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國立中山大學 99 學年度普通物理(二)期中考
100.4.20
答題須知：
一、作答時，請由答題紙第六列開始填寫，並標明題號。考試時間為 19：00~21：30，20：00
以後方可交卷。
二、請詳列相關公式及計算過程，記得寫上單位，並以 M.K.S 制表示。
1. [共 10%]Fig. 1 shows an electric dipole. What are the
(a) magnitude and (b) direction (relative to the
positive direction of the x axis) of the dipole’s electric
field at point P, located at distance r >> d? (c) What's
the electric potential for this dipole at point P?
Fig. 1
y
(a) E  E(  )  E(  )

1
1
q
1 q

rˆ 
rˆ
2 
40 r
40 r2
q
 (2) 
40 ( d ) 2  r 2
2

1
qd
40
d
[( ) 2  r 2 ]3 / 2
2
For r >> d, E  
 The magnitude is
d
2
d
( )2  r 2
2
(1 分)
r
yˆ
x
(1 分)
yˆ (in the  ŷ direction)
r Fig.
1
E
1
qd
yˆ
40 r 3
1
qd
.
40 r 3
(1 分)
【直接以 dipole 計算，扣 1 分】
(b) The direction is  ŷ direction.
(c) V  V  V 
(2 分)
1
q
1 q
q
1 1


(  )  0 【說明 1 分，結果 4 分】
40 r 40 r 40 r r
(∵ r+ = r-, the absolute magnitudes are the same)
E
2. [共 15%]A solid insulating sphere of radius a carries a net
negative charge –Q uniformly distributed throughout its volume.
A conducting spherical hollow shell of inner radius b and outer
radius c is concentric with the solid sphere and carries a net
charge 2Q. (a) Using Gauss’s law, find the electric field in the
regions where the radius r satisfies (i) r < a, (ii) a < r < b, (iii) b
Fig. 2
< r < c, (iv) r > c. (b) With V = 0 at infinity, find the potential in
the regions where the radius r satisfies (i) r < a, (ii) a < r < b, (iii) b < r < c, (iv) r > c.
(a) By Gauss’s Law,

 E  ds 
qenc
0
(i)
4 3
r
Q
1 Q

Q
Q r3
3
r (or  k e 3 r ). (2 分)
For r < a, E  4r2 =
,E=

3
3
40 a
a
 0 4 a 3
0 a
3
(ii)
For a < r < b, E  4r2 = 
(iii)
For b < r < c, no electric field exists inside the conducting spherical shell.  E = 0. (2 分)
2Q  Q
1 Q
For r > c, E  4r2 =
E=
. (2 分)
0
4 0 r 2
(iv)
Q
0
E= 
1
Q
. (2 分)
40 r 2
rf

