Problem Solutions
15.1
Since the charges have opposite signs, the force is one of attraction .
Its magnitude is
F=
15.10
(
r
)(
)
−9
−9

N ⋅ m 2  4.5 × 10 C 2 .8 × 10 C
=  8.99 × 10 9
= 1.1 × 10−8 N

2
2
C
( 3.2 m )


k e q1 q2
2
The forces are as shown in the sketch
at the right.
F1 =
F2 =
F3 =
k e q1 q2
r122
2
6.00 × 10−6 C) ( 1.50× 10−6 C)

9 N⋅m (
=  8.99 × 10
= 89.9 N

2
C2 

( 3.00 × 10-2 m)
k e q1 q3
r132
k e q2 q3
r232
2
6.00 × 10−6 C) ( 2 .00× 10−6 C)

9 N⋅m (
=  8.99 × 10
= 43.2 N

2
C2 

( 5.00 × 10-2 m)
−6
−6

N ⋅ m 2  ( 1.50 × 10 C) ( 2 .00× 10 C)
=  8.99 × 109
= 67.4 N

2
C2 

( 2 .00 × 10-2 m)
The net force on the 6 µ C charge is F6 = F1 − F2 = 46.7 N (to the left)
The net force on the 1.5 µ C charge is F1.5 = F1 + F3 = 157 N (to the right)
The net force on the −2 µ C charge is F−2 = F2 + F3 = 111 N (to the left)
1
2 CHAPTER 15
15.11
In the sketch at the right, FR is the resultant of the
forces F6 and F3 that are exerted on the charge at
the origin by the 6.00 nC and the –3.00 nC charges
respectively.
2
6.00 × 10−9 C) ( 5.00× 10−9 C)

9 N⋅m  (
F6 =  8.99 × 10

2
C2 
( 0.300 m)

= 3.00 × 10−6 N
−9
−9

N ⋅ m 2  ( 3.00 × 10 C) ( 5.00× 10 C)
F3 =  8.99 × 109
= 1.35 × 10−5 N

2
2
C
( 0.100 m)


( F6 )
The resultant is FR =
or
15.13
2
−1  F3 
2
+ ( F3 ) = 1.38 × 10−5 N at θ = tan   = 77.5°
 F6 
FR = 1.38 × 10 −5 N at 77.5° below − x axis
The forces on the 7.00 µC charge are shown at the right.
2
7.00 × 10−6 C) ( 2.00× 10−6 C)

9 N⋅m  (
F1 =  8.99 × 10

2
C2 
( 0.500 m)

= 0.503 N
2
7.00 × 10−6 C) ( 4.00 × 10−6 C )

