2. 2 Chapter 23 Solutions
23.4 We find the equal-magnitude charges on both spheres:
q1q2 q2 F 1.00 × 10 4 N
F = ke 2 = ke 2 so q=r = (1.00 m ) = 1.05 × 10 −3 C
r r ke 8.99 × 10 9 N ⋅ m 2/ C 2
The number of electron transferred is then
(
N xfer = 1.05 × 10 −3 C ) (1.60 × 10 −19
)
C / e − = 6.59 × 1015 electrons
The whole number of electrons in each sphere is
Ntot =
10.0 g
( −
)( 24 −
6.02 × 10 atoms / mol 47 e / atom = 2.62 × 10 e
107.87 g / mol
23
)
The fraction transferred is then
N xfer 6.59 × 1015
f= = = 2.51 × 10–9 = 2.51 charges in every billion
Ntot 2.62 × 1024
( )( ) (6.02 × 10 )
2 23 2
qq 8.99 × 10 9 N ⋅ m 2 C 2 1.60 × 10 −19 C
23.5 F = ke 1 2 2 = = 514 kN
[ ]
2
r 2(6.37 × 106 m)
*23.6 (a) The force is one of attraction. The distance r in Coulomb's law is the distance between
centers. The magnitude of the force is
F=
ke q1q2
= 8.99 × 10 9
( −9
)(
−9
N ⋅ m 2 12.0 × 10 C 18.0 × 10 C
= 2.16 × 10 − 5 N
)
r2 C2 (0.300 m)2
(b) The net charge of − 6.00 × 10 −9 C will be equally split between the two spheres, or
− 3.00 × 10 −9 C on each. The force is one of repulsion, and its magnitude is
F=
ke q1q2
= 8.99 × 10 9 N⋅m
2 ( )(
3.00 × 10 −9 C 3.00 × 10 −9 C
=
) 8.99 × 10 −7 N
r2 C2 (0.300 m)2
4. 4 Chapter 23 Solutions
*23.8 Let the third bead have charge Q and be located distance x from the left end of the rod. This
bead will experience a net force given by
ke ( 3q)Q ke ( q)Q
F= i+ ( −i)
x2 ( d − x )2
3 1 x
The net force will be zero if = , or d − x =
x 2 ( d − x )2 3
This gives an equilibrium position of the third bead of x = 0.634d
The equilibrium is stable if the third bead has positive charge .
k ee 2 (1.60 × 10–19 C)2
*23.9 (a) F= 2 = (8.99 × 109 N ⋅ m2/C 2) = 8.22 × 10–8 N
r (0.529 × 10–10 m)2
(b) We have F =
mv 2
from which v =
Fr
=
(8.22 × 10 −8
)(
N 0.529 × 10 −10 m )= 2.19 × 106 m/s
r −31
m 9.11 × 10 kg
ke qQ
23.10 The top charge exerts a force on the negative charge which is directed upward and
( d 2 )2 + x 2
to the left, at an angle of tan −1 ( d / 2x ) to the x-axis. The two positive charges together exert
force
2 k qQ (− x)i
− 2 ke qQ
2 e = ma or for x << d 2, a≈
( ) (
x
d 4 + x2 d2 4 + x2
)
1/2
md 3 / 8
(a) The acceleration is equal to a negative constant times the excursion from equilibrium, as i n
16 ke qQ
a = −ω 2 x, so we have Simple Harmonic Motion with ω 2 = .
md 3
2π π md 3
T= = , where m is the mass of the object with charge −Q.
