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Fundamentals of Physics, 8th Ed
Principle of Physics, 9th Ed
Halliday & Resnic
東海大學物理系
8th Ed【CH24】Finding the Electric Potential
9th Ed【CH24】Electric Potential
東海大學物理系
⎛ σ ⎞
ΔV = E Δx = ⎜
⎟ Δx .
⎝ 2ε 0 ⎠
8th Ed：Homework of Chapter 24：
3, 5, 7, 9, 13, 17, 19, 23, 25, 27, 29, 35, 37, 39, 41, 43, 45, 49, 53, 55, 63, 65, 67
th
Fundamentals of Physics, 8th Ed
Principle of Physics, 9th Ed
Halliday & Resnic
Thus, Δx =
th
8 Ed【Problem 24-3】：9 Ed【Problem 24-1】
2ε 0 Δ V
σ
=
c
hb g = 8.8 × 10
2 8.85 × 10−12 C 2 N ⋅ m2 50 V
010
. × 10−6 C m2
−3
m.
A particular 12 V car battery can send a total charge of 84A ⋅ h (ampere-hours) through a circuit,
from one terminal to the other. (a) How many coulombs of charge does this represent? (Hint: See
Eq. 21-3.) (b) If this entire charge undergoes a change in electric potential of 12 V, how much
energy is involved?
8th Ed【Problem 24-7】：9th Ed【Problem 24-9】
An infinite nonconducting sheet has a surface charge density σ = +5.8 pC / m 2 . (a) How much
work is done by the electric field due to the sheet if a particle of charge q = +1.6 × 10−19 C is
moved from the sheet to a point P at distance d = 3.56cm from the sheet? (b) If the electric
一顆特別的 12V 車子電池可以發送 84A ⋅ h 的電荷到電路，從一端到另一端，（a）問充了多
少庫侖的電荷？（b）如果充電過程中改變了 12V 的電位，問儲存多少能量？
一無限不導電板子表面電荷密度為 σ = +5.8 pC / m 2 ，
（a）一電荷 q = +1.6 × 10−19 C ，從板子移
動到 P 點，P 點距離板子 d = 3.56cm ，問做了多少功？（b）如果板子電位為 0，問 P 點電
位為何？
<解>：(a) An Ampere is a Coulomb per second, so
FG
H
84 A ⋅ h = 84
C⋅h
s
IJ FG 3600 s IJ = 3.0 × 10
K H hK
5
C.
(b) The change in potential energy is ΔU = qΔV = (3.0 × 105 C)(12 V) = 3.6 × 106 J.
8th Ed【Problem 24-5】：9th Ed【Problem 24-5】
An infinite nonconducting sheet has a surface charge density σ = 0.1μ C / m 2 on one side. How far
apart are equipotential surfaces whose potentials differ by 50 V?
一無限不導電的板子，其一邊的電荷密度為 σ = 0.1μ C / m 2 ，距離板子多遠時，電位差為
50V？
<解>：The electric field produced by an infinite sheet of charge has magnitude
E=
σ
, where σ
2ε 0
is the surface charge density. The field is normal to the sheet and is uniform. Place the origin
of a coordinate system at the sheet and take the x axis to be parallel to the field and positive in
the direction of the field. Then the electric potential is
V = Vs −
z
x
0
potential V is defined to be zero on the sheet, what is V at P?
E dx = Vs − Ex ,
<解>：(a) The work done by the electric field is
G G qσ d
f
q σd (1.60 ×10−19 C)(5.80 ×10−12 C/m 2 )(0.0356 m)
W = ∫ q0 E ⋅ ds = 0 ∫ dz = 0 =
0
i
2ε 0
2ε 0
2(8.85 ×10−12 C 2 /N ⋅ m 2 )
= 1.87 ×10−21 J.
(b) Since V – V0 = –W/q0 = –σz/2ε0, with V0 set to be zero on the sheet, the electric potential
at P is
V =−
σz
(5.80 ×10−12 C/m 2 )(0.0356 m)
=−
= −1.17 ×10−2 V.
2ε 0
2(8.85 ×10−12 C2 /N ⋅ m 2 )
8th Ed【Problem 24-9】：9th Ed【Problem 24-7】
The electric field in a region of space has the components E y = Ez = 0 and Ex = (4.00 N / C ) x .
Point A is on the y axis at y = 3m , and point B is on the x axis at x = 4m .What is the potential
difference VB − VA ?
電場在某一空間的分量為 E y = Ez = 0 和 Ex = (4.00 N / C ) x 。A 點在 y 軸上 y = 3m 的地方。B
where Vs is the potential at the sheet. The equipotential surfaces are surfaces of constant x;
that is, they are planes that are parallel to the plane of charge. If two surfaces are separated by
點在 x 軸上 x = 4m 的地方。問位能差 VB − VA ？
Δx then their potentials differ in magnitude by
<解>：We connect A to the origin with a line along the y axis, along which there is no change of
ௐ 2 ࢱĂВ 29 ࢱ
ௐ 3 ࢱĂВ 29 ࢱ!
