國立中山大學 100 學年度普通物理(二)期中考
101.4.18
答題須知：
一、請由答題紙第六列開始作答，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。
【0 = 4×10-7 C2/Nm2，ke =
1.
1 = 9×109 Nm2/C2，me = 9.1×10-31 kg，mp = 1.67×10-27 kg】
4 0
(10%) Figure 1 shows two concentric rings, of radii R and R = 3.00R, that
lie on the same plane. Point P lies on the central z axis, at distance D = 3 R
from the center of the rings. The smaller ring has uniformly distributed
charge –Q. The larger ring has uniformly distributed charge +3Q. (a) What is

electrostatic field E at position P? (b) What is electrostatic force F on a
particle charge q at position P?
Fig. 1
The electric field due to the smaller ring point to –z
r  R 2  3R 2  2 R
D
3R
3


r
2R
2
1
Q
E1z  E1 cos 
cos
40 R 2  ( 3R) 2
cos 

r


1
3Q
3Q

2
40 8R
320 R 2
【方向及大小 3 分】【方向為向下或-z， ĵ 或 k̂ 皆不正確】
The electric field due to larger ring point to +z
r '  9 R 2  3R 2  2 3 R
3R 1

2 3R 2
1
 3Q
E2 z 
cos
2
40 (3R)  ( 3R) 2
1  3Q 1
1  3Q
Q



2
2
40 12 R 2 40 24 R
320 R 2
cos 
【方向及大小 3 分】【方向為向下或-z， ĵ 或 k̂ 皆不正確】
E z  E1z  E2 z 
1 Q
1
3Q
Q


( 3  1) along –z
2
2
320 R
320 R
320 R 2
【2 分，無方向或無負號扣 1 分】【方向為向下或-z， ĵ 或 k̂ 皆不正確】
1 Qq
( 3  1) to –z.
The force on +q at position P is F 
320 R 2
【2 分，無方向或無負號扣 1 分】【方向為向下或-z， ĵ 或 k̂ 皆不正確】
r
2.
(10%) Figure 2 shows a spherical shell with uniform volume charge density
 = 1.00  10-9 C/m3, inner radius a = 1.0 m, and outer radius b = 3.0 m.
What is the magnitude of the electric field at radial distances (a) r = 0.5 m;
(b) r = 2.0 m, and (c) r = 4.0 m? (d) At what radial distance to the center, it
will have the maximum electric field?
Fig. 2
(a)


 E  da 
qenc
【1 分】
0
For r = 0.5 m

qenc = 0
 
E
  da  0  E = 0
【1 分】
【只寫答案僅給 1 分】
(b) For 1  r  3

qenc =
4
(r 3  a 3 )
3
【1 分】
  qenc 4

3
3
3
3
E
  da   0  3 0  (r  a )  E  3 0 r 2 (r  a )
 = 1.00 nC/m3, r = 2.0 m, a = 1.0 m  E 
(c) For r > 3 qenc =



 E  da 
qenc
0
4
(b 3  a 3 )
3

【1 分】
7
 10-9 (N/C)
12 0
【1 分】
【1 分】
4

 (b 3  a 3 )  E 
(b 3  a 3 )
2
3 0
3 0 r
 = 1.00 nC/m3, r = 4.0 m, a = 1.0 m, b = 3.0 m  E 
【1 分】
13
 10-9 (N/C)
24 0
【1 分】
(d) For 1  r  3 r時 E
For r > 3
r時 E
【1 分】
 r = 3 m 時 E 有最大值 【1 分】
or 畫圖
E
【1 分】
1
3
 r = 3 m 時 E = Emax
r
【未解釋原因或解釋不完整僅給 1 分】
【1 分】
3.
(10%) A certain electric dipole is placed in a uniform electric field of
magnitude 40 N/C. Figure 3 gives the magnitude of the torque || on

the dipole versus the angle  between field and the dipole moment p .
The vertical axis scale is set by s = 80  10-28 Nm. (a) What is the

magnitude of p ? (b) How much work must an external agent do to
rotate this electric dipole from the angle with minimal torque ( = 0o)
to the angle with maximal torque ( = max)?
  