ω 2 ke qQ
ke qQ
(b) vmax = ω A = 4a
md 3
6. 6 Chapter 23 Solutions
*23.15 (a)
ke q
E1 = 2 =
( )(
8.99 × 10 9 7.00 × 10 −6 )
= 2.52 × 10 5 N C
r (0.500) 2
E2 =
ke q
=
( )(
8.99 × 10 9 4.00 × 10 −6 )
= 1.44 × 10 5 N C
r2 (0.500)2
Ex = E2 − E1 cos 60° = 1.44 × 10 5 − 2.52 × 10 5 cos 60.0° = 18.0 × 10 3 N C
Ey = −E1 sin 60.0° = −2.52 × 10 5 sin 60.0° = −218 × 10 3 N C
E = [18.0i − 218 j] × 10 3 N C = [18.0i − 218 j] kN C
(b) ( )
F = qE = 2.00 × 10 −6 C (18.0i − 218 j) × 10 3 N C = ( 36.0i − 436 j) × 10 −3 N = (36.0i − 436 j ) mN
*23.16 (a) E1 =
ke q1
(− j) =
(8.99 × 10 )(3.00 × 10 ) (− j) = − (2.70 × 10
9 −9
3
)
NC j
2
r1 (0.100)2
E2 =
ke q2
(− i) =
(8.99 × 10 )(6.00 × 10 ) (− i) = − (5.99 × 10
9 −9
2
)
NC i
2
r2 (0.300)2
( ) (
E = E 2 + E1 = − 5.99 × 10 2 N C i − 2.70 × 10 3 N C j )
(b) ( )
F = qE = 5.00 × 10 −9 C ( −599i − 2700 j) N C
(
F = − 3.00 × 10 −6 i − 13.5 × 10 −6 j N = ) (− 3.00 i − 13.5 j ) µN
23.17 (a) The electric field has the general appearance shown. It is zero
at the center , where (by symmetry) one can see that the three
charges individually produce fields that cancel out.
(b) You may need to review vector addition in Chapter Three.
The magnitude of the field at point P due to each of the charges
along the base of the triangle is E = ke q a 2 . The direction of the field
in each case is along the line joining the charge in question to point
P as shown in the diagram at the right. The x components add to
zero, leaving
ke q kq ke q
2 (
E= sin 60.0°) j + e2 (sin 60.0°) j = 3 j
a a a2
8. 8 Chapter 23 Solutions
ke q1 k q k q k ( 2q) k ( 3q) k ( 4q)
23.19 (a) E= 2 1
~ + e 22 ~2 + e 23 ~3 = e 2 i + e 2 (i cos 45.0° + j sin 45.0°) + e 2 j
r1 r2 r3 a 2a a
ke q kq kq
E = 3.06 i + 5.06 e2 j = 5.91 e2 at 58.8°
a2 a a
ke q 2
(b) F = qE = 5.91 at 58.8°
a2
23.20 The magnitude of the field at (x, y) due to charge q at (x0 , y0 )
is given by E = ke q r 2 where r is the distance from (x0 , y0 ) to
(x, y). Observe the geometry in the diagram at the right.
From triangle ABC, r 2 = (x − x0 )2 + (y − y0 )2 , or
(y − y0 ) (x − x0 )
r = (x − x0 )2 + (y − y0 )2 , sin θ = , and cos θ =
r r
ke q (x − x0 ) ke q(x − x0 )
Thus, Ex = Ecos θ = =
r 2
r [(x − x0 )2 + (y − y0 )2 ]3/2
ke q (y − y0 ) ke q(y − y0 )
and Ey = Esin θ = =
r2 r [(x − x0 )2 + (y − y0 )2 ]3/2
k eq k eq k eq(4ax)
23.21 The electric field at any point x is E= – =
(x – a)2 (x – (–a))2 (x 2 – a 2)2
(4a)(k eq)
When x is much, much greater than a, we find E≈
x3
ke Q/n
23.22 (a) One of the charges creates at P a field E=
R 2 + x2
at an angle θ to the x-axis as shown.
When all the charges produce field, for n > 1, the components
perpendicular to the x-axis add to zero.
nke (Q/n)i keQxi
The total field is 2 2 cos θ =
R +x (R + x 2)3/2
2
(b) A circle of charge corresponds to letting n grow beyond all bounds, but the result does not
depend on n. Smearing the charge around the circle does not change its amount or its
distance from the field point, so it does not change the field. .