Fundamentals of Physics, 8th Ed
Principle of Physics, 9th Ed
Halliday & Resnic
東海大學物理系
potential (Eq. 24-18:
z
G G
E ⋅ ds = 0 ). Then, we connect the origin to B with a line along the x
z
x=4
0
G G
E ⋅ ds = −4.00
z
4
0
x dx = −4.00
FG 4 IJ
H 2K
東海大學物理系
(a) For 0 < x < d we have d1 = x and d2 = d – x. Let
V ( x) = k
axis, along which the change in potential is
ΔV = −
Fundamentals of Physics, 8th Ed
Principle of Physics, 9th Ed
Halliday & Resnic
FG q
Hd
1
+
1
IJ
K
q2
q
=
d2
4 pε 0
FG 1 + −3 IJ = 0
H x d − xK
2
⇒
which yields VB – VA = –32.0 V.
1
−3
+
=0
x d−x
⇒ x=
8th Ed【Problem 24-13】：9th Ed【Problem 24-13】
What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius
0.15 m whose potential is 200 V (with V = 0 at infinity)?
d 24cm
=
= 6cm
4
4
(b) Similarly, for x < 0 the separation between q1 and a point on the x axis whose coordinate
is x is given by d1 = –x; while the corresponding separation for q2 is d2 = d – x. We set
V ( x) = k
半徑 0.15m 導體球，其位能 200V（無窮遠處 V = 0 ）
，問（a）電荷（b）電荷密度。
⇒
<解>：(a) The charge on the sphere is
(200 V)(0.15 m)
q = 4πε 0 VR =
= 3.3 ×10−9 C.
8.99 ×109 N ⋅ m 2 C 2
FG q
Hd
1
1
+
IJ
K
q2
q
=
d2
4 pε 0
FG 1 + −3 IJ = 0
H −x d − xK
1
−3
+
=0
−x d − x
⇒ x=−
d
24cm
=−
= −12cm
2
2
(b) The (uniform) surface charge density (charge divided by the area of the sphere) is
8th Ed【Problem 24-19】：9th Ed【Problem 24-15】
A spherical drop of water carrying a charge of 30 pC has a potential of 500 V at its surface (with
q
3.3 × 10 −9 C
. × 10 −8 C / m 2 .
σ=
=
= 12
2
2
4 pR
. m
4 p 015
b
g
V = 0 at infinity). (a) What is the radius of the drop? (b) If two such drops of the same charge and
radius combine to form a single spherical drop, what is the potential at the surface of the new drop?
8th Ed【Problem 24-17】：9th Ed【Problem 24-19】
In Fig. 24-33, particles of the charges q1 = +5e and q2 = −15e are fixed in place with a separation
of d = 24cm . With V = 0 at infinity, what are the finite (a) positive and (b) negative values of x at
一顆球型水滴帶有 30 pC 的電荷，表面位能 500V（無窮遠處 V = 0 ）（a）水滴半徑？（b）
which the net electric potential on the x axis is zero?
兩顆相同水滴，帶有一樣電荷，有一樣半徑，合併組合成一顆球型水滴，問新水滴的表面
電位？
圖 24-33，粒子電荷 q1 = +5e 和 q2 = −15e ，被固定在相距 d = 24cm 的地方，無窮遠處 V = 0 ，
<解>：(a) The electric potential V at the surface of the drop, the charge q on the drop, and the radius
問在（a）正（b）負 x 軸上，哪一各地方是電位為 0？
R of the drop are related by V = q/4πε0R. Thus
R=
（圖24-33）
<解>：First, we observe that V (x) cannot be equal to zero for x > d. In fact V (x) is always negative
for x > d. Now we consider the two remaining regions on the x axis: x < 0 and 0 < x < d.
ௐ 4 ࢱĂВ 29 ࢱ
c
hc
h
8.99 × 10 9 N ⋅ m 2 / C 2 30 × 10 −12 C
q
=
= 5.4 × 10 −4 m.
4 pε 0V
500 V
(b) After the drops combine the total volume is twice the volume of an original drop, so the
radius R' of the combined drop is given by (R')3 = 2R3 and R' = 21/3R. The charge is twice
the charge of original drop: q' = 2q. Thus,
1 q′
1
2q
V′ =
=
= 2 2 / 3V = 2 2 / 3 (500 V) ≈ 790 V.
4 pε 0 R ′ 4 pε 0 2 1/ 3 R
ௐ 5 ࢱĂВ 29 ࢱ!
Fundamentals of Physics, 8th Ed
Principle of Physics, 9th Ed
Halliday & Resnic
東海大學物理系
Fundamentals of Physics, 8th Ed
Principle of Physics, 9th Ed
Halliday & Resnic
東海大學物理系
(a) Figure 24-39a shows a nonconducting rod of length L = 6cm and uniform linear charge density
λ = +3.68 pC / m .Take V = 0 at infinity. What is V at point P at distance d = 8cm along the rod’s
8th Ed【Problem 24-23】：9th Ed【Problem 24-25】
A plastic rod has been bent into a circle of radius R = 8.2cm . It has a charge Q1 = +4.2 pC
uniformly distributed along one-quarter of its circumference and a charge Q2 = −6Q1 uniformly
perpendicular bisector? (b) Figure 24-39b shows an identical rod except that one half is now
negatively charged. Both halves have a linear charge density of magnitude 3.68 pC / m . With
distributed along the rest of the circumference (Fig. 24-37).With V = 0 at infinity, what is the
V = 0 at infinity, what is V at P?
electric potential (a) at the center C of the circle and (b) at point P, which is on the central axis of
the circle at distance D = 6.71cm from the center?