(b) It’s good to integrate step by step from  to r, Vf － Vi =   E  dr
ri
r

(iv) For r > c, V－0 =   E  dr

V=
Q
40
r
1
Q
1 Q
Q 1
 r 2 dr = 40 ( r )  = 40 r (at r = c, V = 40 c ) (2 分)
r
r

1 Q
(iii) For b < r < c, V－V(r = c) =   E  dr  V－
=0
c
40 c
V=
1 Q
40 c
(2 分)
r
r
1 Q
Q 1
Q 1
Q
Q 1
)dr = 
(ii) For a < r < b, V－
=   (
+
( ) =
2
b
40 r
40 r 4 0 b
40 c
40 r b
1
V=
(i) For r < a, V－
Q
1 1 1
Q 1 1 1
(   ) [at r = a, V =
(   ) ] (2 分)
40 b c r
40 b c a
r
1 1 1
1 Q
Q
(   )=  (
r )dr =
3
a
40 b c a
40 a
40 a 3
Q
V=
Q
80 a
3
(r 2  a 2 ) +
Q
1 1 1
(   ) (1 分)
40 b c a
r
Q
 rdr = 8 a
a
0
3
(r 2  a 2 )
3. [共 10%]As shown in Fig. 3, a cylindrical capacitor of length L
formed by two coaxial cylinder of radii a and b. We assume L >>
b, so that we can neglect the fringing of electric field. We assume
r
+q on the surface of radius a, and –q on the surface of radius b.
a
(1) (2%)Draw a cylindrical Gaussian surface of radius r that
b
encloses the +q plate a, and use Gauss’ law to calculate the
Fig. 3
electric field at any point between a and b.
(2) (2%)What is the energy density between the plate of a and b.
(3) (3%)Find the total potential energy stored in the capacitor.
(4) (3%)Show that half the stored potential energy lies within the cylinder whose radius
is r  ab .
(1) 設圓柱內徑 a，表面帶+q，外徑 b 之表面帶-q 之電荷。取一圓柱高斯面，如圖 3(a)、
3(b)所示。 (1%)
－
－
－
r
－
a b
全部電荷-q
－
E
＋＋
a ＋
＋
＋＋ ＋
b
r
高斯面
全部電荷+q
－
－
圖 3(a)
－
圖 3(b)
圓柱電容器及高斯面圖形
圓柱電容器之高斯面及電場電力線截面圖
由高斯定律：  E  d A 
q
0
(2) 在半徑 r 之能量密度為
E(2r)L =
q
0
E 
q
20 Lr
(1%)
1
q2
u   0 E 2  2 2 2 (2%)
2
8  0 L r
(3) 圓柱電容器全部電位能
b
1
q2
q 2 b dr
q2
b
U B   udv    0 E 2 dv  
(
2

rLdr
)


ln
(3%)
2
2
2

a 8  L r
2
40 L a r 40 L a
0
(4) 在半徑 r 之內所含之位能為 Ur，
r
r dr
1
q2
q2
r
U r   udv    0 E 2 dv 

ln

a
2
40 L a r 40 L a
1
U r  U B
2
1 b
r
 ln  ln
2 a
a
r
b
  ( )1 / 2
a
a
 r  ab (3%)
4. [共 10%]Fig. 4 shows a parallel plate capacitor of plate area
A and plate separation d. A dielectric slab of thickness b and
dielectric constant  is placed between the plates. (If the free
charge q is maintained on the plate of the capacitor)
(1) (5%)What is the capacitance with the slab in place
between the plates of the capacitor?
(2) (5%)What is the ratio of the energy before the slab is
inserted to that after the slab is inserted?
Fig. 4
(1) 設未插入介質於二平行板時，二板間之電位差為 V0，電場強度為 E0，電容則為 C0 
插入介質後，介質內之電場強度 E 
VA－VB = E0(d－b)＋Eb = E0(d－b)＋
故電容為 C 
q

V A  VB
E0

C0V0
E0 ( d  b) 
E0

E0

b
，電容器二板 A、B 間之電位差為
b

0 A
(2%)
( 0
E0

A
)( E0 d )
d
 0 A

 ( d  b)  b
[ (d  b)  b]
(3%)
(另解)
1
1
1
1



(2%)
C C1 C S C 2
∵ C1 
CS  
0 A
x1
0 A
b
C2 
0 A
x2
∴
x x
1
1
d b

 1 2 
(1%)
C1 C 2
0 A
0 A
C1
CS
C2
(1%)
1
1
1
1 d b
b
 ( d  b)  b






C C1 C S C2  0 A  0 A
 0 A
(2) 未插入介質前電容器之能量 U 0 
1 2
q (1%)
2C 0
插入介質後電容器之儲藏能量 U  
1 2
q (1%)
2C
1
 0 A
U
C
C
d
 ( d  b)  b
 0  0  k 