10. 10 Chapter 23 Solutions
ke Q x
23.28 E=
(x + a2 )3/2
2
dE 1 3x 2
For a maximum, = Qke 2 − 2 2 5/2
=0
(x + a ) (x + a )
dx 2 3/2
a
x 2 + a2 − 3x 2 = 0 or x=
2
Substituting into the expression for E gives
ke Qa k Q 2ke Q Q
E= = e = =
2( 2 a2 )3/2 3 3 a2
3 3 3 a2 6 3 π e0 a2
2
x
23.29 E = 2 π ke σ 1 −
2
x +R
2
( )(
E = 2 π 8.99 × 10 9 7.90 × 10 −3 1 −
) x
x 2 + (0.350)
2
= 4.46 × 108 1 −
x
x 2 + 0.123
(a) At x = 0.0500 m, E = 3.83 × 108 N C = 383 MN C
(b) At x = 0.100 m, E = 3.24 × 108 N C = 324 MN C
(c) At x = 0.500 m, E = 8.07 × 107 N C = 80.7 MN C
(d) At x = 2.00 m, E = 6.68 × 108 N C = 6.68 MN C
x
23.30 (a) From Example 23.9: E = 2 π ke σ 1 −
x2 + R2
Q
σ= = 1.84 × 10 −3 C m 2
πR 2
E = (1.04 × 108 N C)(0.900) = 9.36 × 107 N C = 93.6 MN/C
appx: E = 2 π ke σ = 104 MN/C (about 11% high)
30.0 cm
(b) E = (1.04 × 108 N / C) 1 − = (1.04 × 10 N C)(0.00496) = 0.516 MN/C
8
30.0 + 3.00 cm
2 2
Q 5.20 × 10 −6
appx: E = ke = (8.99 × 10 9 ) = 0.519 MN/C (about 0.6% high)
r2 (0.30)2
12. 12 Chapter 23 Solutions
23.34 (a) We define x = 0 at the point where we are to find the field. One ring, with thickness dx, has
charge Qdx/h and produces, at the chosen point, a field
ke x Q dx
dE = 2 2 3/2
i
(x + R ) h
The total field is
d+h keQ x dx k Qi d + h 2
E= ∫ dE =∫ 2 2 3/2
i = e ∫ (x + R 2 )− 3/2 2x dx
all charge
d h(x + R ) 2h x = d
k Q i (x 2 + R 2 )− 1/2 d+h keQ i 1 1
E= e = −
(d 2 + R 2 )1/2
( )
2h (− 1/ 2) h 1/2
x=d
(d + h)2 + R 2
(b) Think of the cylinder as a stack of disks, each with thickness dx, charge Q dx/h, and charge-
per-area σ = Q dx / πR 2 h. One disk produces a field
2 π keQ dx x
dE =
π R2h 1 − (x 2 + R 2 )1/2 i
d+h 2 keQ dx x
So, E= ∫ dE = ∫ 1 − (x 2 + R 2 )1/2 i
all charge x=d R2 h
2 keQ i d + h d+ h 1 (x 2 + R 2 )1/2 d + h
1 d + h(x 2 + R 2 )− 1/2 2x dx = 2 keQ i x
R 2 h ∫d
E= dx − 2 ∫ −
x=d R2h d
2 1/ 2 d
2 keQ i
(
d + h − d − (d + h)2 + R 2
) + (d 2 + R 2 )1/2
1/2
E=
R 2h
(
h + (d 2 + R 2 )1/2 − (d + h)2 + R 2
)
2 keQ i 1/2
E=
R 2h
ke dq
23.35 (a) The electric field at point P due to each element of length dx, is dE = and is directed
(x + y 2 ) 2
along the line joining the element of length to point P . By symmetry,
Ex = ∫ dEx = 0 and since dq = λ dx,
y
E = Ey = ∫ dEy = ∫ dE cos θ where cos θ =
(x + y 2 )1 2
2
dx 2ke λ sinθ 0
Therefore, =
(x + y 2 )3 2
2
y
2ke λ
(b) For a bar of infinite length, θ → 90° and Ey =
y
16. 16 Chapter 23 Solutions
a=
qE
=
(
1.602 × 10 −19 (640) )
= 6.14 × 1010 m/s2
( )
23.43 (a)
m 1.67 × 10 −27
(b) v = v i + at
1.20 × 106 = (6.14 × 1010)t
t = 1.95 × 10-5 s
(c) x − xi = 2 ( vi + v )t
1
( )( )
x = 2 1.20 × 106 1.95 × 10 −5 = 11.7 m
1
(d) K = 2 mv 2 = 1 (1.67 × 10 − 27 kg)(1.20 × 106 m / s)2 = 1.20 × 10-15 J
1
2
23.44 The required electric field will be in the direction of motion . We know that Work = ∆K
1 2
So, –Fd = – 2 m v i (since the final velocity = 0)
1 2
mv i
1 2
This becomes Eed = mvi 2 or E=
2 ed
1.60 × 10–17 J
E = = 1.00 × 103 N/C (in direction of electron's motion)
(1.60 × 10–19 C)(0.100 m)
