INSTRUCTOR’S SOLUTIONS MANUAL FOR_SERWAY AND VUILLE’S_COLLEGE PHYSICS NINTH EDITION, VOLUME 2

Choice (b). Object A must have a net charge because two neutral objects do not attract each other.
Since object A is attracted to positively-charged object B, the net charge on A must be negative.
2. Choice (b). By Newton’s third law, the two objects will exert forces having equal magnitudes but
opposite directions on each other.
3. Choice (c). The electric fi eld at point P is due to charges other than the test charge. Thus, it is
unchanged when the test charge is altered. However, the direction of the force this fi eld exerts on
the test change is reversed when the sign of the test charge is changed.
4. Choice (a). If a test charge is at the center of the ring, the force exerted on the test charge by
charge on any small segment of the ring will be balanced by the force exerted by charge on the
diametrically opposite segment of the ring. The net force on the test charge, and hence the electric
fi eld at this location, must then be zero.
5. Choices (c) and (d). The electron and the proton have equal magnitude charges of opposite signs.
The forces exerted on these particles by the electric fi eld have equal magnitude and opposite
directions. The electron experiences an acceleration of greater magnitude than does the proton
because the electron’s mass is much smaller than that of the proton.
6. Choice (a). The fi eld is greatest at point A because this is where the fi eld lines are closest together.
The absence of lines at point C indicates that the electric fi eld there is zero.
7. Choice (c). When a plane area A is in a uniform electric fi eld E, the fl ux through that
area is ΦE = EAcosq , where q is the angle the electric fi eld makes with the line normal
to the plane of A. If A lies in the xy-plane and E is in the z-direction, then q = 0° and
ΦE = EA = (5.00 N C) 4.00 m( 2 ) = 20.0 N⋅m2 C.
8. Choice (b). If q = 60° in Quick Quiz 15.7 above, then ΦE = EAcosq which yields
ΦE = (5.00 N C) 4.00 m( 2 )cos(60°) = 10.0 N⋅m2 C.
9. Choice (d). Gauss’s law states that the electric fl ux through any closed surface is equal to the net
enclosed charge divided by the permittivity of free space. For the surface shown in Figure 15.28,
the net enclosed charge is Q = −6 C, which gives ΦE = Q ∈0 = −(6 C) ∈0.
10. Choices (b) and (d). Since the net fl ux through the surface is zero, Gauss’s law says that the net
change enclosed by that surface must be zero as stated in (b). Statement (d) must be true because
there would be a net fl ux through the surface if more lines entered the surface than left it (or
vise-versa).
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2 Chapter 15
ANSWERS TO MULTIPLE CHOICE QUESTIONS
1. To balance the weight of the ball, the magnitude of the upward electric force must equal the magnitude
of the downward gravitation force, or qE = mg, which gives
E = mg
q
=
(5.0 × 10−3 kg) 9.80 m s( 2 )
4.0 × 10−6 C
= 1.2 × 104 N C
and the correct choice is (b).
2. The magnitude of the electric fi eld at distance r from a point charge q is E = keq r2, so
E =
8.99 ×109 N ⋅m2 C( 2 ) 1.60 ×10−19 ( C)
5.29 ×10−11 ( m)2 = 5.14 ×1011 N C∼ 1012 N C
making (e) the best choice for this question.
3. The magnitude of the electric force between two protons separated by distance r is F = kee2 r2,
so the distance of separation must be
r = kee2
F
=
8.99 × 109 N ⋅m2 C( 2 ) 1.60 × 10−19 ( C)2
2.3 × 10−26 N
= 0.10 m
and (a) is the correct choice.
4. The ball is made of a metal, so free charges within the ball will very quickly rearrange themselves
to produce electrostatic equilibrium at all points within the ball. As soon as electrostatic equilibrium
exists inside the ball, the electric fi eld is zero at all points within the ball. Thus, the correct
choice is (c).
5. Choosing the surface of the box as the closed surface of interest and applying Gauss’s law, the net
electric fl ux through the surface of the box is found to be
ΦE = Qinside
∈0
= (3.0 − 2.0 − 7.0 +1.0)×10−9 C
8.85 ×10−12 C2 N⋅m2 = −5.6 ×102 N ⋅m2 C
meaning that (b) is the correct choice.
6. From Newton’s second law, the acceleration of the electron will be
ax = Fx
m
= qEx
m
=
−1.60 × 10−19 ( C) 1.00 × 103 ( N C)
9.11× 10−31 kg
= −1.76 × 1014 m s2
The kinematics equation vx
2 = v0 x
2 + 2ax (Δx), with vx = 0, gives the stopping distance as
Δx =
−v0 x
2
2ax
=
− 3.00 ×106 ( m s)2
2 −1.76 ×1014 m s( 2 ) = 2.56 ×10−2 m = 2.56 cm
so (a) is the correct response for this question.
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