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國立中山大學 102 學年度普通物理(二)期中考
103.04.23
答題須知：
一、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、請詳述相關物理觀念及計算過程，記得寫上單位，建議以M.K.S制表示。
e
1. (12%) An electron is constrained to the central axis of the ring of
q
charge of radius R in Fig. 1, with z << R. Find (a) the electrostatic
force on the electron and (b) the period of the electron oscillate
through ring center with simple harmonic motion. (If q is the ring’s
charge, m is the electron mass, and e is the electron charge.)
Fig. 1
(a) 在 ring 上取一基素 ds，含電量 dq，dq =  ds
dE
P
其中 = charges per unit length
dq
ds
ds
 dE 

【2%】 
2
2
40 r
40 r
40 ( R 2  z 2 )
1
-
e
-

q
r
z
在 ring 上之電荷於 P 點所產生之電場：
ds
垂直於 z 軸之總和為零。
【1%】
平行於 z 軸為 E z   dE cos  

ds
z
 ,
40 ( R  z ) r
2
z
40 ( R  z )
2
2 3/ 2

2R
0
2
ds 
r = (z2 + R2)1/2
2Rz
qz

【2%】
2
2 3/ 2
40 ( R  z )
40 ( R 2  z 2 ) 3 / 2
其中，q = 2Rz.
故在 P 點之電子受力 F 
 eqz
(在－z 方向) 【1%】
40 ( R 2  z 2 ) 3 / 2
(b) 當 R >> z
F
 eqz
d 2z

m
40 R 3
dt 2
2 
【3%】
eq
2
 ( )2
3
T
40 mR
40 mR3
 T  2
eq
【2%】
【1%】
2. (12%) (a) The nucleus in an iron atom has a radius of about 4.0  10-15 m and contains 26
protons. (i) What is the magnitude of the repulsive electrostatic force between two of the protons
that are separated by 4.0  10-15 m? (ii) What is the magnitude of the gravitational force between
those same two protons? [ G = 6.67×10-11 Nm2/kg2; mp = 1.67×10-27 kg; 0 = 8.854×10-12 F/m]
(b) Charge is distributed uniformly throughout the volume of an infinitely long solid cylinder of
r
radius R. Show that, at a distance r < R from the cylinder axis, E 
, where  is the volume
2 0
charge density.
(a) (i) Charge of proton is ＋e
F 
19 2
1 e2
)
9 (1.6 10
【2%】
 14 Nts

9

10
2
15 2
40 r
(4.0 10 )
【1%】
(ii) Mass of proton mp = 1.67×10-27 kg
F G
m 2p
r2
【2%】  6.67 10 11
(1.67 10 27 ) 2
 1.2  10-35 Nts 【1%】
(4.0 10 15 ) 2
(b) 設高斯面為圓柱形，半徑為 r，長度 l，所含之電荷為 qenc。
由 Gauss’ law
  qenc
E
  dA   0 【2%】, qenc = ( r2 l) 
E (2rl ) 
高斯面
r
(r 2l ) 
0
r
E 
【4%】
2 0
R
Cross section of
charged cylinder
(end view)
3. (12%) Fig. 2 is showing a charged particle (either an electron or a proton) moving
rightward between two parallel plates separated by distance d  2(mm) . The
potential are V1 = 70.0(V) and V2 = 50.0(V). The particle is accelerating from an
initial speed of 90.0(m/s) at the left plate. (a) Is the particle an electron or a proton?
(b) Find the electric field E at a point between two parallel plates. (c) What is its
speed just as it reaches plate 2. The mass of electron is 9.10 1031 (kg) , and the
mass of proton is 1.67 1027 (kg) .
V1  70(V ), V2  50(V )
(a)
∴
V1  V2
正電荷從高電位移向低電荷
負電荷從低電位移向高電荷
今質點從 V1 移向 V2
故質點為電子
(b) 由
V2
2
V1  V2    dV   E  ds
V1
1
2
d
  E  dx   Ex dx  Ex d
1
Ex 
0
V1  V2 (70  50)(V)

 104 (V/m)
d
2 103 (m)
ˆ x  iˆ104 (V/m)
E  iE
(c) 電子從左極板至右極板所降低之電能為
W  q(V1  V2 )  e(70  50)(V )  20(eV )
降低之電能化成電子增加之動能
W  Ek 
v
1
me (v 2  vo2 )
2
2W
2(20  1.6  1019 J )
 vo2 
 (90m/s) 2
31
me
9.10  10 (kg)
 2.61106 (m/s)
Fig. 2
4. (12%) Charge Q is uniformly distributed inside a spherical region of
radius R in Fig. 3. Find both (a) electric field, and (b) electric potential at
point P a distance r ( r  R and r  R ) from the center of the sphere. (c)
What is the electric potential at the center of the sphere?
Fig. 3
(a) 根據應用高斯定律的經驗得知以下之結論:
(1) 球狀對稱分佈之電荷, 在電荷分佈範圍以外之區域產生之電場, 如同全部電荷全部
集中於球心形成點電荷所產生之電場. 如此, 則於下左圖中, P 點之電場為
Q
E
rˆ, (r  R)
4 o r 2
(2) 下圖中之 P 點在球狀對稱分佈電荷範圍以內. 位於半徑為 r 之球狀範圍(如虛線所
示)以外之電荷, 在 P 點產生之總電場為零，而位於半徑為 r 之球狀範圍以內之總電
Qr 3
像是集中於球心之點電荷, 在 P 點產生電場, 其值為
R3
Qr
E
rˆ, (r  R)
4 o R 3
荷Q' 
(b) 在空間任意取兩點 a 及 b , 此二點與球心之距離分別為 ra 及 rb ,二點之電位分別為 Va 及
rb
Vb . 由電位差與電場之關係 Va  Vb   E  ds
ra
當電荷分佈在有限區域時, 無窮遠處之電場與電位皆趨於零. 故可設定
rb   時, Vb  0 以及 ra  r 時, Va  V .