(a)   P  E  PE sin 
Fig. 3
【2 分，未列出此式者不給分】
∵ = 90 (or  = 90) for max ∴max = PE
 80  10-28 = P  (40) 【1 分】
【1 分】
 P = 2  10-28 (Cm) 【1 分】
【答案為
2 10 28
，僅給 2 分】
sin 
(b) Wext = U 【1 分】
= U( = 90 or  = 90)  U( = 0) =  PE cos 90  (PE cos 0)
= PE 【1 分】
= 80  10-28 (J)
4.
【2 分】
【1 分】
【只寫答案僅給 1 分】
(10%) A charge q is distributed uniformly throughout a spherical volume of radius R. Let V
(potential) = 0 at infinity. What is the potential at radial distance r from the center of the sphere if r
< R?
By Gauss’ law
  q' 1 r 3
E
  dA   0   0 ( R 3 )q (q為半徑 r 內之電荷) 【1 分】
1 r3
( )q
 0 R3
Er ( 4 r2 ) =
Er =
q
40 R 3
r
【2 分】
R 
R

VR  Vr =   Er  dr    (
r
但 VR =
r
q
40 R 3
) r dr
【3 分】
1
q
40 R
 Vr = VR 

R
r
(
q
40 R
3
) r dr 
q
40 R

q
1 2
q
(R  r 2 ) 
(3R 2  r 2 )
3
40 R 2
80 R
3
【4 分】
5.
(10%) Figure 4 shows, in cross section, a cylindrical capacitor of
length L formed by two coaxial cylinders of radii a and b. We assume
that L  b so that we can neglect the fringing of the electric field
that occurs at the ends of the cylinders. If the inner cylinder has a
charge +q, (a) derive that the capacitance of this cylindrical capacitor
L
is C  20
; (b) how much potential energy is stored in the
ln( b / a)
electric field of this charged cylindrical capacitor? And (c) what is the
energy density at r = a? (d) If this capacitor is filled with dielectric
materials with the dielectric constant , what is the capacitance of
this cylindrical capacitor?
(a) Use Gauss’s Law  
2rLE 
q
0
 E (r ) 
qenc
0
,
Fig. 4
【寫出高斯定律 1 分，向量或內積標示不清者扣 0.5 分】
q
20 rL
【導出電場 2 分】【只寫出電場僅給 1 分】
 
E  ds  Eds cos180  Eds

  
V    E  ds   Eds and ds  dr


【寫出電位積分公式 1 分，上下限錯誤或向量標示不清者扣 0.5 分】

a
a
q
q
1
q
q
b
V   Eds  
dr  
dr  
(ln a  ln b) 
ln( )

2rL 0
20 L b r
20 L
20 L a

b
【導出電位 2 分，直接寫出電位僅給 1 分】
q

V
C
q
q
b
ln( )
20 L a

20 L
b
ln( )
a
【算出電容 1 分】
(b) W 
(c) u 
q 2 CV 2 qV
q2
b



ln( )
2C
2
2
40 L a
0E
2
2
0(

(d) C '  C0 
q
20 aL
2
20 L
b
ln( )
a
)2

1
q 2
(
)
8 0 aL
【1 分】
【1 分，只寫出公式僅給 0.5 分】
【1 分，只寫出公式僅給 0.5 分】
6.
(10%) Switch S in Fig. 5 is closed at time t = 0, to begin charging an
initially uncharged capacitor of capacitance C = 20.0 F through a
resistor of resistance R = 50.0  and a battery with emf . (a) What are
the potential differences across the resistor and the capacitor at time t =
10-3 sec, respectively? (b) At what time is the potential across the
capacitor just half of that across the resistor?
Fig. 5
In the RC circuits (charging),
and thus the current is given by
.
(a) The potential differences across the capacitor and the resistor are
and
Therefore, at t = 10-3 sec,
, respectively.
and
【表示式 4 分, 數值 1 分】
(b) The requirement that the potential across the capacitor is half of that across the resistor leads to
∴
, that is,
.
【表示式 4 分, 數值 1 分】
.
7.
(10%) In Figure 6, a charge particle moves into a region of uniform
magnetic field B, goes through half a circle, and then exits that
region. The particle is either a proton or an electron (you must
decide which). It spends 160 ns in the region.
(a) What is the particle? A proton or an electron?
(b) What is the magnetic field?
(c) If the particle is sent back through the magnitude (along the
Fig. 6
same initial path) but with 2.00 times its previous kinetic energy, how much time does it spend
in the field during this trip?