(3%)
1
0 A
U
C0
 ( d  b)  b
C
d
 C 
 0 A
 ( d  b)  b
(1%)
x1
x2
d
。
i2
5. [共 10%]In Fig. 5, 1 = 8.00 V, 2 = 12.0 V, R1 = 100 ,
i3
R2 = 200 , and R3 = 300 . One point of the circuit is
grounded (V = 0). Calculate (a) the currents i1 , i2 , i3 ,
i1
and (b) the electric potential V at point A.
Fig. 5
(a) 將電流 i1 , i2 及 i3 假設之方向如圖所示. 各電流經過電阻器所產生的壓降方向畫出如下圖所示.
1  8.00 V,  2  12.0 V, R1  100 , R2  200 , R3  300 .
應用克希荷夫電流定律(KCL)列方程式:
在節點 A:
i1  i2  i3  0
(1)
應用克希荷夫電壓定律(KVL)列方程式:
從 A 點循迴路 I 回到 A 點的壓降與壓升之總和為零:
i2 R2  1  i1R1  0
i2 (200 )  8 V  i1 (100 )  0
(25 )i1  (50 )i2  2 V
(2)
未畫線路圖扣 3 分
未標明 i1 , i2 , i3 扣 1 分
從 A 點循迴路 II 回到 A 點的壓降與壓升之總和為零:
 i1 R1   2  i3 R3  0
 i1 (100)  12V  i3 (300)  0
 (25)i1  (75)i3  3V
(3)
未標示各元件之極性扣 1 分
未畫(或說明迴路方向)扣 1 分
未列(1)(2)(3)式各扣 1 分
未說明電流方向扣 1 分
解(1), (2), (3)三式之聯立方程得
方向與假設者相同
i1  0.0436 A
i2  0.0182 A 負號表示方向與假設者相反
i3  0.0254 A 負號表示方向與假設者相反
(b) 令 A 點的電位為 VA  V , D 點的電位為 VD  0 , 則
沿 ABD 路線, V A 與 VD 關係為:
VA  i2 R2  1  VD
V  (0.0182 A)(200 )  8 V  0
V  4.36 V
或沿 AD 路線, V A 與 VD 關係為:
VA  i1R1  VD
V  (0.0436 A)(100 )  0
V  4.36 V
或沿 ACD 路線, V A 與 VD 關係為:
VA  i3 R3   2  VD
V  (0.0255 A)(300 )  12 V  0
V  4.35 V
未寫此式扣 1 分
三路線任選一條路線即可
答案錯扣 2 分
i1
6. [共 10%]In the circuit of Fig. 6,   1.2kV , C  6.5  F ,
i3
R1 = R2 = R3 = 0.73 M. With C completely uncharged,
switch S is suddenly closed (at t = 0). What are the
currents i1 , i2 , i3 , and the potential difference V2
across resistor R2 at (a) t  0 , and (b) t   .
i2
Fig. 6
  1.2 kV , C  6.5  F , 令 R  R1  R2  R3  0.73 M
(a) 電鍵 S 接通之瞬間, 即 t  0 時, 電容器可視為短路. 此瞬間電路的等效電路如圖 6-1 所示:
令 Re 為 R2 及 R3 二並聯電阻的等效電阻, 則
A
1
1
1
1


, 得 Re  R
2
Re R2 R3
故圖 6-1 又可畫成圖 6-2 之串聯等效電路
由圖 6-2 可解出電流 i1 為
i1 

R1  Re




1
3
R R
R
2
2
1.2 10 3V

 1.096 10 3 A
3
(0.73 10 6 )
2
圖 6-1
無此等效電路圖扣 1 分
i1
i1
R1
在由圖 6-1 中之節點 A 可知
i1  i2  i3
R2 及 R3 二並聯電阻兩端的電壓相等可得
i2 R2  i3 R3
(1)
Re
(2)
圖 6-2
解(1), (2)二式可得
1
1
i2  i3  i1  (1.096 103 A)  5.48  104 A ( i1 , i2 , i3 解錯各扣 1 分)
2
2
R2 電阻兩端的電壓為
V  i2 R2  (0.548 103 A)(0.73 106 )  4 102 V (未寫式子扣 1 分)
(b) 電鍵 S 接通後許久, 即 t  0 時, 電容器可視為斷路. 此瞬間電路的等效電路如圖 6-3 所示:
i3  0 ,
此時
i1  i2 

