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Physics 122
Solutions to Chapter 25
25.1. Model: Use the charge model.
Solve: (a) In the process of charging by rubbing, electrons are removed from one material and transferred to
the other because they are relatively free to move. Protons, on the other hand, are tightly bound in nuclei. So,
electrons have been removed from the glass rod to make it positively charged.
(b) Because each electron has a charge of 1.60  1019 C , the number of electrons removed is
5  109 C
 3.13  1010
1.60  10 19 C
25.6. Model: Use the charge model.
Solve: (a) No, we cannot conclude that the wall is charged. Attractive electric forces occur between (i) two
opposite charges, or (ii) a charge and a neutral object that is polarized by the charge. Rubbing the balloon does
charge the balloon. Since the balloon is rubber, its charge is negative. As the balloon is brought near the wall, the
wall becomes polarized. The positive side of the wall is closer to the balloon than the negative side, so there is a
net attractive electric force between the wall and the balloon. This causes the balloon to stick to the wall, with a
normal force balancing the attractive electric force and an upward frictional force balancing the very small
weight of the balloon.
(b)
25.11. Model: Model the charged masses as point charges.
Visualize:
Solve: (a) The charge q1 exerts a force F1 on 2 on q2 to the right, and the charge q2 exerts a force F2 on 1 on q1 to
the left. Using Coulomb’s law,
F1 on 2  F2 on 1 
K q1 q2
2
12
r

 9.0  10
9

N m2 /C2 1.0 C 1.0 C
1.0 m 
2
 9.0  109 N
(b) Newton’s second law on either q1 or q2 is
F1 on 2  m1a1  a1 
9.0  109 N
 9.0  109 m/s2
1.0 kg
Assess: Coulomb is a pretty “big” unit. That is why F1 on 2  or F2 on 1  is such a large force.
25.20. Model: The gravitational field of an object depends on its mass and extends through all of space.
Solve: (a) The gravitational field strength due to a planet at the radius of its satellite’s orbit is 12 N/kg. That is,
 G mplanet

gorbit  
, toward planet   12 N/kg, toward planet 
2
 rorbit

When the radius of the orbit is doubled,
G m
 1Gm

planet
planet

gnew orbit  
,
toward
planet
, toward planet    3 N/kg, toward planet 
2
2
  2rorbit 
  4 rorbit



(b) When the planet’s density is doubled, then mnew  newV  2V  2mplanet. Thus, assuming that V remains the
same,
 G2mplanet

gorbit  
, toward planet    24 N/kg, toward planet 
2
 rorbit

(c) gorbit does not depend on the satellite’s mass. Thus, gorbit  12 N/kg, toward planet  .
25.26. Model: A field is the agent that exerts an electric force on a charge.
Visualize:
Solve: (a) To balance the weight of a proton   Fnet  y  Fon p  w  0 N. This means
Fon p  w  q E  mg  E 


1.67  1027 kg  9.8 N/kg
mg

 1.02  107 N/C
q
1.60  1019 C
Because Fon p must be upward and the proton charge is positive, the electric field at the location of the proton
must also be pointing upward. Thus E  1.02  10 7 N/C, downward .

(b) In the case of the electron,



9.11  1031 kg  9.8 N/kg
mg
E

 5.58  1011 N/C
q
1.60  1019 C
Because Fon e must be upward and the electron has a negative charge, the electric field at the location of the
electron must be pointing downward. Thus E  5.58  10 11 N/C, downward .


25.54. Model: The charges are point charges.
Visualize:
We must first identify the region of space where the third charge q3 is located. You can see from the figure that
the forces can’t possibly add to zero if q3 is above or below the axis or outside the charges. However, at some
point on the x-axis between the two charges the forces from the two charges will be oppositely directed.
Solve: The mathematical problem is to find the position for which the forces F1 on 3 and F2 on 3 are equal in
magnitude. If q3 is the distance x from q1, it is the distance L  x from q2. The magnitudes of the forces are
F1 on 3 
K q1 q3
2
13
r

Kq q3
x
F2 on 3 
2
K q2 q3
2
23
r

K  4q  q3
L  x
2
Equating the two forces,
Kq q3
x
2

K  4q  q3
L  x
2
  L  x   4x 2  x 
2
L
and  L
3
The solution x  L is not allowed as you can see from the figure. To find the magnitude of the charge q3, we
apply the equilibrium condition to charge q1:
F2 on 1  F3 on 1 
K q2 q1
L2

K q3 q1
 L
1
3
2
 4q  9 q3  q3 
4
q
9
We are now able to check the static equilibrium condition for the charge 4q (or q2):
F1 on 2  F3 on 2  K
q1 q2
2
L

K q3 q2
L  x

2
q

L2
4
9
q
 L
2
3
2

q
L2
The sign of the third charge q3 must be negative. A positive sign on q3 will not have a net force of zero either on
the charge q or the charge 4q. In summary, a charge of  94 q placed x  13 L from the charge q will cause the 3charge system to be in static equilibrium.
25.58. Model: The charged plastic beads are point charges and the spring is an ideal spring that obeys Hooke’s
law.
Solve: Let q be the charge on each plastic bead. The repulsive force between the beads pushes the beads apart.
The spring is stretched until the restoring spring force on either bead is equal to the repulsive Coulomb force. That
is,
k x r 2
K
Kq2
 k x  q 
r2
The spring constant k is obtained by noting that the weight of a 1.0 g mass stretches the spring 1.0 cm. Thus
mg  k(1.0  102 m)  k 
q
1.0  10
3
1.0  10 2 m
 0.98 N/m   4.5  102 m  4.0  102 m  4.5  102 m 
9.0  109 N m2 /C2

kg  9.8 N/kg 
 0.98 N/m
2
 33.2 nC
25.59. Model: The charged spheres are point charges.
Visualize:
Each sphere is in static equilibrium and the string makes an angle  with the vertical. The three forces acting on
each sphere are the electric force, the weight of the sphere, and the tension force.
Solve: In static equilibrium, Newton’s first law is Fnet  T  w  Fe  0 . In component form,
 Fnet x  Tx  wx   Fe x  0 N
 T sin  0 N 
 T sin 
 Fnet y  Ty  wy   Fe y  0 N
Kq2
0N
d2
T cos  mg  0 N  0 N
Kq2
Kq2

