Electric Forces and Electric Fields
15.4
(a)
15
The two ions are both singly charged ( q = 1e ) , one positive and one negative. Thus,
F =
)
9
2
2
−19
ke q1 q2
ke e2 ( 8.99 × 10 N ⋅ m C (1.60 × 10 C
=
=
2
r2
r2
( 0.50 × 10−9 m
)
)
2
= 9.2 × 10 −10 N
Since these are unlike charges, the force is attractive .
(b)
15.5
(a)
(b)
No. The electric force depends only on the magnitudes of the two charges and the distance between them.
)
)
The mass of an alpha particle is m = 4.002 6 u, where 1 u = 1.66 × 10 −27 kg is the unified
mass unit. The acceleration of either alpha particle is then
a=
15.6
)
2
2
( 8.99 × 109 N ⋅ m 2 C2  4 (1.60 × 10−19 C 
ke ( 2 e )
F=
=
= 36.8 N
2
r2
( 5.00 × 10−15 m
F
36.8 N
=
= 5.54 × 10 27 m s 2
m 4.002 6 (1.66 × 10 −27 kg
)
The attractive force between the charged ends tends to compress the molecule. Its magnitude is
F=
2
2
1.60 × 10 −19 C

ke (1e )
9 N⋅m  (
=
8
.
99
×
10

r2
C2  ( 2 .17 × 10 −6 m
)
)
2
2
= 4.89 × 10 −17 N
The compression of the “spring” is
)
x = ( 0.010 0 ) r = ( 0.010 0 ) ( 2.17 × 10 −6 m = 2.17 × 10 −8 m,
so the spring constant is k =
15.7
F 4.89 × 10 −17 N
=
= 2.25 × 10 −9 N m .
x 2.17 × 10 −8 m
1.00 g of hydrogen contains Avogadro’s number of atoms, each containing one proton and one
electron. Thus, each charge has magnitude q = N A. The distance separating these charges is
r = 2 RE , where RE is Earth’s radius. Thus,
F=
(
ke N A e
)
2
( 2 RE )2
)
)
23
−19

N ⋅ m 2  ( 6.02 × 10 (1.60 × 10 C 
=  8.99 × 10 9
= 5.12 × 10 5 N
2
6
C2 

4 ( 6.38 × 10 m
56157_15_ch15_p001-038.indd 15
)
2
3/15/08 3:15:24 PM
16
15.8
Chapter 15
Refer to the sketch at the right. The magnitudes of the
F1 = F2 = ke
q ( 2q )
q ( 3q )
q2
q2
= 2 ke 2 and F3 = ke
2 = 1.50 ke
2
a
a2
a
a 2
)
(
The components of the resultant force on charge q are
Fx = F1 + F3 cos 45° = ( 2 + 1.50 cos 45° ) ke
and
q2
q2
= 3.06 ke 2
2
a
a
Fy = F1 + F3 sin 45° = ( 2 + 1.50 sin 45°) ke
q2
q2
=
3
.
06
k
e 2
a
a2

q2
q2 
The magnitude of the resultant force is Fe = Fx2 + Fy2 = 2  3.06 ke 2  = 4.33ke 2
a
a 

 Fy 
and it is directed at θ = tan −1   = tan −1 (1.00 ) = 45° above the horizontal .
 Fx 
15.9
(a)
The spherically symmetric charge distributions behave as if all charge was located at the
centers of the spheres. Therefore, the magnitude of the attractive force is
F=
(b)
)
)
When the spheres are connected by a conducting wire, the net charge
qnet = q1 + q2 = −6.0 × 10 −9 C will divide equally between the two identical spheres.
Thus, the force is now
2
−6.0 × 10 −9 C
ke ( qnet 2 )

9 N⋅m  (
=
8
.
99
×
10

r2
C2  4 ( 0.30 m )2
2
F=
or
56157_15_ch15_p001-038.indd 16
)
12 × 10 −9 C (18 × 10 −9 C
ke q1 q2
9
2
2 (
=
8
.
99
×
10
N
⋅
m
C
= 2.2 × 10 −5 N
(
r2
( 0.30 m )2
)
2
F = 9.0 × 10 −7 N (repulsion)
3/15/08 3:16:31 PM
Electric Forces and Electric Fields
15.10
17
The forces are as shown in the sketch
at the right.
)
)
F1 =
2
6.00 × 10 −6 C (1.50 × 10 −6 C
ke q1q2 
9 N⋅m  (
=
8
.
99
×
10
= 89.9 N
2

