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CHAPTER 18 ELECTRIC FORCES AND
ELECTRIC FIELDS
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
1. 1.9 × 10^13
- (d) The fact that the positive rod repels one object indicates that that object carries a net positive charge. The fact that the rod repels the other object indicates that that object carries a net negative charge. Since both objects are identical and made from conducting material, they share the combined net charges equally after they are touched together. Since the rod repels each object after they are touched, each object must then carry a net positive charge. But the net electric charge of any isolated system is conserved, so the total net charge initially must also have been positive. This means that the initial positive charge had the greater magnitude.
- (e) Coulomb’s law states that the magnitude of the force is given by 1 22
q q F k r
=. The force
is directed along the line between the charges and is an attraction for unlike charges and a repulsion for like charges. Charge B is attracted by charge A with a force of magnitude
2
q q k d
and repelled by charge C with a force of the same magnitude. Since both forces
point to the left, the net force acting on B has a magnitude of (^2 )
q q k d
. Charge A is attracted
by charge B with a force of (^2)
q q k d
and also by charge C with a force of ( )
2 2
q q k d
. Since both
forces point to the right, the net force acting on A has a magnitude of (^) (1.25 ) (^2)
q q k d
. Charge
C is pushed to the right by B with a force of (^2)
q q k d
and pulled to the left by A with a force
of ( )
2 2
q q k d
. Since these two forces have different directions, the net force acting on C has a
magnitude of (^) ( 0.75) (^2)
q q k d
Chapter 18 Answers to Focus on Concepts Questions 943
- (b) According to Coulomb’s law, the magnitude of the force that any one of the point charges
exerts on another point charge is given by (^2)
q q F k d
= , where d is the length of each side of
the triangle. The charge at B experiences a repulsive force from the charge at A and an attractive force from the charge at C. Both forces have vertical components, but one points in the + y direction and the other in the − y direction. These vertical components have equal magnitudes and cancel, leaving a resultant that is parallel to the x axis.
- (c) The electric field created by a point charge has a magnitude (^2)
k q E r
= and is inversely
proportional to the square of the distance r. If r doubles, the charge magnitude must increase by a factor of 2^2 = 4 to keep the field the same.
- (a) The electric field has a greater magnitude where the field lines are closer together. They are closest together at B and farthest apart at A. Therefore, the field has the greatest magnitude at B and the smallest magnitude at A.
- (d) In a conductor electric charges can readily move in response to an electric field. In A, B, and C the electric charges experience an electric field and, hence, a force from neighboring charges and will move outward, away from each other. They will rearrange themselves so that the electric field within the metal is zero at equilibrium. This means that they will reside on the outermost surface. Thus, only D could represent charges in equilibrium.
Chapter 18 Problems 945
2 F = k q 1 (^) q 2 (^) / r (see Coulomb's law, Equation 18.1) pulling to the right, and the tension T
in the wire pulling up and to the left at an angle θ with respect to the vertical as shown in the
drawing in the problem statement. We can analyze the forces to determine the desired
quantities θ and T.
SOLUTION.
a. We can see from the diagram given with the problem statement that
Tx = F which gives T sin θ = k q 1 (^) q 2 / r^2
g
and
T y = W which gives T cos θ = m
Dividing the first equation by the second yields
2 sin (^) tan 1 2 / cos
T^ k q^ q^ r T m
θ θ θ
g
Solving for θ, we find that
–1^1 2
9 2 2 –6 –
tan
(8.99 10 N m /C )(0.600 10 C)(0.900 10 C) tan 15. (8.00 10 kg)(9.80 m/s )(0.150 m)
k q q mgr
θ
⎡ × ⋅ × × ⎤
⎣ × ⎦
b. Since T cos^ θ^ =^^ mg , the tension can be obtained as follows:
(8.00 10 2 kg) (9.80 m/s )^2 0.813 N cos cos 15.
mg T
× −
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35. REASONING AND SOLUTION
a. In order for the field to be zero, the point cannot be between the two charges. Instead, it must be located on the line between the two charges on the side of the positive charge and away from the negative charge. If x is the distance from the positive charge to the point in question, then the negative charge is at a distance (3.0 m + x ) meters from this point. For the field to be zero here we have
( ) ( )
3.0 m 2 2 or 3.0 m^2
k q k q q q x x^ x x^2
946 ELECTRIC FORCES AND ELECTRIC FIELDS
Solving for the ratio of the charge magnitudes gives
2 2 2 2
16.0 C 3.0 m^ 3.0 m or 4. 4.0 C
q (^) x x q (^) x x
− = = +^ = +
Suppressing the units for convenience and rearranging this result gives
2 2 2 2 2 4.0 x = 3.0 + x or 4.0 x = 9.0 + 6.0 x + x or 3 x − 6.0 x − 9.0 = 0
Solving this quadratic equation for x with the aid of the quadratic formula (see Appendix C.4) shows that x = 3.0 m or x = −1.0 m
We choose the positive value for x , since the negative value would locate the zero-field spot between the two charges, where it can not be (see above). Thus, we have x = 3.0 m.