（a）長 L = 6cm 的非導體棒子，如圖 24-39a，均勻線電荷密度為 λ = +3.68 pC / m ，無窮遠處
一根塑膠棒被彎曲成一個半徑 R = 8.2cm 的圓，電荷 Q1 = +4.2 pC 均勻分佈在四分之一圓，另
外電荷 Q2 = −6Q1 均勻分佈在其他四分之三圓上。無窮遠處 V = 0 （a）圓的中心點 C 的電位
V = 0 ，P 點在沿著棒子的垂直平分線上，距離 d = 8cm ，問電位 V？（b）圖 24-39b 為一相
同棒子，只是現在的一半帶負電，兩邊的線電荷密度為 3.68 pC / m ，無窮遠處 V = 0 ，問 P
點的電位 V？
為何？（b）離中心 D = 6.71cm 在中心軸上的 P 點，電位為何？
（圖 24-39）
（圖 24-37）
1
<解>：(a) All the charge is the same distance R from C, so the electric potential at C is
V=
1 ⎛ Q1 6Q1 ⎞
⎜ −
⎟
4πε 0 ⎝ R
R ⎠
=−
dq
<解>： dV =
4πε 0 r
5Q1
5(8.99 ×109 N ⋅ m 2 C2 )(4.20 ×10−12 C)
=−
= −2.30 V,
4πε 0 R
8.20 ×10−2 m
L
2
L
−
2
λ dx
( x 2 + d 2 )1/ 2
=
λ
4πε 0
∫
=
1/ 2 2
λ
ln ⎡⎢ x + ( x 2 + d 2 ) ⎤⎥ L
⎣
⎦−
4πε 0
2
=
λ ⎧⎪ ⎡ L ⎛ L2
2⎞
⎨ln ⎢ + ⎜ + d ⎟
4πε 0 ⎪ ⎢ 2 ⎝ 4
⎠
⎣
L
2
L
2
−
∫
dx
( x 2 + d 2 )1/ 2
（由積分表查詢： ∫
R 2 + D 2 , so the electric
potential at P is
1/ 2
Q1
6Q1 ⎤
5Q1
1 ⎡
−
⎢
⎥ =−
2
4πε 0 ⎣ R 2 + D 2
R2 + D2 ⎦
4πε 0 R 2 + D
5(8.99 × 10 N ⋅ m C )(4.20 ×10
9
2
2
−12
C)
(8.20 × 10−2 m) 2 + (6.71× 10−2 m) 2
dx
x2 + a2
= ln( x + x 2 + a 2 ) ）
L
(b) All the charge is the same distance from P. That distance is
=−
dq
1
=
4πε 0
r
4πε 0
where the zero was taken to be at infinity.
V=
1
∫
V = ∫ dV =
⎩
1/ 2
= −1.78 V.
=
λ
4πε 0
ln
⎞
L ⎛ L2
+⎜ + d2 ⎟
2 ⎝ 4
⎠
1/ 2
−
⎞
L ⎛ L2
+⎜ + d2 ⎟
2 ⎝ 4
⎠
1/ 2
⎤
⎡ L ⎛ L2
⎞ ⎤ ⎫⎪
⎥ − ln ⎢ − + ⎜ + d 2 ⎟ ⎥ ⎬
⎥⎦
⎢⎣ 2 ⎝ 4
⎠ ⎥⎦ ⎭⎪
1/ 2
⎡ L ⎛ L2
⎞
⎢ +⎜ + d2 ⎟
2 ⎝ 4
λ
⎠
ln ⎢
=
1/ 2
4πε 0 ⎢ L ⎛ L2
⎢ − + ⎜ + d 2 ⎞⎟
⎠
⎣⎢ 2 ⎝ 4
⎤
⎥
⎥
1/ 2 ⎥
⎞
L ⎛ L2
+⎜ + d2 ⎟ ⎥
2 ⎝ 4
⎠ ⎦⎥
1/ 2
⎞
L ⎛ L2
+⎜ + d2 ⎟
2 ⎝ 4
⎠
8th Ed【Problem 24-25】：9th Ed【Problem 24-23】
ௐ 6 ࢱĂВ 29 ࢱ
ௐ 7 ࢱĂВ 29 ࢱ!