r122
C2 
( 3.00 × 10−2 m
F2 =
2
6.00 × 10 −6 C ( 2 .00 × 10 −6 C
ke q1 q3 
9 N⋅m  (
=
8
.
99
×
10
= 43.2 N
2

r132
C2 
( 5.00 × 10−2 m
F3 =
2
1.50 × 10 −6 C ( 2 .00 × 10 −6 C
ke q2 q3 
9 N⋅m  (
=
8
.
99
×
10
= 67.4 N
2

r232
C2 
( 2 .00 × 10−2 m
)
)
)
)
)
)
)
The net force on the 6 µC charge is F6 = F1 − F2 = 46.7 N (to the left)
The net force on the 1.5 µC charge is F1.5 = F1 + F3 = 157 N (to the right)
The net force on the −2 µC charge is F−2 = F2 + F3 = 111 N (to the left)
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.
)
)
)
)
−9
−9

N ⋅ m 2  ( 6.00 × 10 C ( 5.00 × 10 C
F6 =  8.99 × 10 9
2
2
C 

( 0.300 m )
= 3.00 × 10 −6 N
−9
−9

N ⋅ m 2  ( 3.00 × 10 C ( 5.00 × 10 C
= 1.35 × 10 −5 N
F3 =  8.99 × 10 9
2
2
C 

( 0.100 m )
The resultant is FR =
or
56157_15_ch15_p001-038.indd 17
( F6 )2 + ( F3 )2
F 
= 1.38 × 10 −5 N at θ = tan −1  3  = 77.5°
 F6 
!
FR = 1.38 × 10 −5 N at 77.5° below − x - axis
3/15/08 11:20:16 AM
18
Chapter 15
15.12
Consider the arrangement of charges shown in the sketch
at the right. The distance r is
( 0.500 m )2 + ( 0.500 m )2
r=
= 0.707 m
The forces exerted on the 6.00 nC charge are
)
)
)
)
−9
−9

