Download Prob. 23.5 (a) Identify: Use conservation of energy: U for the pair of

Prob. 23.5
(a) IDENTIFY: Use conservation of energy:
Ka  U a  Wother  Kb  U b
U for the pair of point charges is given by Eq.(23.9).
SET UP:
Let point a be where q2 is 0.800 m
from q1 and point b be where q2 is
0.400 m from q1, as shown in Figure
23.5a.
Figure 23.5a
EXECUTE: Only the electric force does work, so Wother  0 and U 
1 q1q2
.
4 P0 r
K a  12 mva2  12 (1.50  103 kg)(22.0 m/s) 2  0.3630 J
Ua 
1 q1q2
(2.80 106 C)(7.80 106 C)
 (8.988 109 N  m2/C2 )
 0.2454 J
4 P0 ra
0.800 m
K b  12 mvb2
Ub 
1 q1q2
(2.80 106 C)(7.80 106 C)
 (8.988 109 N  m2 /C2 )
 0.4907 J
4 P0 rb
0.400 m
The conservation of energy equation then gives Kb  Ka  (U a  Ub )
1
2
mvb2  0.3630 J  (0.2454 J  0.4907 J)  0.1177 J
vb 
2(0.1177 J)
 12.5 m/s
1.50  103 kg
EVALUATE: The potential energy increases when the two positively charged spheres get closer together, so the
kinetic energy and speed decrease.
(b) IDENTIFY: Let point c be where q2 has its speed momentarily reduced to zero. Apply conservation of energy to
points a and c: Ka  U a  Wother  Kc  U c .
SET UP: Points a and c are shown in Figure 23.5b.
EXECUTE: K a  0.3630 J (from part
(a))
U a  0.2454 J (from part (a))
Figure 23.5b
Kc  0 (at distance of closest approach the speed is zero)
Uc 
Thus conservation of energy K a  U a  U c gives
rc 
EVALUATE:
1 q1q2
4 P0 rc
1 q1q2
 0.3630 J  0.2454 J  0.6084 J
4 P0 rc
1
q1q2
(2.80 106 C)(7.80 106 C)
 (8.988 109 N  m2 /C2 )
 0.323 m.
4 P0 0.6084 J
0.6084 J
U   as r  0 so q2 will stop no matter what its initial speed is.
Prob. 23.9
y
q2 = -3 nC
q1 = 4 nC
q3 = 2 nC
20.0 cm
(a) What is the potential energy of the system with q3 at x = 10.0 cm?
(b) Where should q3 be placed for zero potential energy?
IDENTIFY:
SET UP:
qq
qq
qq 
U  k 1 2  1 3  2 3 
r13
r23 
 r12
In part (a), r12  0.200 m , r23  0.100 m and r13  0.100 m. In part (b) let particle 3 have coordinate x, so
r12  0.200 m , r13  x and r23  0.200  x.
 (4.00 nC)(3.00 nC) (4.00 nC)(2.00 nC) ( 3.00 nC)(2.00 nC) 
7


  3.60  10 J
(0.200 m)
(0.100 m)
(0.100 m)


EXECUTE: (a) U  k 
 q1q2 q1q3
qq 

 2 3  . Solving for x we find:
x
r12  x 
 r12
(b) If U  0 , then 0  k 
0   60 
8
6

 60 x 2  26 x  1.6  0  x  0.074 m, 0.360 m. Therefore, x  0.074 m since it is the only value
x 0.2  x
between the two charges.
EVALUATE: U13 is positive and both U 23 and U12 are negative. If U  0 , then U13  U 23  U12 . For x  0.074 m ,
U13  9.7  107 J , U 23  4.3  107 J and U12  5.4  107 J. It is true that U  0 at this x.
Prob. 23.15
B
q1
A
q1 = -5uC
mass m = 2 x 10-4 kg
velocity at point A: vA = 5 m/s
VA = + 200 volts
VB = +800 volts
What is the speed at point B?
Is it moving faster at point B than A? Explain?
IDENTIFY and SET UP: Apply conservation of energy to points A and B.
EXECUTE: K A  U A  K B  U B
U  qV , so K A  qVA  K B  qVB
K B  K A  q(VA  VB )  0.00250 J  (5.00  106 C)(200 V  800 V)  0.00550 J
vB  2K B /m  7.42 m/s
EVALUATE: It is faster at B; a negative charge gains speed when it moves to higher potential.
Prob. 23.21
Charges: q1 = +2.40 nC
E = 0 at infinity
q2 = -6.50 nC
(a) V = ? at point A
(b) V = ? at point B
(c) Work on a +2.50 nC charge moved from point B to A ?
IDENTIFY:
V
1
q
 i
4 P0 i ri
SET UP: The locations of the changes and points A and B are sketched in Figure 23.21.
Figure 23.21
EXECUTE: (a) VA 
1  q1 q2 
 