9 N⋅m  (
F2 =  8.99 × 10

C2 
( 0.500 m) 2

= 1.01 N
Thus, ΣFx = ( F1 + F2 ) cos 60.0° = 0.755 N
and
ΣFy = ( F1 − F2 ) sin 60.0° = −0.436 N
The resultant force on the 7.00 µC charge is
FR =
or
( ΣFx )
2
(
+ ΣFy
)
2
Σ Fy 
−1 
= 0.872 N at θ = tan 
 = − 30.0°
 Σ Fx 
FR = 0.872 N at 30.0° below the +x axis
Electric Forces and Electric Fields
15.15
Consider the free-body diagram of one of the spheres given
at the right. Here, T is the tension in the string and Fe is the
repulsive electrical force exerted by the other sphere.
mg
cos 5.0 °
ΣFy
= 0 ⇒ T cos 5.0° = mg , or T =
ΣFx
= 0 ⇒ Fe = T sin 5.0° = mg tan 5.0°
At equilibrium, the distance separating the two spheres is r = 2 L sin 5.0° .
Thus, Fe = mg tan 5.0° becomes
q = ( 2 L sin 5.0 ° )
2
= mg tan 5.0° and yields
( 0.20 × 10
−3
kg) ( 9.80 m s2 ) tan 5.0°
8.99 × 109 N ⋅ m2 C2
= 7.2 nC
For the object to “float” it is necessary that the electrical force support the weight, or
qE = mg
15.20
( 2 L sin 5.0° )
mg tan 5.0°
ke
= 2 ( 0.300 m ) sin 5.0° 
15.17
k eq 2
or
m=
qE
g
( 24 × 10
=
−6
C) ( 610 N C)
9.8 m s 2
= 1.5× 10−3 kg
(a) The magnitude of the force on the electron is F = q E = eE , and the acceleration is
−19
F
eE ( 1.60 × 10 C) ( 300 N C)
a=
=
=
= 5.27 × 1013 m s2
me me
9.11× 10−31 kg
(b) v = v0 + at = 0 + ( 5.27 × 1013 m s 2 ) ( 1.00 × 10 −8 s ) = 5.27 × 10 5 m s
3
4 CHAPTER 15
15.24
The altitude of the triangle is
h = ( 0.500 m ) sin 60.0° = 0.433 m
and the magnitudes of the fields due to
each of the charges are
9
2
2
−9
ke q1 ( 8.99 × 10 N ⋅ m C ) ( 3.00× 10 C)
E1 = 2 =
2
h
( 0.433 m)
= 144 N C
E2 =
and
k e q2
2
2
r
( 8.99× 10
=
E3 =
9
k e q3
2
3
r
=
N ⋅ m2 C2 ) ( 8.00× 10−9 C)
( 0.250 m)
( 8.99 × 10
9
2
N ⋅ m 2 C2
= 1.15 × 10 3 N C
) ( 5.00× 10
( 0.250 m)
−9
) = 719 N C
C
2
3
Thus, ΣEx = E2 + E3 = 1.87 × 10 N C and ΣEy = −E1 = −144 N C
giving
ER =
( ΣEx )
2
(
+ ΣEy
)
2
= 1.88 × 10 3 N C
and
θ = tan −1 ( ΣEy ΣEx ) = tan −1 ( −0.0769 ) = −4.40°
Hence E R = 1.88 × 10 3 N C at 4.40° below the +x axis
Electric Forces and Electric Fields
15.27
If the resultant field is zero, the
contributions from the two charges must
be in opposite directions and also have
equal magnitudes. Choose the line
connecting the charges as the x-axis, with
the origin at the –2.5 µC charge. Then, the
two contributions will have opposite
directions only in the regions x < 0 and
x > 1.0 m . For the magnitudes to be equal, the point must be nearer the smaller charge.
Thus, the point of zero resultant field is on the x-axis at x < 0 .
Requiring equal magnitudes gives
Thus,
( 1.0 m + d )
k e q1
r12
=
k e q2
r22
or
2.5 µ C
6.0 µ C
=
2
d
( 1.0 m + d ) 2
2.5
=d
6.0
Solving for d yields
d = 1.8 m ,
15.28
or
1.8 m to the left of the − 2.5 µ C charge
The magnitude of q2 is three times the magnitude of q1 because 3 times as many lines
emerge from q2 as enter q1 . q2 = 3 q1
(a) Then,
(b)
q1 q2 = − 1 3
q2 > 0 because lines emerge from it,
and q1 < 0 because lines terminate on it.
15.30
Rough sketches for these charge configurations are shown below.
5
6 CHAPTER 15
15.32
(a) In the sketch for (a) at
the right, note that there
are no lines inside the
sphere. On the outside
of the sphere, the field
lines are uniformly
spaced and radially
outward.
(b) In the sketch for (b) above,
note that the lines are
perpendicular to the surface
at the points where they emerge. They should also be symmetrical about the
symmetry axes of the cube. The field is zero inside the cube.
15.36
If the weight of the drop is balanced by the electric force, then mg = q E = eE or the
mass of the drop must be
m=
−19
4
eE ( 1.6 × 10 C) ( 3 × 10 N C)
=
≈ 5 × 10 −16 kg
g
9.8 m s2
 4 3
 3m 
But, m = ρV = ρ  π r  and the radius of the drop is r = 

3


 4πρ 
 3 ( 5 × 10 −16 kg ) 

r=
3
 4π ( 858 kg m ) 
15.38
13
13
= 5.2 × 10 −7 m
or
r~ 1 µ m
The flux through an area is Φ E = EA cos θ , where θ is the angle between the direction of
the field E and the line perpendicular to the area A.
(a) Φ E = EA cos θ = ( 6.2 × 10 5 N C )( 3.2 m 2 ) cos0° = 2.0 × 10 6 N ⋅ m 2 C
(b) In this case, θ = 90° and Φ E = 0
15.43
We choose a spherical gaussian surface, concentric with the charged spherical shell and
of radius r. Then, ΣEA cos θ = E ( 4π r 2 ) cos0° = 4π r 2 E .
(a) For r > a (that is, outside the shell), the total charge enclosed by the gaussian
surface is Q = + q − q = 0 . Thus, Gauss’s law gives 4π r 2 E = 0, or E = 0 .
Electric Forces and Electric Fields
7
(b) Inside the shell, r < a , and the enclosed charge is Q = + q .
2
Therefore, from Gauss’s law, 4π r E =
ke q
directed radially outward .
r2
The field for r < a is E =
15.46
q
q
kq
= e2
, or E =
2
∈0
4π ∈0 r
r
Choose a very small cylindrical gaussian surface with one end inside the conductor.
Position the other end parallel to and just outside the surface of the conductor.
Since, in static conditions, E = 0 at all points inside a conductor, there is no flux through
the inside end cap of the gaussian surface. At all points outside, but very close to, a
conductor the electric field is perpendicular to the conducting surface. Thus, it is parallel
to the cylindrical side of the gaussian surface and no flux passes through this cylindrical
side. The total flux through the gaussian surface is then Φ = EA , where A is the crosssectional area of the cylinder as well as the area of the end cap.
The total charge enclosed by the cylindrical gaussian surface is Q = σ A , where σ is the
charge density on the conducting surface. Hence, Gauss’s law gives
EA =
F=
15.48
(a)
σA
σ
or E =
∈0
∈o
k e q1 q 2
r2
k ee 2
= 2
r
( 8.99× 10
=
9
N ⋅ m2 C2 )( 1.60× 10−19 C)
( 0.53 × 10
−10
m)
2
2
= 8.2 × 10−8 N
(b) F = me ac = me ( v2 r ) , so
v=
r ⋅F
=
me
( 0.53 × 10
−10
m )( 8.2 × 10−8 N)
9.11 × 10-31 kg
= 2 .2 × 10 6 m s