23.45 The required electric field will be in the direction of motion .
1 2
Work done = ∆K so, –Fd = – 2 m v i (since the final velocity = 0)
K
which becomes eEd = K and E=ed
18. 18 Chapter 23 Solutions
x 0.0500
23.47 (a) t= = = 1.11 × 10-7 s = 111 ns
v 4.50 × 10 5
qE (1.602 × 10 −19 )(9.60 × 10 3 )
(b) ay = = = 9.21 × 1011 m / s 2
m (1.67 × 10 − 27 )
y − yi = v y i t + 2 ay t 2
1
y = 2 (9.21 × 1011 )(1.11 × 10 −7 )2 = 5.67 × 10-3 m = 5.67 mm
1
(c) v x = 4.50 × 105 m/s
vy = vy i + ay = (9.21 × 1011)(1.11 × 10-7) = 1.02 × 105 m/s
qE (1.602 × 10 −19 )(390)
23.48 ay = = − 31
= 6.86 × 1013 m / s 2
m (9.11 × 10 )
2vi sin θ
(a) t= from projectile motion equations
ay
2(8.20 × 10 5 )sin 30.0°
t= = 1.20 × 10-8 s) = 12.0 ns
6.86 × 1013
vi 2 sin 2 θ (8.20 × 10 5 )2 sin 2 30.0°
(b) h= = = 1.23 mm
2ay 2(6.86 × 1013 )
vi 2 sin 2 θ (8.20 × 10 5 )2 sin 60.0°
(c) R= = = 4.24 mm
2ay 2(6.86 × 1013 )
23.49 vi = 9.55 × 103 m/s
eE (1.60 × 10 −19 )(720)
(a) ay = = = 6.90 × 1010 m s 2
m (1.67 × 10 −27 )
vi 2 sin 2θ (9.55 × 10 3 )2 sin 2θ
R= = 1.27 × 10-3 m so that = 1.27 × 10 −3
ay 6.90 × 1010
sin 2θ = 0.961 θ = 36.9° 90.0° – θ = 53.1°
R R
(b) t= = If θ = 36.9°, t = 167 ns If θ = 53.1°, t = 221 ns
vix vi cos θ
20. 20 Chapter 23 Solutions
(b) The distance from the positive plate to where the meeting occurs equals the distance the
sodium ion travels (i.e., dNa = 2 aNat 2 ). This is found from:
1
1 eE 2 1 eE 2
4.00 cm = 2 aNat 2 + 2 aClt 2 : 4.00 cm =
1 1 t + t
2 22.99 u 2 35.45 u
This may be written as 1 1
(
4.00 cm = 2 aNat 2 + 2 (0.649aNa )t 2 = 1.65 2 aNat 2
1
)
4.00 cm
so dNa = 2 aNat 2 =
1 = 2.43 cm
1.65
23.52 From the free-body diagram shown, ∑Fy = 0
and T cos 15.0° = 1.96 × 10–2 N
So T = 2.03 × 10–2 N
From ∑Fx = 0, we have qE = T sin 15.0°
T sin 15.0° (2.03 × 10–2 N) sin 15.0°
or q= = = 5.25 × 10–6 C = 5.25 µC
E 1.00 × 103 N/C
23.53 (a) Let us sum force components to find
∑Fx = qEx – T sin θ = 0, and ∑Fy = qEy + T cos θ – mg = 0
Combining these two equations, we get
mg (1.00 × 10-3)(9.80)
q= = = 1.09 × 10–8 C = 10.9 nC
(Ex cot θ + Ey) (3.00 cot 37.0° + 5.00) × 105
Free Body Diagram
(b) From the two equations for ∑Fx and ∑Fy we also find
for Goal Solution
qEx
sin 37.0° = 5.44 × 10 N = 5.44 mN
–3
T=
22. 22 Chapter 23 Solutions
Goal Solution
A charged cork ball of mass m is suspended on a light string in the presence of a uniform electric field, as
shown in Figure P23.53. When E = ( Ai + Bj) N / C , where A and B are positive numbers, the ball is i n
equilibrium at the angle θ . Find (a) the charge on the ball and (b) the tension in the string.