4 P0  rA1 rA 2 
 2.40  109 C 6.50  109 C 
VA  (8.988  109 N  m 2 /C2 ) 

  737 V
0.050 m 
 0.050 m
(b) VB 
1  q1 q2 



4 P0  rB1 rB 2 
 2.40  109 C 6.50  109 C 
VB  (8.988  109 N  m 2 /C2 ) 

  704 V
0.060 m 
 0.080 m
(c) IDENTIFY and SET UP: Use Eq.(23.13) and the results of parts (a) and (b) to calculate W.
EXECUTE: WB  A  q(VB  VA )  (2.50  109 C)(704 V  (737 V))  8.2  108 J
EVALUATE: The electric force does positive work on the positive charge when it moves from higher potential (point
B) to lower potential (point A).
Prob. 23.33
Ring radius = 15.0 cm
Q = + 24.0 nC
An electron is placed and constrained at x = 30.0 cm (P);
The electron is then released.
(a)What is the motion of the electron?
(b) What is the speed of the electron when it reaches the center of the ring?
.
(a) IDENTIFY and SET UP: The electric field on the ring’s axis is calculated in Example 21.10. The force on the electron
exerted by this field is given by Eq.(21.3).
EXECUTE: When the electron is on either side of the center of the ring, the ring exerts an attractive force directed
toward the center of the ring. This restoring force produces oscillatory motion of the electron along the axis of the
ring, with amplitude 30.0 cm. The force on the electron is not of the form F  kx so the oscillatory motion is not
simple harmonic motion.
(b) IDENTIFY: Apply conservation of energy to the motion of the electron.
SET UP: Ka  U a  Kb  Ub with a at the initial position of the electron and b at the center of the ring. From Example
23.11, V 
EXECUTE:
1
4 P0
Q
, where R is the radius of the ring.
x  R2
xa  30.0 cm, xb  0.
2
K a  0 (released from rest), K b  12 mv 2
Thus 12 mv 2  U a  U b
And U  qV  eV so v 
2e(Vb  Va )
.
m
Va 
1
Q
24.0 109 C
 (8.988 109 N  m2 / C2 )
2
2
4 P0 xa  R
(0.300 m)2  (0.150 m)2
Va  643 V
Vb 
v
1
Q
24.0 109 C
 (8.988 109 N  m2 / C2 )
 1438 V
4 P0 xb2  R 2
0.150 m
2e(Vb  Va )
2(1.602 1019 C)(1438 V  643 V)

