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Physics II
Homework VI
CJ
Chapter 25; 6, 14, 25, 41, 54, 61, 68, 75
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)
25.14. Model: Charges A, B, and C are point charges.
Visualize: Please refer to Figure Ex25.14. Charge A experiences an electric force FB on A due to charge B and
an electric force FC on A due to charge C. The force FB on A is directed to the right and the force FC on A is directed to the left.
Solve: Coulomb’s law yields:
 9  10 N m /C 1  10 C1  10
1  10 m 
9
FB on A 
FC on A 
2
9
2
2
2
 9  10
9

9
C
  9.0  10

  9.0  10
 
 
N m2 /C2 1  10 9 C 4  10 9 C
 2  10
2
m

5
2
5
N
N
The net force on A is

Fon A  FB on A  FC on A  9.0  105 N iˆ  9.0  105 N iˆ  0 N
25.25. Model: A field is the agent that exerts an electric force on a charge.
Visualize:
 
Solve: Newton’s second law on the plastic ball is  Fnet
y
 Fon q  w . To balance the weight with the electric force,


1.0  103 kg  9.8 N/kg 
mg

 3.27  106 N/C
q
3.0  109 C
Fon q  w  q E  mg  E 
Because Fon q must be upward and the charge is negative, the electric field at the location of the plastic ball must be
pointing downward. Thus E  3.27  106 N/C, downward .

Assess:

F  qE means the sign of the charge q determines the direction of F or E . For positive q, E and F are pointing
in the same direction. But E and F point in opposite directions when q is negative.
25.41. Model: The charges are point charges.
Visualize:
Solve:
The electric force on charge q1 is the vector sum of the forces F2 on 1 and F3 on 1 . We have
 K q1 q2

F2 on 1  
, toward q2 
 r2







 9.0  10 9 N m 2 /C2 10  10 9 C 5  10 9 C



,
toward
q
2
2

2

1.0  10 m


3
3 ˆ
 4.5  10 N, toward q  4.5  10 j N



2

 K q1 q3

F3 on 1  
, toward q3 
2
 r







 9.0  10 9 N m 2 /C2 10  10 9 C 15  10 9 C



,
toward
q
3
2


3.0  10 2 m


 1.5  10 3 N, toward q  1.5  10 3iˆ N


3




 Fon 1  F2 on 1  F3 on 1  1.5  103 iˆ  4.5  103 ˆj N
The magnitude and direction of the resultant force vector are
Fon 1 
tan 
 1.5  10
3
 
2
N  4.5  103 N

2
 4.74  103 N
4.5  103 N
   tan1  3  71.6 above the  x axis
1.5  103 N
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

2
L
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 3-charge system to be in static equilibrium.
25.61. Model: The electric field is that of a positive point charge located at the origin.
Visualize: Please refer to Figure P25.61. Place the 10 nC charge at the origin.
Solve:
The electric field is



9
2
2
9

 1 q
  9.0  10 N m /C 10  10 C

E
,
away
from
q

, away from q 

2
2

r
 4 0 r
 

 90.0 N m2 /C


, away from q 
2
r


At each of the three points,

E1  



E2  



  1.0  10 5 ˆj N/C
,
away
from
q
2

2
3.0  10 m


90.0 N m 2 /C
  3.6  10 4 N/C cos iˆ  sin ˆj
,
away
from
q
2

5.0  10 2 m

4
3 ˆ
4 ˆ
 3.6  10 N/C 5 i  5 j  2.88  10 4iˆ  2.16  10 4 ˆj N/C

90.0 N m 2 /C







 




90.0 N m 2 /C
  5.63  10 4iˆ N/C
E3  
,
away
from
q
2


2
4.0  10 m




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
25.75. Model: The charged spheres are point charges.
Visualize:
The figure shows the free-body diagram of the forces on the sphere with the negative
charge that is shown in Figure CP25.75. The force FE is due to the external electric field.
The force Fe is the attractive force between the positive and the negative spheres. The
tension in the string and the weight are the remaining two forces on the spheres.
Solve: The two electrical forces are calculated as follows:



FE  q E  100  10 9 C 105 N/C  1.0  10 2 N
Fe 
K q1 q2
r
2
 9.0  10

9

N m2 /C2 100  10 9 C
r
2

2

9.0  10 5 N m2 /C
r2
From the geometry of Figure CP25.75,
r  2 L sin10  2 0.50 m sin10  0.1736 m  Fe 
9.0  105 N m2 /C
 0.1736 m 
From the free-body diagram,
T cos10  mg
T sin10  Fe  FE
Rearranging and dividing the two equations,
F  Fe
sin10
 tan10  E
cos10
mg
m
FE  Fe
1.0  102 N  0.2986  102 N

 0.406  102 kg  4.06 g
g tan10
 9.8 N/kg tan10
2
 2.986  103 N