Fundamentals of Physics, 8th Ed
Principle of Physics, 9th Ed
Halliday & Resnic
東海大學物理系
2
1/ 2
1/ 2
⎡ L ⎛ L2
⎡ L ⎛ L2
⎞ ⎤
⎞ ⎤
⎢ +⎜ + d2 ⎟ ⎥
⎢ +⎜ + d2 ⎟ ⎥
⎢2 ⎝ 4
⎢2 ⎝ 4
⎠ ⎥⎦
⎠ ⎥⎦
λ
λ
=
=
ln ⎣ 2
ln ⎣
2
2
πε
4πε 0
4
d
⎛L
0
2⎞ ⎛ L⎞
⎜ + d ⎟−⎜ ⎟
⎝ 4
⎠ ⎝2⎠
Fundamentals of Physics, 8th Ed
Principle of Physics, 9th Ed
Halliday & Resnic
2
=
東海大學物理系
⎛ L
⎞⎛ L
⎞ ⎫⎪
λ ⎧⎪
L2
L2
+ d 2 ⎟⎜ +
+ d 2 ⎟⎬
⎨2 ln d − ln ⎜⎜ − +
⎟
⎜
⎟
4πε 0 ⎪
4
4
⎝ 2
⎠⎝ 2
⎠⎪
⎩
⎭
λ
=
{2 ln d − ln d 2} = 4πελ {2 ln d − 2 ln d } = 0 …#
4πε 0
0
2
1/ 2
1/ 2
⎡ L ⎛ L2
⎞ ⎤
⎞
L ⎛ L2
+⎜ + d2 ⎟
⎢ +⎜ + d2 ⎟ ⎥
2 ⎝ 4
λ
⎠ ⎥ = 2λ ln 2 ⎝ 4
⎠ …#
=
ln ⎢
⎥
4πε 0 ⎢
d
4πε 0
d
⎢
⎥
⎢⎣
⎥⎦
8th Ed【Problem 24-27】：9th Ed【Problem 24-29】
In Fig. 24-40, what is the net electric potential at the origin due to the circular arc of charge
Q1 = +7.21 pC and the two particles of charges Q2 = 4Q1 and Q3 = −2Q1 ? The arc’s center of
curvature is at the origin and its radius is R = 2m ; the angle indicated is θ = 200 .
(a) From Eq. 24-35, we find the potential to be
V =2
⎡ L / 2 + ( L2 / 4) + d 2
λ
ln ⎢
4πε 0
d
⎢⎣
圖 24-40，圓弧電荷 Q1 = +7.21 pC ，兩個粒子分別帶電荷為 Q2 = 4Q1 和 Q3 = −2Q1 ，問原點淨
⎤
⎥
⎥⎦
電位？圓弧曲率中心在原點，半徑為 R = 2m ，圖中 θ = 200 。
⎡ (0.06 m / 2) + (0.06 m) 2 / 4 + (0.08 m) 2
= 2(8.99 ×109 N ⋅ m 2 C2 )(3.68 ×10−12 C/m) ln ⎢
0.08 m
⎢⎣
⎤
⎥
⎥⎦
= 2.43×10−2 V.
(b) The potential at P is V = 0 due to superposition.
V = ∫ dV =
L
1 ⎡ 0
(λ )dx
(−λ )dx ⎤
+∫ 2 2
⎢ ∫− L 2
⎥
2 1/ 2
0
4πε 0 ⎣ 2 ( x + d )
( x + d 2 )1/ 2 ⎦
=
L
⎤
λ ⎡ 0
dx
dx
2
−
⎢ ∫− L 2
⎥
4πε 0 ⎣ 2 ( x + d 2 )1/ 2 ∫0 ( x 2 + d 2 )1/ 2 ⎦
（由積分表查詢： ∫
（圖 24-40）
<解>：Since the charge distribution on the arc is equidistant from the point where V is evaluated, its
dx
x2 + a2
= ln( x + x + a ) ）
2
2
=
0
⎫
λ ⎧⎪ ⎡
2
2 1/ 2
2
2 1/ 2 2 ⎪
⎨ln x + ( x + d ) ⎤⎦⎥ L − ln ⎡⎣⎢ x + ( x + d ) ⎤⎦⎥ ⎬
4πε 0 ⎪ ⎣⎢
−
0
2
⎩
⎭⎪
=
⎤ ⎫⎪
⎞
⎛ L
⎞⎤ ⎡ ⎛ L
λ ⎧⎪ ⎡
L2
L2
+ d 2 ⎟ − ln d ⎥ ⎬
+ d 2 ⎟ ⎥ − ⎢ ln ⎜ +
⎨ ⎢ln d − ln ⎜⎜ − +
⎟
⎟⎥ ⎢ ⎜ 2
4πε 0 ⎪ ⎢
4
4
⎥⎪
⎠
⎝
⎠
⎝ 2
L
⎩⎣
=
⎦ ⎣
⎦⎭
⎫⎪
⎛ L
⎞
⎛L
⎞
λ ⎧⎪
L2
L2
+ d 2 ⎟ − ln ⎜ +
+ d 2 ⎟ + ln d ⎬
⎨ln d − ln ⎜⎜ − +
⎟
⎜2
⎟
4πε 0 ⎪
2
4
4
⎠
⎝
⎠
⎝
⎩
⎭⎪
contribution is identical to that of a point charge at that distance. We assume V → 0 as r →
∞ and apply Eq. 24-27:
1 +Q1
1 +4Q1
1 −2Q1
V=
+
+
4πε 0 R 4πε ο 2 R 4πε 0 R
=
1 Q1
4πε 0 R
=
(8.99 ×109 N ⋅ m 2 C2 )(7.21×10−12 C)
2.00 m
= 3.24 ×10−2 V.