 

(a) By F  qV  B , 令 B 為 Z 方向 V 為 X 方向
 
當 q 為正值時,利用右手定則知 V  B 為 –Y 方向 與圖相符合
所以知道 q 為正值 (proton)
【2 分，有解釋且表示為質子得 2 分，如直接寫下答案為質子得 1 分】


 

(b) F  qV  B 且因為 V 和 B 垂直，所以粒子在磁場內作圓周運動
2r
mV 2
，V 
代入
T
r
粒子進入磁場只有經過 T/2 = 160 ns
F
【2 分】
…………..[Eq. 1]
【2 分，沒寫單位扣 1 分】
(c)
【4 分】
Z
X
Y
8.
(10%) In Figure 7, a metal wire of mass m = 24.1  10-3 g can
slide with negligible friction on two horizontal parallel rails
separated by distance d = 2.56 cm. The track lies in a vertical
uniform magnetic field of magnitude 735 gauss. At time t = 0,
the speed of the wire is zero and device G is connected to the
Fig. 7
rails, producing a constant current i = 9.13  10-3 A in the wire
and rails (even as the wire moves). At t = 61.1  10-3 s, what are the wire’s (a) speed and (b)
direction of motion (left or right) ?
(a) 金屬條所受磁力，大小為
【1 分】
由右手定則可知受力方向向左
【1 分】
By
【3 分】
(b) 利用右手定則可知方向向左
9.
【5 分】
(10%) In Fig. 8, two circular arcs have radii a = 2R and b = R, subtend angle
60°, carry current i, and share the same center of curvature P. What are the (a)
magnitude and (b) direction (into or out of the page) of the net magnetic field
at P?
Fig. 8
(a)   60 

3
The direction of magnetic field due to the smaller arc is out of page. And the magnitude of
 0i
 i   0i
 0

【方向及大小 4 分】
4R
4R 3 12 R
The direction of magnetic field due to the larger arc is into the page. And the magnitude of
 0i
 0i 
i
magnetic field is Bin 
【方向及大小 4 分】

 0
4 (2 R)
4 (2 R) 3 24 R
magnetic field is Bout 
The net magnetic field is B  Bin  Bout 
The magnitude of magnetic field is B 
(b) The direction is out of the page.
 0i

 0i
,
【1 分】
24 R 12 R
 0i
24 R
【1 分】

 0i
24 R
.
i1
10. (10%) In Figure 9, a long straight wire carries a current i1 and a
rectangular loop carries current i2 = 2 i1. take a = R, b = 8 R, and L
a
= 5 R. In unit-vector notation, what is the net force on the loop due
to i1?
b
L
Fig. 9
i1 為無限長直導線
由安培定律得知：
 0 i1
【1 分】
2r

 
 ii L
因此 F  i2 L  B1  0 1 2
【2 分】
2r
線圈左右所造成的磁力利用右手定則知互相抵消 【2 分】
線圈上下兩段分別受力：

 0i1i2 L
 0i1  2i1 5R
5 0i12
ˆ
ˆ
F上 
( y) 
( y) 
( yˆ ) 【1 分】
2a
2R


 ii L
 i  2i 5R
5 i 2
F下  0 1 2 ( yˆ )  0 1 1 ( yˆ )  0 1 ( yˆ ) 【1 分】
2 (a  b)
2  9 R
9
2
2



5 i
40 0i1
1
Fnet  F上  F下  0 1 (1  )( yˆ ) 
( yˆ ) 【3 分】

9
9
B1 