N ⋅ m 2  ( 6.00 × 10 C ( 2 .00 × 10 C
F2 =  8.99 × 10 9
2
2
C 

( 0.707 m )
= 2 .16 × 10 −7 N
and
−9
−9

N ⋅ m 2  ( 6.00 × 10 C ( 3.00 × 10 C
F3 =  8.99 × 10 9
= 3.24 × 10 −7 N
2
2
C 

( 0.707 m )
Thus, ΣFx = ( F2 + F3 ) cos 45.0° = 3.81 × 10 −7 N
and
ΣFy = ( F2 − F3 ) sin 45.0° = −7.63 × 10 −8 N
The resultant force on the 6.00 nC charge is then
FR =
or
15.13
F1 =
2
 ΣFy 
= − 11.3°
= 3.89 × 10 −7 N at θ = tan −1 
 ΣFx 
!
FR = 3.89 × 10 −7 N at 11.3° below + x -axis
Please see the sketch at the right.
( 8.99 × 10
or
F2 =
( ΣFx )2 + ( ΣFy )
9
( 0.500 m )
2
)
)
)
)
F1 = 0.288 N
( 8.99 × 10
or
)
N ⋅ m 2 C2 ( 2.00 × 10 −6 C ( 4.00 × 10 −6 C
9
)
N ⋅ m 2 C2 ( 2.00 × 10 −6 C ( 7.00 × 10 −6 C
( 0.500 m )
2
F2 = 0.503 N
The components of the resultant force acting on the 2.00 µC charge are:
Fx = F1 − F2 cos 60.0° = 0.288 N − ( 0.503 N ) cos 60.0° = 3.65 × 10 −2 N
and
Fy = − F2 sin 60.0° = − ( 0.503 N ) sin 60.0° = − 0.436 N
The magnitude and direction of this resultant force are
F = Fx2 + Fy2 =
at
56157_15_ch15_p001-038.indd 18
( 0.0365 N )2 + ( 0.436 N )2
= 0.438 N
 Fy 
 − 0.436 N 
θ = tan −1   = tan −1 
= −85.2° or 85.2° below + x -axis
 0.0365 N 
F
 x
3/15/08 11:20:17 AM
Electric Forces and Electric Fields
15.14
19
If the forces exerted on the
positive third charge by the two
negative charges are to be in
opposite directions as they must,
the third charge must be located
on the line between the two
negative charges as shown at the
right.
If the two forces are to add to zero, their magnitudes must be equal. These magnitudes are
F1 = ke
where
q1 q3
x2
F2 = ke
and
q2 q3
( 50.0 cm − x )2
q1 = −3.00 nC and q2 = −5.80 nC.
( 50.0 cm − x )2 =
Equating and canceling common factors gives
q2 2
x
q1
or
50.0 cm − x = x
5.80 nC
= 1.39 x
3.00 nC
giving 2.39 x = 50.0 cm
and
x = 20.9 cm
Thus, the third charged should be located 20.9 cm to the right of the − 3.00 nC charge or
50.0 cm − 20.9 cm = 29.1 cm to the left of the −55.80 nC charge .
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.
∑ Fy = 0 ⇒ T cos 5.0° = mg , or T =
mg
cos 5.0°
∑ Fx = 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 L sin 5.0°)2
= mg tan 5.0° and yields
mg tan 5.0°
ke
=  2 ( 0.300 m ) sin 5.0° 
56157_15_ch15_p001-038.indd 19
ke q 2
( 0.20 × 10
−3
)
)
kg ( 9.80 m s 2 tan 5.0°
8.99 × 10 N ⋅ m C
9
2
2
= 7.2 nC
3/15/08 11:20:18 AM
20
15.16
Chapter 15
In the sketch at the right,
rBC =
( 4.00 m )2 + ( 3.00 m )2
= 5.00 m
and
 3.00 m 
θ = tan −1 
= 36.9°
 4.00 m 
(a)
( FAC )x =
0
(b)
( FAC )y =
FAC = ke
qA qC
2
rAC
( FAC )y = ( 8.99 × 10 9 N ⋅ m 2
15.17
C2
)
( 3.00 × 10
q q
= ke B 2 C = ( 8.99 × 10 9 N ⋅ m 2 C2
rBC
)
−4
)
C (1.00 × 10 −4 C
( 3.00 m )
2
( 6.00 × 10
(c)
FBC
(d)
( FBC )x =
(e)
( FBC )y = − FBC sin θ = − ( 21.6 N ) sin ( 36.9°) =
(f)
( FR )x = ( FAC )x + ( FBC )x = 0 + 17.3 N = 17.3 N
(g)
( FR )y = ( FAC )y + ( FBC )y = 30.0 − 13.0 N = 17.0 N
(h)
FR =
−4
)
) = 30.0 N
C (1.00 × 10 −4 C
( 5.00 m )
2
) = 21.6 N
FBC cos θ = ( 21.6 N ) cos ( 36.9°) = 17.3 N
( FR )2x + ( FR )2y
=
(17.3 N )2 + (17.0 N )2
−13.0 N
= 24.3 N
and
 ( FR )y 
−1  17.0 N 
ϕ = tan −1 
= 44.5°
 = tan 
 17.3 N 
F
(
)
 R x 
or
!
FR = 24.3 N at 44.5° above the + x direction
In order for the object to “float” in the electric field, the electric force exerted on the object by
the field must be directed upward and have a magnitude equal to the weight of the object. Thus,
Fe = qE = mg , and the magnitude of the electric field must be
E=
(
)
)
3.80 g ( 9.80 m s 2  1 kg 
mg
3
=
 10 3 g  = 2.07 × 10 N C
18 × 10 −6 C
q


The electric force on a negatively charged object is in the direction opposite to that of the
electric field. Since the electric force must be directed upward, the electric field must be
directed downward.
56157_15_ch15_p001-038.indd 20
3/15/08 3:20:24 PM
Electric Forces and Electric Fields
15.18
(a)
21
Taking to the right as
positive, the resultant
electric field at point P
is given by
ER = E1 + E3 − E2
=
ke q1 ke q3 ke q2
+ 2 − 2
r12
r3
r2

N ⋅ m 2   6.00 × 10 −6 C 2 .00 × 10 −6 C 1.50 × 10 −6 C 
=  8.99 × 10 9
+
−


C2   ( 0.020 0 m )2 ( 0.030 0 m )2 ( 0.010 0 m )2 

This gives ER = + 2 .00 × 10 7 N C
!
or E R = 2 .00 × 10 7 N C to the right
(b)
!
!
F = qE R = ( −2 .00 × 10 −6 C ( 2 .00 × 10 7 N C = − 40.0 N
)
)
!
or F = 40.0 N to the left
15.19
We shall treat the concentrations as point charges. Then, the resultant field consists of two contributions, one due to each concentration.
The contribution due to the positive charge at 3 000 m altitude is
E+ = ke
q 
N ⋅ m 2  ( 40.0 C )
=  8.99 × 10 9
= 3.60 × 10 5 N C ( downward )
2
r
C2  (1 000 m )2