G: This is the general version of the preceding problem. The known quantities are A , B, m, g , and θ .
The unknowns are q and T .
O: The approach to this problem should be the same as for the last problem, but without numbers to
substitute for the variables. Likewise, we can use the free body diagram given in the solution to
problem 53.
A: Again, Newton's second law: −T sin θ + qA = 0 (1)
and + T cos θ + qB − mg = 0 (2)
qA qA cos θ
(a) Substituting T = , into Eq. (2), + qB = mg
sin θ sin θ
mg
Isolating q on the left, q=
( A cot θ + B)
mgA
(b) Substituting this value into Eq. (1), T=
( A cos θ + Bsin θ )
L : If we had solved this general problem first, we would only need to substitute the appropriate values
in the equations for q and T to find the numerical results needed for problem 53. If you find this
problem more difficult than problem 53, the little list at the Gather step is useful. It shows what
symbols to think of as known data, and what to consider unknown. The list is a guide for deciding
what to solve for in the Analysis step, and for recognizing when we have an answer.
ke q1q2 15.0
23.55 F= tan θ = 60.0 θ = 14.0°
r2
(8.99 × 109)(10.0 × 10–6)2
F1 = = 40.0 N
(0.150)2
(8.99 × 109)(10.0 × 10–6)2
F3 = = 2.50 N
(0.600)2
(8.99 × 109)(10.0 × 10–6)2
F2 = = 2.35 N
(0.619)2
Fx = –F3 – F2 cos 14.0° = –2.50 – 2.35 cos 14.0° = – 4.78 N
Fy = –F1 – F2 sin 14.0° = – 40.0 – 2.35 sin 14.0° = – 40.6 N
2 2
Fnet = Fx + Fy = (– 4.78)2 + (– 40.6)2 = 40.9 N
Fy – 40.6
tan φ = F = – 4.78 φ = 263°
x
24. 24 Chapter 23 Solutions
*23.59 According to the result of Example 23.7, the lefthand rod
creates this field at a distance d from its righthand end:
k eQ
E = d(2a + d)
keQQ dx
dF = 2a d(d + 2a)
k eQ 2 b
k eQ 2 1 2a + x b
∫ x = b – 2a x(x + 2a)
dx
F= = – ln
2a 2a 2a x
b – 2a
+k eQ 2 2a + b b k eQ 2 b2 k e Q 2 b2
F = – ln + ln = 2 ln (b – 2a)(b + 2a) = 2 ln 2 2
4a 2 b b – 2a 4a 4a b – 4a
*23.60 The charge moves with acceleration of magnitude a given by ∑F = ma = q E
qE 1.60 × 10–19 C (1.00 N/C)
(a) a= m = = 1.76 × 1011 m/s2
9.11 × 10–31 kg
v 3.00 × 107 m/s
Then v = v i + at = 0 + at gives t=a = = 171 µs
1.76 × 1011 m/s2
v vm (3.00 × 107 m/s)(1.67 × 10–27 kg)
(b) t = a = qE = = 0.313 s
(1.60 × 10–19 C)(1.00 N/C)
vm
(c) From t = qE , as E increases, t gets shorter in inverse proportion.
