(t x u ，从 而速度为t u ，加速度为tt u . （2）立假设：①弦振动是微小的，1<<α，因此，sin tan ααα≈≈，1cos ≈α，又 tan u x αα?=≈?，1< 2007年12月大学英语六级真题 Part ⅢListening Comprehension (35 minutes) 本题Section A&B共计25分，每小题1分。 Section A Directions: In this section, you will hear 8 short conversations and 2 long conversation. At the end of each conversation, one or more questions will be asked about what was said. Both the conversation and the questions will be spoken only once. After each question there will be a pause. During the pause, you must read the four choices marked A)，B)，C) and D)，and decide which is the best answer. Then mark the corresponding letter on Answer Sheet 2 with a single line through the centre. 11. A) Proceed in his own way. B) Stick to the original plan. C) Compromise with his colleague. D) Try to change his colleague’s mind. 12. A) Mary has a keen eye for style. B) Nancy regrets buying the dress. C) Nancy and Mary went shopping together in Rome. D) Nancy and Mary like to follow the fashion. 13. A) Wash the dishes. B) Go to the theatre. C) Pick up George and Martha. D) Take her daughter to hospital. 14. A) She enjoys making up stories about other people. B) She can never keep anything to herself for long. C) She is eager to share news with the woman. D) She is the best informed woman in town. 15. A) A car dealer. B) A mechanic. C) A driving examiner. D) A technical consultant. 16. A) The shopping mall has been deserted recently. B) Shoppers can only find good stores in the mall. C) Lots of people moved out of the downtown area. D) There isn’t much business downtown nowadays. 17. A) He will help the woman with her reading. B) The lounge is not a place for him to study in. C) He feels sleepy whenever he tries to study. 第七章波浪理论 Water Wave Theory）Chapter 7. Water Wave Theory （Chapter 7. 讨论重力场中具有自由面的水波（ （水表面波本章内容：讨论重力场中具有自由面的水波 运动， ，重点讲述线性小振幅简谐波的数学描）运动 或重力波） 或重力波 为研究非线性波打下基础。 。 。为研究非线性波打下基础 述、运动特性和能量概念 运动特性和能量概念。 采用方法----速度势方法，即根据边界条件和初始条件求解拉普拉斯方程，求出波浪运动的速度势，进而解出波面方程式和速度分布与压力分布，并建立波动参数之间的关系式。 波动的物理本质： 恢复力与惯性力的动态平衡。 波浪要素和波浪分类 波浪运动的有关概念 微幅波方程质点轨迹和压力分布波浪微幅波方程、质点轨迹和压力分布 波群与波群速 船波 波能传递与兴波阻力 课堂提问：“无风不起浪”“无风三尺浪” 课堂提问：无风不起浪无风三尺浪 船舶与海洋工程中： 船舶摇摆和拍击，船舶稳性，兴波阻力。 沿岸工程中：波浪对港口、防波堤的作用。 离岸工程中：钻井平台，海工建筑、海底油管等水波起制约作用的物理因素是重力，粘性 力可略而不计，因此可用理想流体的势流理论 来研究波浪运动的规律。 来研究波浪运动的规律 波浪分类7.1 波浪要素和波浪分类干扰与恢复力 海域水深运动状态破碎与否波浪形态深水波振荡波未破碎波日月运动、风 干扰力： 规则波有限水深波 推进波 破碎波暴、地震、风恢复力：随机波（不规则波、方向谱）混浅水波立波破后波柯氏力、重力、 表面张力合浪 周期： 潮汐12~24hours 2i ~10h 界限： h/L 0.05---0.5微小振幅波线性微分方程波浪理论研究方法内波2min 10hrs 海啸10min~>1hour 风暴潮> 5min 风浪、涌浪1~30 s 微小振幅波---线性微分方程有限振幅波---非线性微分方程 风浪涌浪船行波 1.5~3s 表面张力波<0.1s2016年12月英语六级听力原文(含翻译)第二套
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