8th Ed【Problem 24-29】：9th Ed【Problem 24-31】
A plastic disk of radius R = 64cm is charged on one side with a uniform surface charge density
σ = 7.73 fC / m 2 , and then three quadrants of the disk are removed. The remaining quadrant is
ௐ 8 ࢱĂВ 29 ࢱ
ௐ 9 ࢱĂВ 29 ࢱ!
Fundamentals of Physics, 8th Ed
Principle of Physics, 9th Ed
Halliday & Resnic
東海大學物理系
shown in Fig. 24-42.With V = 0 at infinity, what is the potential due to the remaining quadrant at
point P, which is on the central axis of the original disk at distance D = 25.9cm from the original
Fundamentals of Physics, 8th Ed
Principle of Physics, 9th Ed
Halliday & Resnic
東海大學物理系
= 4.71×10−5 V.
8th Ed【Problem 24-35】：9th Ed【Problem 24-35】
center?
一塑膠盤半徑 R = 64cm ，四分之一的部分帶有均勻電荷密度 σ = 7.73 fC / m 2 的電荷，另外四
分之三被移除，如圖 24-42。無窮遠處 V = 0 。離中心 D = 25.9cm 在圓盤中心軸上的 P 點，
The electric potential at points in an xy plane is given by V = (2V / m 2 ) x 2 − (3V / m 2 ) y 2 . In
unit-vector notation, what is the electric field at the point (3m, 2m) ?
在 xy 平面上，電位為 V = (2V / m 2 ) x 2 − (3V / m 2 ) y 2 ，問 (3m, 2m) 這一點的電場？
問剩下四分之一圓盤對 P 點的電位？
<解>：We use Eq. 24-41:
∂V
∂
(2.0V / m2 ) x 2 − 3.0V / m2 ) y 2 = −2(2.0V / m2 ) x;
=−
∂x
∂x
∂
∂V
=−
(2.0V / m2 ) x 2 − 3.0V / m2 ) y 2 = 2(3.0V / m2 ) y .
E y ( x, y) = −
∂y
∂y
E x ( x, y) = −
（圖 24-42）
<解>：The disk is uniformly charged. This means that when the full disk is present each quadrant
c
c
h
h
We evaluate at x = 3.0 m and y = 2.0 m to obtain
G
E = (−12 V/m)i + (12 V/m)j .
contributes equally to the electric potential at P, so the potential at P due to a single quadrant
is one-fourth the potential due to the entire disk. First find an expression for the potential at
P due to the entire disk. We consider a ring of charge with radius r and (infinitesimal) width
8th Ed【Problem 24-37】：9th Ed【Problem 24-39】
dr. Its area is 2πr dr and it contains charge dq = 2πσr dr. All the charge in it is a
unit-vector notation, what is the electric force on the electron?
distance r 2 + D 2 from P, so the potential it produces at P is
An electron is placed in an xy plane where the electric potential depends on x and y as shown in Fig.
24-45 (the potential does not depend on z).The scale of the vertical axis is set by Vs = 500V . In
Xy 平面上，一個電子的電動勢決於 x 和 y，如圖 24-45，（電位和 z 無關）
。圖中垂直軸
Vs = 500V ，問電子電力？
1 2pσ rdr
σ rdr
=
.
dV =
4pε 0 r 2 + D 2 2ε 0 r 2 + D 2
The total potential at P is
σ R rdr
V=
2ε 0 ∫0 r 2 + D 2
R
σ
r 2 + D2
ε
2 0
0
σ ⎡ 2
2
R + D − D⎤ .
=
⎦
2ε 0 ⎣
=
The potential Vsq at P due to a single quadrant is
V σ ⎡ 2
Vsq = =
R + D2 − D⎤
⎦
4 8ε 0 ⎣
（圖 24-45）
<解>：The electric field (along some axis) is the (negative of the) derivative of the potential V with
respect to the corresponding coordinate.
(7.73 ×10−15 C/m 2 ) ⎡
=
(0.640 m) 2 + (0.259 m) 2 − 0.259 m ⎤
⎦
8(8.85 ×10−12 C 2 /N ⋅ m 2 ) ⎣
In this case, the derivatives can be read off of the
graphs as slopes (since the graphs are of straight lines). Thus,
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Ex = −
∂V
⎛ −500 V ⎞
= −⎜
⎟ = 2500 V/m = 2500 N/C
∂x
⎝ 0.20 m ⎠
Ey = −
∂V
⎛ 300 V ⎞
= −⎜
⎟ = −1000 V/m = −1000 N/C.
∂y
⎝ 0.30 m ⎠
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圖 24-46 中， q = 2.3 pC ， a = 64cm ，如果粒子最初是無限遠且靜止的，要做多少功才能有
圖 24-46 的排列？
These components imply the electric field has a magnitude of 2693 N/C and a direction of
−21.80 C (with respect to the positive x axis). The force on the electron is given by
G
G
F = qE where q = −e . The minus sign associated with the value of q has the implication
G
G
that F points in the opposite direction from E (which is to say that its angle is found by
G
adding 180º to that of E ). With e = 1.6 × 10−19 C , we obtain
G
F = (−1.60 ×10−19 C)[(2500 N/C)i?− (1000 N/C)j] = (−4.0 ×10−16 N)i?+ (1.60 ×10−16 N)j .