 1.67 107 m/s
m
9.109 1031 kg
EVALUATE: The positively charged ring attracts the negatively charged electron and accelerates it. The electron has
its maximum speed at this point. When the electron moves past the center of the ring the force on it is opposite to its
motion and it slows down.
Prob. 23.41
++++++++++++++++++++
360 V
45.0 mm
Two Charged Plates
-----------------(a)
(b)
(c)
(d)
E=?
Force on + 2.40 nC charge?
Work from higher potential plate to lower plate?
Use electric potential to compute work?
IDENTIFY and SET UP: Use the result of Example 23.9 to relate the electric field between the plates to the potential
difference between them and their separation. The force this field exerts on the particle is given by Eq.(21.3). Use the
equation that precedes Eq.(23.17) to calculate the work.
V
360 V
EXECUTE: (a) From Example 23.9, E  ab 
 8000 V/m
d 0.0450 m
(b) F  q E  (2.40 109 C)(8000 V/m)  1.92 105 N
(c) The electric field between the plates is shown in Figure 23.41.
Figure 23.41
The plate with positive charge (plate a) is at higher potential. The electric field is directed from high potential toward
low potential (or, E is from + charge toward  charge), so E points from a to b. Hence the force that E exerts on the
positive charge is from a to b, so it does positive work.
b
W   F  dl  Fd , where d is the separation between the plates.
a
W  Fd  (1.92  105 N)(0.0450 m)  8.64  107 J
(d) Va  Vb  360 V (plate a is at higher potential)
U  U b  U a  q (Vb  Va )  (2.40  109 C)(360 V)  8.64  107 J.
EVALUATE: We see that Wa b  (Ub  U a )  U a  Ub .
Prob. 23.63
Electron at initial velocity: V0 = 6.50 x 106 m/s
Plates electric field: E =1.10 x 103 V/m
(a) What is the force on the electron between the plates?
(b) What is the acceleration from part (a)?
(c) What is the vertical drop at the end of the plates?
(d) What is the angle at the end of the plates?
(e) What is the vertical position at S?
IDENTIFY and SET UP: Use Eq.(21.3) to calculate F and then F = ma gives a.
EXECUTE: (a) FE  qE. Since q  e is negative FE and E are in opposite directions; E is upward so FE is
downward. The magnitude of FE is
FE  q E  eE  (1.602 1019 C)(1.10 103 N/C)  1.76 1016 N.
(b) Calculate the acceleration of the electron produced by the electric force:
a
F
1.76 1016 N

 1.93 1014 m/s2
m 9.109 1031 kg
EVALUATE: This is much larger than g  9.80 m/s 2 , so the gravity force on the electron can be neglected. FE is
downward, so a is downward.
(c) IDENTIFY and SET UP: The acceleration is constant and downward, so the motion is like that of a projectile. Use
the horizontal motion to find the time and then use the time to find the vertical displacement.
EXECUTE: x-component
v0 x  6.50  106 m/s; ax  0; x  x0  0.060 m; t  ?
x  x0  v0 xt  12 axt 2 and the ax term is zero, so
x  x0
0.060 m

 9.231 109 s
v0 x
6.50  106 m/s
y-component
v0 y  0; ay  1.93 1014 m/s2 ; t  9.231109 m/s; y  y0  ?
t
y  y0  v0 yt  12 ayt 2
y  y0  12 (1.93  1014 m/s 2 )(9.231  109 s) 2  0.00822 m  0.822 cm
(d) The velocity and its components as the electron leaves the plates are sketched in Figure 23.63.
vx  v0 x  6.50  106 m/s (since ax  0 )
v y  v0 y  a y t
vy  0  (1.93 1014 m/s2 )(9.231109 s)
vy  1.782 106 m/s
Figure 23.63
1.782 106 m/s
 0.2742 so   15.3.
vx 6.50 106 m/s
EVALUATE: The greater the electric field or the smaller the initial speed the greater the downward deflection.
(e) IDENTIFY and SET UP: Consider the motion of the electron after it leaves the region between the plates. Outside
the plates there is no electric field, so a  0. (Gravity can still be neglected since the electron is traveling at such high
speed and the times are small.) Use the horizontal motion to find the time it takes the electron to travel 0.120 m
horizontally to the screen. From this time find the distance downward that the electron travels.
EXECUTE: x-component
v0 x  6.50  106 m/s; ax  0; x  x0  0.120 m; t  ?
tan  
vy

x  x0  v0 xt  12 axt 2 and the ax term is term is zero, so
x  x0
0.120 m

 1.846  108 s
v0 x
6.50  106 m/s
y-component
v0 y  1.782 106 m/s (from part (b)); a y  0; t  1.846 108 m/s; y  y0  ?
t
y  y0  v0 yt  12 ayt 2  (1.782 106 m/s)(1.846 108 s)  0.0329 m  3.29 cm
EVALUATE: The electron travels downward a distance 0.822 cm while it is between the plates and a distance 3.29 cm
while traveling from the edge of the plates to the screen. The total downward deflection is
0.822 cm  3.29 cm  4.11 cm.
The horizontal distance between the plates is half the horizontal distance the electron travels after it leaves the plates.
And the vertical velocity of the electron increases as it travels between the plates, so it makes sense for it to have
greater downward displacement during the motion after it leaves the plates.