2
2
d
 2L sin 
T cos  mg
Dividing the two equations,



2
9.0  10 9 N m 2 /C2 100  10 9 C
Kq2
sin  tan  2

 4.59  10 4
4L mg 4 1.0 m 2 5.0  10 3 kg  9.8 N/kg 
2


For small-angles, tan  sin . With this approximation we obtain sin  0.07714 rad and   4.42.
25.63. Model: The electric field is that of a positive point charge located at the origin.
Visualize: Please refer to Figure P25.63. Place the 5 nC charge at the origin.
Solve: The electric field is



9
2
2
9

 1 q
  9  10 N m / C 5  10 C

E 
,
away
from
q

, away from q 

2
2

r
 4 0 r
 

 45.0 N m 2 /C


, away from q 
2
r


At each of the three points,


45.0 N m 2 /C
  9.0  10 4 N/C cos iˆ  sin ˆj
E1  
,
away
from
q
2
2


2.0  10 2 m  1.0  10 2 m



 
 1 ˆ 2
 9.0  10 4 N/C 
i
5
 5






ˆj   4.02  10 4 iˆ  8.05  10 4 ˆj N/C






45.0 N m2 /C
  4.5  105iˆ N/C
E2  
,
away
from
q
2


2
1.0  10 m




45.0 N m2 /C
  4.02  104 iˆ  8.05  104 ˆj N/C
E3  
,
away
from
q
2
2


2.0  10 2 m  1.0  10 2 m





 



25.66. Model: The electric field is that of three point charges.
Visualize:
Solve: (a) In the figure, the distances are r1  r3 
tan
1
1 cm    3 cm 
2
1/ 3  18.43 . Using the equation for the field of a point charge,
2
 3.162 cm and the angle is  
E1  E3 
K q1
2
1
r

 9.0  10
9

N m2 /C2 1.0  10 9 C
 0.03162 m 
2

9000 N/C
We now use the angle  to find the components of the field vectors:

 N/C
E1  E1 cos iˆ  E1 sin ˆj  8540iˆ  2840 ˆj

E3  E3 cos iˆ  E3 sin ˆj  8540iˆ  2840 ˆj
 N/C
E2 is easier since it has only an x-component. Its magnitude is
E2 
K q2
r22

 9.0  10
9

N m2 /C2 1.0  10 9 C
 0.0300 m 
  10,000 N/C  E
2
2
 E2iˆ  10,000iˆ N/C
(b) The electric field is defined in terms of an electric force acting on charge q: E  F q . Since forces obey a
principle of superposition ( Fnet  F1  F2 
principle of superposition.
) it follows that the electric field due to several charges also obeys a
(c) The net electric field at a point 3 cm to the right of q2 is Enet  E1  E2  E3  27,100iˆ N/C . The y-components
of E1 and E2 cancel, giving a net field pointing along the x-axis.
25.68. Model: The charged ball attached to the string is a point charge.
Visualize:
The ball is in static equilibrium in the external electric field when the string makes an angle   20 with the
vertical. The three forces acting on the charged ball are the electric force due to the field, the weight of the ball,
and the tension force.
Solve: In static equilibrium, Newton’s second law for the ball is Fnet  T  w  Fe  0 . In component form,
 Fnet x  Tx  0 N  qE  0 N
 Fnet y  Ty  mg  0 N  0 N
The above two equations simplify to
T cos  mg
T sin  qE
Dividing both equations, we get
tan 


5.0  103 kg  9.8 N/kg tan20
qE
mg tan
q

 1.78  107 C  178 nC
mg
E
100,000 N/C
25.76. Solve: (a) Kinetic energy is K  21 mv2 , so the velocity squared is v2  2K/m. From kinematics, a
particle moving through distance x with acceleration a, starting from rest, finishes with v2  2ax. To gain
K  2  1018 J of kinetic energy in x  2.0 m requires an acceleration
a
v2
2K / m
K
2.0  10 18 J



 1.10  1018 m/s2
2x
2x
mx
9.11  10 31 kg 2.0  10 6 m



(b) The force that produces this acceleration is



F  ma  9.11  10 31 kg 1.10  1018 m/s2  1.0  10 12 N
(c) The electric field is
E
F 1.0  1012 N

 6.25  106 N/C
e 1.6  1019 C
(d) The force on an electron due to charge q is F  K q e r 2 . To have a breakdown, the force on the electron
must be at least 1.0  1012 N. The minimum charge that could cause a breakdown will be the charge that causes
exactly a force of 1.0  1012 N:
F 
Kqe
r2
 1.0  10


 0.01 m  1.0  10 12 N
r2F
N q 

 6.9  10 8 C  68 nC
9
2
2
19
Ke
9.0  10 N m / C 1.6  10 C
2
12