（圖 24-46）
<解>：We choose the zero of electric potential to be at infinity. The initial electric potential energy
Ui of the system before the particles are brought together is therefore zero. After the system
is set up the final potential energy is
8th Ed【Problem 24-39】：9th Ed【Problem 24-37】
What is the magnitude of the electric field at the point (3i − 2 j + 4k ) m if the electric potential is
given by V = 2 xyz 2 , where V is in volts and x, y, and z are in meters?
如果電位為 V = 2 xyz 2，其中 V 單位為伏特，xyz 單位為米，問在 (3i − 2 j + 4k )m 的電場為何？
q2 ⎛ 1 1
1
1 1
1 ⎞
− − +
− − +
⎟
4pε 0 ⎜⎝ a a
2a a a
2a ⎠
2q 2 ⎛ 1
⎞
=
− 2 ⎟.
4pε 0 a ⎜⎝ 2
⎠
Thus the amount of work required to set up the system is given by
Uf =
2q 2 ⎛ 1
⎞
− 2⎟
4πε 0 a ⎜⎝ 2
⎠
2(8.99 ×109 N ⋅ m 2 C2 )(2.30 ×10−12 C) 2 ⎛ 1
⎞
=
− 2⎟
⎜
0.640 m
⎝ 2
⎠
= −1.92 ×10−13 J.
W = ΔU = U f − U i = U f =
<解>： E = −∇V ⇒ Ex = −
∂V
= −2.00 yz 2 = −2(−2)(42 ) = 64V / m
∂x
Ey = −
∂V
= −2.00 xz 2 = −2(3)(42 ) = −96V / m
∂y
Ez = −
∂V
= −4.00 xyz = −4(3)(−2)(4) = 96V / m
∂z
8th Ed【Problem 24-43】：9th Ed【Problem 24-41】
A particle of charge +7.5μ C is released from rest at the point x = 60cm on an x axis. The
G
E = Ex2 + E y2 + Ez2
particle begins to move due to the presence of a charge Q that remains fixed at the origin. What is
the kinetic energy of the particle at the instant it has moved 40 cm if (a) Q = +20μ C and (b)
Q = −20μ C ?
= (64V / m) 2 + (−96V / m) 2 + (96V / m) 2
= 150 V m = 150 N C.
粒子帶電 +7.5μ C ，在 x = 60cm 處，從靜止釋放。因為固定在原點的電荷 Q，粒子持續移動，
問如果（a） Q = +20μ C （b） Q = −20μ C ，粒子瞬間移動 40 公分，問動能？
8th Ed【Problem 24-41】：9th Ed【Problem 24-43】
How much work is required to set up the arrangement of Fig. 24-46 if q = 2.3 pC , a = 64cm , and
the particles are initially infinitely far apart and at rest?
<解>：We apply conservation of energy for the particle with q = 7.5 ×10−6 C (which has zero initial
kinetic energy):
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where U =
4πε 0 r
東海大學物理系
(b) Charge q1 is a distance w from point b and charge q2 is a distance A, so the electric
U0 = K f + U f ,
qQ
Fundamentals of Physics, 8th Ed
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potential at B is
VB =
.
(a) The initial value of r is 0.60 m and the final value is (0.6 + 0.4) m = 1.0 m (since the
particles repel each other). Conservation of energy, then, leads to K f = 0.9 J .
(b) Now the particles attract each other so that the final value of r is 0.60 − 0.40 = 0.20 m.
Use of energy conservation yields K f = 4.5 J in this case.
8th Ed【Problem 24-45】：9th Ed【Problem 24-51】
In the rectangle of Fig. 24- 48, the sides have lengths 5.0 cm and 15 cm, q1 = −5μ C , and
q2 = +2μ C .With V = 0 at infinity, what is the electric potential at (a) corner A and (b) corner B?
(c) How much work is required to move a charge q3 = +3μ C from B to A along a diagonal of the
rectangle? (d) Does this work increase or decrease the electric potential energy of the three-charge
system? Is more, less, or the same work required if q3 is moved along a path that is (e) inside the
−6
1 ⎛ q1 q2 ⎞
2.0 ×10−6 C ⎞
9
2
2 ⎛ −5.0 × 10 C
+
⎟
⎜ + ⎟ = (8.99 × 10 N ⋅ m / C ) ⎜
4πε 0 ⎝ w A ⎠
0.15 m ⎠
⎝ 0.050 m
= −7.8 × 105 V.
(c) Since the kinetic energy is zero at the beginning and end of the trip, the work done by an
external agent equals the change in the potential energy of the system. The potential
energy is the product of the charge q3 and the electric potential. If UA is the potential
energy when q3 is at A and UB is the potential energy when q3 is at B, then the work done
in moving the charge from B to A is
W = U A − U B = q3 (VA − VB ) = (3 × 10−6 C )(6 × 104 V + 7.8 × 105V ) = 2.5 J
(d) The work done by the external agent is positive, so the energy of the three-charge system
increases.