90.0° 90.0°
23.61 Q = ∫ λ dl = ∫ –90.0° λ 0 cos θ Rd θ = λ 0 R sin θ –90.0° = λ 0 R [1 – (–1)] = 2λ 0 R
Q = 12.0 µC = (2λ 0 )(0.600) m = 12.0 µC so λ 0 = 10.0 µC/m
dFy =
1 ( 3.00 µ C)(λ dl)
cos θ =
1
( 3.00 µ C) λ 0 cos 2 θ Rdθ
( )
4 π e0 R2 4 π e0
R2
90.0°
N · m2 (3.00 × 10–6 C)(10.0 × 10–6 C/m)
Fy = ∫ –90.0° 8.99 × 109 cos2 θ d θ
C2 (0.600 m)
π /2
8.99( 30.0)
Fy =
0.600
(
10 −3 N )∫( 1
2
1
+ cos 2θ dθ
2 )
− π /2
Fy = (0.450 N ) ( 1
2
π
1 π /2
)
+ sin 2θ − π /2 = 0.707 N
4
Downward.
Since the leftward and rightward forces due to the two halves of the
semicircle cancel out, Fx = 0.
26. 26 Chapter 23 Solutions
23.64 At an equilibrium position, the net force on the charge Q is zero. The equilibrium position
can be located by determining the angle θ corresponding to equilibrium. In terms of lengths s,
1 a 3 , and r, shown in Figure P23.64, the charge at the origin exerts an attractive force
2
keQ q (s + 2 a 3)2 . The other two charges exert equal repulsive forces of magnitude keQq r 2 .
1
The horizontal components of the two repulsive forces add, balancing the attractive force,
2 cos θ
1
Fnet = keQq − 1 a 3)2
=0
r 2
(s + 2
1a
From Figure P23.64, r= 2 s = 2 a cot θ
1
sin θ
4 1
The equilibrium condition, in terms of θ , is Fnet = k Qq 2 cos θ sin 2θ − =0
a2 e ( 3 + cot θ )2
Thus the equilibrium value of θ is 2 cos θ sin 2 θ ( 3 + cot θ )2 = 1.
One method for solving for θ is to tabulate the left side. To three significant figures the value
of θ corresponding to equilibrium is 81.7°. The distance from the origin to the equilibrium
position is x = 2 a( 3 + cot 81.7°) = 0.939a
1
θ 2 cos θ sin 2 θ ( 3 + cot θ )2
60° 4
70° 2.654
80° 1.226
90° 0
81° 1.091
81.5° 1.024
81.7° 0.997
23.65 (a) The distance from each corner to the center of the square is
( L 2 )2 + ( L 2 )2 = L 2
The distance from each positive charge to −Q is then z 2 + L2 2 .
Each positive charge exerts a force directed along the line joining
q and −Q, of magnitude keQq
z + L2 2
2
z
The line of force makes an angle with the z-axis whose cosine is
z + L2 2
2
The four charges together exert forces whose x and y components 4keQ q z
add to zero, while the z-components add to F= − k
(z )
32
2
+ L2 2
28. 28 Chapter 23 Solutions
(b) If Fe = 0, then E at P must equal zero. In order for the field to cancel at P , the − 4q must be
above + q on the y-axis.
ke q ke (4q)
Then, E=0=− + , which reduces to y 2 = 4.00 m 2 .
(1.00 m) 2
y2
Thus, y = ± 2.00 m . Only the positive answer is acceptable since the − 4q must be located
above + q. Therefore, the − 4q must be placed 2.00 meters above point P along the + y − axis .
23.68 The bowl exerts a normal force on each bead, directed along
the radius line or at 60.0° above the horizontal. Consider the
free-body diagram of the bead on the left:
ΣFy = nsin 60.0° −mg = 0 ,
mg
or n=
sin 60.0°
Also, ΣFx = −Fe + ncos 60.0° = 0,
n
ke q 2 mg mg
or 2
= ncos 60.0° = = Fe 60.0˚
R tan 60.0° 3
12 mg
mg
Thus, q= R
ke 3
23.69 (a) There are 7 terms which contribute:
3 are s away (along sides)
1
3 are 2 s away (face diagonals) and sin θ = = cos θ
2
1
1 is 3 s away (body diagonal) and sin φ =
3
The component in each direction is the same by symmetry.
ke q 2 2 1 ke q 2
F= 1+ 2 2 + 3 3 (i + j + k) = (1.90)(i + j + k)
s2 s2
ke q 2
(b) F= F2 +F2 +F2 =
x y z 3.29 away from the origin
s2