R1  R2


2R
1.2 10 V
 8.2 104 A
6
2(0.73 10 )
( i1 , i2 , i3 解錯各扣 1 分)
R2 電阻兩端的電壓為

3
圖 6-3
無此等效電路圖扣 1 分
V  i2 R2  (0.82 103 A)(0.73 106 )  6 102 V (未寫式子扣 1 分)
7. [共 10%]In Fig. 7, an electron
accelerated from rest through potential
difference V1 = 1.00 kV enters the gap
between two parallel plates having
separation d = 15.0 mm and potential
Fig. 7
difference V2 = 100 V. The lower plate
is at the lower potential. Neglect fringing and assume that the electron’s velocity vector is
perpendicular to the electric field vector between the plates. In unit-vector notation, what
uniform magnetic field allows the electron to travel in a straight line in the gap?
(Elementary charge e = 1.6×10-19 C; Electron mass me = 9.1×10-31 kg)
1
mv2
v = 2qV1 / m (2 分)
2


B q E = 0 (4 分)
Ue = qV1 =


F= qv

B=

E
=
v
V2
v
d =
V2

d
(E 的代換 2 分)= (3.56 10-4 k ) (T) (數值 1 分，方向及單位 1 分)
2qV1 / m
8. [共 10%]In Fig. 8, the current in the long, straight wire is I1 =
5.00 A, and the wire lies in the plane of the rectangular loop,
which carries 10.0 A. The dimensions shown are c = 0.100 m, a
= 0.150 m, and l = 0.500 m. Find the magnitude and direction
of the net force exerted by the magnetic field due to the straight
wire on the loop. (0 = 4×10-7 Tm/A)
I1
I2
l
c
a
Fig. 8
電流方向相同會相吸、相反會相斥

 0 I1
I1 所造成的磁場 B1 =
(進入紙面) (2 分)
2R


F1a = I2 l


F1c = I2 l

 IIl
B1 = 0 1 2 (←) (2 分)
2 c

 0 I1 I 2l
B1 =
(→) (2 分)
2 (c  a )


F1b 跟 F1d 大小相等，方向相反 ∴互相抵消 (2 分)



 IIl
∴ Fnet = F1a － F1c = 0 1 2
2
1
1
( 
) (←) = 3 10-5 (N) (←) (數值 1 分，方向及單位 1 分)
c ca
9. [共 15%]Figure 9 shows a wood cylinder of mass m =
0.250 kg and length L = 0.100 m, with N = 20.0 turns of
wire wrapped around it longitudinally, so that the plane of
the wire coil contains the long central axis of the cylinder.
The cylinder is released on a plane inclined at an angle  to
the horizontal, with the plane of the coil parallel to the
incline plane. If there is a vertical uniform magnetic field of
magnitude 0.500 T, what is the least current in the coil that
enables the sphere to rest in equilibrium on the incline
plane?
Fig. 9
(The coefficients of static and kinetic friction are s and k between the cylinder and the
incline plane.)
The sphere is in translational equilibrium, thus
F  0
fs－mgsin = 0
(1) (3 分)
【如只有 F  0 無(1)式，只給 1 分】
The sphere is in rotational equilibrium. If the radius of the cylinder is R and torques are
taken about the center of the cylinder (or are taken at the contact point between the
cylinder and the incline plane),
  0
fs R－Bsin = 0 (or mgRsin－Bsin = 0) (2) (3 分)
【如只有   0 無(2)式，只給 1 分】
From (1): fs = mgsin. Substituting this in (2) and canceling out sin, we can obtains
B = mgR
(3)
【如無(1, 2)式，直接得(3)式，只給 1 分】
Now  = N×I×2R×L (3 分)
Thus (3) gives I =
mg
(3 分)=1.225 (A) (錯數值扣 2 分，錯方向及單位扣 1 分)
N  2L  B