(e) and (f) The electrostatic force is conservative, so the work is the same no matter which
path is used.
8th Ed【Problem 24-49】：9th Ed【Problem 24-47】
rectangle but not on a diagonal and (f) outside the rectangle?
What is the escape speed for an electron initially at rest on the surface of a sphere with a radius of
圖 24-48，一長方形，邊長為 5cm 和 15cm， q1 = −5μ C 且 q2 = +2μ C 。無窮遠處 V = 0 。問（a）
A 端點（b）B 端點位能？（c）要將一電荷 q3 = +3μ C ，從 B 沿著長方形對角線移到 A，問
需做多少功？（d）在這三個電荷系統中，功是增加還是減少的？如果 q3 沿著（e）長方形
1.0 cm and a uniformly distributed charge of 1.6 ×10−15 C ? That is, what initial speed must the
electron have in order to reach an infinite distance from the sphere and have zero kinetic energy
when it gets there?
裡面，非對角線（f）長方形外面，功會增加、減少還是一樣？
半徑 1 公分的球表面，均勻分佈電荷 1.6 × 10−15 C ，電荷一開始靜止，問脫離速度？亦即，電
子必須有一初速度可以從球到達無限遠的距離，當它到那邊時，動能為 0。
<解>：The escape speed may be calculated from the requirement that the initial kinetic energy (of
（圖 24-48）
<解>：(a) Let A = 015
. m be the length of the rectangle and w = 0.050 m be its width. Charge q1 is a
distance A from point A and charge q2 is a distance w, so the electric potential at A is
1 ⎛ q1 q2 ⎞
VA =
⎜ + ⎟
4πε 0 ⎝ A w ⎠
⎛ −5.0 × 10−6 C 2.0 ×10−6 C ⎞
= (8.99 × 109 N ⋅ m 2 / C2 ) ⎜
+
⎟
0.050 m ⎠
⎝ 0.15 m
launch) be equal to the absolute value of the initial potential energy (compare with the
gravitational case in chapter 14). Thus,
1 2
eq
mv =
2
4πε 0 r
where
m = 9.11× 10−31 kg , e = 1.6 × 10−19 C , q = 10000e , and r = 0.01m .
This yields
v = 22490m / s ≈ 2.2 × 104 m / s
8th Ed【Problem 24-53】：9th Ed【Problem 24-49】
Two electrons are fixed 2cm apart. Another electron is shot from infinity and stops midway between
the two. What is its initial speed?
= 6.0 ×104 V.
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<解>：(a) The electric field between the plates is leftward in Fig, 24-52 since it points towards lower
兩個電子被固定相距 2 公分，另一顆電子從無限遠處被射擊出來，並停止在兩個電子中間，
問初始速度？
<解>：We use conservation of energy, taking the potential energy to be zero when the moving
electron is far away from the fixed electrons. The final potential energy is then
U f = 2e 2 / 4πε 0 d , where d is half the distance between the fixed electrons. The initial kinetic
energy is Ki = 21 mv 2 ,
⇒
4e 2
=
4 pε 0 dm
G
G
右移動的粒子），因此可以得到粒子帶正電（ q > 0 ）（確保力 F 平行電場 E ），所以
得知它是一個質子（質子帶正電）而不是電子（電子帶負電）。
where m is the mass of an electron and v is the initial speed of the
K0 + U 0 = K + U
2
1
1
m p v02 + qV1 = m p v 2 + qV2
2
2
V2 = −50V ,
. × 10
c8.99 × 10 N ⋅ m C hb4gc160
b0.010 mgc9.11 × 10 kgh
9
⇒
Using q = +1.6 × 10−19 C , m p = 1.67 ×10−27 kg , v0 = 90 × 103 m / s , V1 = −70V and
1 2
mv = 2e 2 / 4πε 0 d .
2
Hence,
v=
leftward, from the problem description (indicating deceleration of the rightward moving
G
G
particle), so that q > 0 (ensuring that F is parallel to E ); it is a proton.
V2 電位較高，因此圖 24-52 中，板子間的電場是向左。從問題的敘述（一個減速向
(b) We use conservation of energy（能量守恆）：
moving electron. The final kinetic energy is zero. Thus,
Ki = Uf
values of potential. The force (associated with the field, by Eq. 23-28) is evidently
2
−31
−19
C
h
2
= 3.2 × 102 m s.
v = v02 + 2q
V1 − V2
−70 − (−50)
= (90 × 103 ) 2 + 2(1.6 × 10−19 )
1.67 × 10−27
mp
= (81× 108 ) − (38.32 × 108 )
8th Ed【Problem 24-55】：9th Ed【Problem 24-59】
In Fig. 24-52, a charged particle (either an electron or a proton) is moving rightward between two
parallel charged plates separated by distance d = 2mm . The plate potentials are V1 = −70V and
V2 = −50V . The particle is slowing from an initial speed of 90km / s at the left plate. (a) Is the
= 6.53 × 104 m / s
we obtain the final speed v = 6.53 ×104 m / s . We note that the value of d is not used in the
solution.
particle an electron or a proton? (b) What is its speed just as it reaches plate 2?