r
r
V   E  ds =  Edr 又因電位有連續性
故上式可寫成
r  R 之區域
V 

r
Q
4 o r
2
dr 
Q
4 o


r
1
Q
dr 
2
r
4 o r
r  R 之區域
V 
R
r

Qr
Q
Q( R 2  r 2 )
Q
dr
+
dr


3
2
3

R 4 r
4 o R
8 o R
4 o R
o
(c) 在球心, 即 r  0 處, 電位為
V=
3Q
8 o R
5. (12%) A solid cylindrical conductor of radius a and charge Q is coaxial with a
cylindrical shell of negligible thickness, radius b > a, and charge –Q (Fig. 4).
Find the capacitance of this cylindrical capacitor if its length is ℓ >> a and b.
Please answer the capacitance in terms of ℓ, a and b.
Fig. 4
Using Gauss law to find electric field:


 E  dA 
qenc
0

1 Q
E
rˆ 【5%】
2rl  0
a
 
1 Q
Q
b
Electrical potential difference from b to a: V   E  dr  
【5%】
dr 
ln
2rl  0
20l a
b
b
a
Capacitance: C 
Q 20l

【2%】
b
V
ln
a
6. (10%) Two capacitors C1 and C2 (where C1 > C2)
are charged to the same initial potential difference
Vi. The charged capacitors are removed from the
battery, and their plates are connected with
opposite polarity as in Fig. 5(a). The switches S1
Fig. 5
and S2 are then closed as in Fig. 5(b). (a) Find the
final potential difference Vf between a and b after the switches are closed. (b) Determine the
ratio of the final energy (switches closed) to the initial energy (switches open).
(a) total charge Qi  Q1i  Q2i  (C1  C2 )Vi before S1 and S2 are closed. 【2%】
Charge redistribution after S1 and S2 are closed, Q f  Q1 f  Q2 f  (C1  C2 )V f 【2%】
Charge conservation: Qi  Q f  (C1  C2 )V  (C1  C2 )V f
V f 
(b) U i 
C1  C2
Vi .【2%】
C1  C2
1
1
1
C1Vi 2  C2 Vi 2  (C1  C2 )Vi 2
2
2
2
1
1
1
1
C  C2
1 (C1  C2 ) 2
U f  C1V f2  C2 V f2  (C1  C2 )V f2  (C1  C2 )[ 1
Vi ]2 
Vi 2
2
2
2
2
C1  C2
2 C1  C2
U
C  C2 2
 f ( 1
) 【4%】
Ui
C1  C2
7. (16%) In the circuit of Fig. 6, ε = 1.2 kV, C = 8.5 F, R1 = R2 =
R3 = 0.55 MΩ. With C completely uncharged, switch S is
suddenly closed (at t = 0). At t = 0, what are (a) current I1 in

resistor 1, (b) current I2 in resistor 2, and (c) current I3 in
resistor 3? At t = ∞ (that is, after many time constants), what
are (d) I1,(e) I2, and (f) I3? What is the potential difference V3
across resistor 3 at (g) t = 0 and (h) t = ∞?
(a) t  0
－I1R－I2R = 0
－I1R－I3R = 0

 = 3IR  I1  2 I 
(b) I 2  I 

3R
I2 = I3 = I  I1 = 2I
2
2 1200

 1.45 10 3 (A) 【2%】
6
3R 3  0.55 10
 7.3 10 4 (A) 【2%】
(c) I 3  I 2  7.3 10 4 (A) 【2%】
(d) t  
I3 = 0  I1 = I2
－2I1R = 0  I1 

2R
 1.09  10 3
(A)
【2%】
(e) I 2  I1  1.09 10 3 (A) 【2%】
(f) I3 = 0 【2%】
(g) I1 = I2＋I3
－I1R－I2R = 0  －2I2R－I3R = 0 
I2 R  I3R 
I3 
dQ
dt
Q
C



2

I3R
Q
 I3R 
2
C
3 dQ Q 
R
 
2 dt C 2

t
C
(1  e 3 RC )
 Q(t ) 
2
2
dQ   3 RC t

e
 I3 
dt 3R
2
t0
V3 

3R
R

3
 400 (V)
(h) t  
I3 = 0 => V3 = 0
【2%】
【2%】

I2 

  I3R
2R
3
Q
 I3R   0
2 2
C
Fig. 6
8. (14%) A circular loop of wire having a radius of 8.0 cm carries a current of 0.20 A. A vector of
unit length and parallel to the dipole moment
of the loop is given by 0.60iˆ  0.80 ˆj . (This
unit vector gives the orientation of the magnetic dipole moment vector.) If the loop is located in

a uniform magnetic field given by B  (0.25T )iˆ  (0.30T )kˆ , find (a) the torque on the loop (in
unit-vector notation) and (b) the orientation energy of the loop.



The magnetic dipole moment is    (0.6i  0.8 j ) , where
 = NiA = Nir2 = 1(0.20 A) (0.080 m)2 = 4.02  10–3 A·m2.
  
(a) The torque is     B ……. 【2%】





……. 【4%】



  4.02 10 3 (0.6i  0.8 j )  (0.25i  0.3k )  4.02 10 3 (0.18 j  0.2k  0.24i )




  (9.7 10 4 i  7.2 10 4 j  8.0 10 4 k ) N  m …………. 【3%】
 
(b) The orientation energy of the dipole is given by U     B ……. 【2%】




U  4.02 10 3 (0.6i  0.8 j )  (0.25i  0.3k )  6.0 10 4 J
……. 【3%】