b. Since the field is zero at this point, the force acting on a charge at that point is 0 N.
- SSM REASONING Since the charged droplet (charge = q ) is suspended motionless in
the electric field E , the net force on the droplet must be zero. There are two forces that act on the droplet, the force of gravity W = m g , and the electric force F = q E due to the electric field. Since the net force on the droplet is zero, we conclude that mg = q E. We can use this reasoning to determine the sign and the magnitude of the charge on the droplet.
SOLUTION a. Since the net force on the droplet is zero, and the weight of magnitude W points downward, the electric force of magnitude F = q E must point upward. Since the electric field points upward, the excess charge on the droplet must be positive in order for the force F to point upward. m g
F
b. Using the expression mg = q E , we find that the magnitude of the excess charge on the droplet is –9 2 (3.50 10 kg)(9.80 m/s ) (^) 4.04 10 –12 C 8480 N/C
mg q E
×
= = = ×
The charge on a proton is 1.60 × 10 –19^ C, so the excess number of protons is
–12 7
1 proton 4.04 10 C 2.53 10 protons 1.60 10 C
× ⎛^ ⎞= ×
⎝ × ⎠
948 ELECTRIC FORCES AND ELECTRIC FIELDS
b. Φ (^) E =
−2.3 × 10 –^6 C
8.85 × 10 −^12 C 2 /(N ⋅ m 2 )
= –2.6 × 10 5 N ⋅ m 2 /C
c. Φ (^) E =
(3.5 × 10 –^6 C) + (−2.3 × 10 –^6 C)
8.85 × 10 −^12 C 2 /(N ⋅ m 2 )
= 1.4 × 10 5 N ⋅ m 2 /C
- REASONING AND SOLUTION Since the electric field is uniform, its magnitude and direction are the same at each point on the wall. The angle φ between the electric field and the normal to the wall is 35°. Therefore, the electric flux is
ΦE = ( E cos φ) A = (150 N/C)(cos 35°)[(5.9 m)(2.5 m)] = 1.8 × 10 3 N ⋅ m 2 /C
- SSM REASONING AND SOLUTION Before the spheres have been charged, they exert
no forces on each other. After the spheres are charged, each sphere experiences a repulsive force F due to the charge on the other sphere, according to Coulomb's law (Equation 18.1). Therefore, since each sphere has the same charge, the magnitude F of this force is
9 2 2 6 2 1 2 2 2
(8.99 10 N m /C )(1.60 10 C) 2.30 N (0.100 m)
k q q F r
× ⋅ ×^ −
The repulsive force on each sphere compresses the spring to which it is attached. The magnitude of this repulsive force is related to the amount of compression by Equation 10.1: F x Applied= kx. Setting Fx Appled^ = F and solving for k , we find that
2.30 N 92.0 N/m 0.0250 m
F
k x
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- REASONING AND SOLUTION In order for the net force on any charge to be directed inward toward the center of the square, the charges must be placed with alternate + and – signs on each successive corner. The magnitude of the force on any charge due to an adjacent charge located at a distance r is
( )( ) ( )
2 9 2 2 6 2 2 2
8.99 10 N m / C 2.0 10 C 0.40 N 0.30 m
k q F r
× ⋅ ×^ −
The forces due to two adjacent charges are perpendicular to one another and produce a resultant force that has a magnitude of
Chapter 18 Problems 949
( ) 2 2 F adjacent (^) = 2 F = 2 0.40 N =0.57 N
The magnitude of the force due to the diagonal charge that is located at a distance of r 2 is
( )
diagonal
2 2 2 2
0.40 N
0.20 N
k q k q F r r
since the diagonal distance is r 2. The force F diagonal is directed opposite to F adjacent (since the diagonal charges are of the same sign). Therefore, the net force acting on any of the charges is directed inward and has a magnitude
F net = F adjacent – F diagonal = 0.57 N – 0.20 N = 0.37 N