8th Ed【Problem 24-63】：9th Ed【Problem 24-65】
What is the excess charge on a conducting sphere of radius r = 0.15m if the potential of the sphere
圖 24-52，一帶電粒子（如果不是電子，就是質子），向右移動在兩個平行帶電平板間，平
版相距 d = 2mm ，平板電位分別是 V1 = −70V 和 V2 = −50V 。粒子最初在左邊板子以初速
is 1500V and V = 0 at infinity?
（a）該粒子是電子還是質子？（b）當它到達第二片板子時，速率為
90km / s 向右減速移動，
何？
半徑 r = 0.15m 的導體球，如果電位為 1500V，無窮遠處 V = 0 ，問剩餘電荷為何？
<解>：If the electric potential is zero at infinity, then the potential at the surface of the sphere is
1 q
, where q is the charge on the sphere and r is its radius. Thus,
given by V =
4πε 0 r
q = 4 pε 0 rV =
b
gb
g
015
. m 1500 V
= 2.5 × 10 −8 C.
8.99 × 109 N ⋅ m2 C 2
8th Ed【Problem 24-65】：9th Ed【Problem 24-63】
Two metal spheres, each of radius 3.0 cm, have a center-to- center separation of 2.0 m. Sphere 1 has
（圖 24-52）
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charge +1×10−8 C ; sphere 2 has charge −3 ×10−8 C . Assume that the separation is large enough for
us to assume that the charge on each sphere is uniformly distributed (the spheres do not affect each
other).With V = 0 at infinity, calculate (a) the potential at the point halfway between the centers
一半徑 15cm 的金屬球，帶有電荷 3 ×10−8 C ，
（a）球表面的電場（b）無窮遠處 V = 0 ，球表
面的電位？（c）距離球表面多少距離時？電位會下降 500V。
and the potential on the surface of (b) sphere 1 and (c) sphere 2.
<解>：(a) The magnitude of the electric field is
兩個金屬球，半徑均為 3cm，中心點對中心點的距離為 2m，球 1 帶電 +1× 10−8 C ，球 2 帶電
，無窮遠處 V = 0 ，
−3 ×10−8 C。假設分開距離夠大，電荷在球上是均勻分佈的（球不互相影響）
問（a）兩中心點的中心位置的電位？（b）球 1 表面的電位（c）球 2 表面的電位？
<解>：(a) The electric potential is the sum of the contributions of the individual spheres. Let q1 be
the charge on one, q2 be the charge on the other, and d be their separation. The point
E=
( 3.0 ×10−8 C )(8.99 ×109 N ⋅ m2 C2 ) = 1.2 ×104 N C.
σ
q
=
=
2
ε 0 4πε 0 R 2
( 0.15 m )
(b) V = RE = (0.15 m)(1.2 × 104 N/C) = 1.8 × 103 V.
(c) Let the distance be x. Then
1
1
q
ΔV = V x − V =
−
= −500 V,
4 pε 0 R + x R
halfway between them is the same distance d/2 (= 1.0 m) from the center of each sphere,
which gives
so the potential at the halfway point is
x=
(8.99 ×10 N ⋅ m C )(1.0 ×10 C − 3.0 ×10 C ) = −1.8 ×102 V.
q1 + q2
=
4pε 0 d 2
1.0 m
9
V=
2
2
−8
−8
FG
H
bg
b
gb
IJ
K
g
015
. m −500 V
R ΔV
. × 10−2 m.
=
= 58
−V − ΔV −1800 V + 500 V
(b) The distance from the center of one sphere to the surface of the other is d – R, where R is
the radius of either sphere. The potential of either one of the spheres is due to the charge
on that sphere and the charge on the other sphere. The potential at the surface of sphere 1
is
V1 =
1 ⎡ q1
q ⎤
+ 2
4πε 0 ⎢⎣ R d − R ⎥⎦
⎡1.0 ×10−8 C
3.0 ×10−8 C ⎤
3
= ( 8.99 ×109 N ⋅ m 2 C 2 ) ⎢
−
⎥ = 2.9 × 10 V.
0.030
m
2.0
m
0.030
m
−
⎣
⎦
(c) The potential at the surface of sphere 2 is
1 ⎡ q1
q ⎤
V2 =
+ 2
4πε 0 ⎢⎣ d − R R ⎥⎦
⎡ 1.0 ×10−8 C
3.0 × 10−8 C ⎤
3
= ( 8.99 × 109 N ⋅ m 2 C2 ) ⎢
−
⎥ = −8.9 × 10 V.
2.0
m
0.030
m
0.030
m
−
⎣
⎦
8th Ed【Problem 24-67】：9th Ed【Problem 24-67】
A metal sphere of radius 15 cm has a net charge of 3 ×10−8 C . (a) What is the electric field at the
sphere’s surface? (b) If V = 0 at infinity, what is the electric potential at the sphere’s surface? (c)
At what distance from the sphere’s surface has the electric potential decreased by 500 V?
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