The contribution due to the negative charge at 1 000 m altitude is
E− = ke
q 
N ⋅ m 2  ( 40.0 C )
=  8.99 × 10 9
= 3.60 × 10 5 N C ( downward )
2
r
C2  (1 000 m )2

The resultant field is then
! !
!
E = E+ + E− = 7.20 × 10 5 N C ( downward )
15.20
(a)
The magnitude of the force on the electron is F = q E = eE , and the acceleration is
a=
(b)
56157_15_ch15_p001-038.indd 21
)
−19
F eE (1.60 × 10 C ( 300 N C )
=
=
= 5.27 × 1013 m s 2
me me
9.11 × 10 −31 kg
(
)(
)
v = v0 + at = 0 + 5.27 × 1013 m s 2 1.00 × 10 −8 s = 5.27 × 10 5 m s
3/15/08 11:20:20 AM
22
15.21
Chapter 15
Note
!" that at the point midway between the two charges (x = 2.00 m,
!" y = 0), the field contribution
E1 (due to the negative charge at the origin) and the contribution E2 (due to the positive charge
at x = 4.00 m) are both in the negative x-direction. The magnitude of the resultant field at this
point is therefore
 5.00 × 10 −9 C 7.00 × 10 −9 C 
q
q 
Enet = E1 + E2 = ke  21 + 22  = ( 8.99 × 10 9 N ⋅ m 2 C2 
+
 = 27.0 N C
2
r2 
 r1
( 2.00 m )2 
 ( 2.00 m )
!"
Thus, Enet = 27.0 N C in the negative x direction .
)
15.22
The force an electric field exerts on a positive change is in the direction of the field. Since this
force must serve as a retarding force and bring the proton to rest, the force and hence the field
must be in the direction opposite to the proton’s velocity .
The work-energy theorem, Wnet = KE f − KEi , gives the magnitude of the field as
− ( qE ) ∆x = 0 − KEi
15.23
15.24
E=
or
KEi
3.25 × 10 −15 J
=
= 1.63 × 10 4 N C
q ( ∆ x ) (1.60 × 10 −19 C (1.25 m )
)
)
(a)
−19
F qE (1.60 × 10 C ( 640 N C )
a= =
=
= 6.12 × 1010 m s 2
m mp
1.673 × 10 −27 kg
(b)
t=
(c)
∆x =
(d)
KE f =
(a)
Please refer to the solution of Problem 15.13 earlier in this chapter. There it is shown
that the resultant electric force experienced by the 2.00 µC located at the origin is
!"
F = 0.438 N at 85.2° below the + x-axis . Since the electric field at a location is defined as
the force per unit charge experienced by a test charge placed in that location, the electric
field at the origin in the charge configuration of Figure P15.13 is
∆v
1.20 × 10 6 m s
= 1.96 × 10 −5 s = 19.6 µs
=
6.12 × 1010 m s 2
a
v 2f − v02
2a
(1.20 × 10 m s) − 0 =
2 ( 6.12 × 10 m s )
2
6
=
10
2
11.8 m
1
1
m p v 2f = (1.673 × 10 −27 kg (1.20 × 10 6 m s
2
2
)
)
2
= 1.20 × 10 −15 J
!"
!" F
0.438 N
at − 85.2° = 2.19 × 10 5 N C at 85.2° below + x -axis
E=
=
q0 2.00 × 10 −6 C
(b)
The electric force experienced by the charge at the origin is directly proportional to
the magnitude of that charge. Thus, doubling the magnitude of this charge would
double the magnitude of the electric force . However, the electric field is the force per unit
charge and the field would be unchanged if the charge was doubled . This is easily seen in
the calculation of part (a) above. Doubling the magnitude of!"the charge at the origin would
double both the numerator and the denominator of the ratio F q0 , but the value of the ratio
(i.e., the electric field) would be unchanged.
56157_15_ch15_p001-038.indd 22
3/15/08 3:22:12 PM
Electric Forces and Electric Fields
15.25
23
The alpha particle (with a charge +2e and mass of 6.63 × 10 −27 kg) will experience a constant
electric force in the −y-direction (the direction of the electric field) once it enters the tube. This
force gives the alpha particle a constant acceleration with components
ax = 0
)
)
−19
−4
Fe qEy ( 3.20 × 10 C ( − 4.50 × 10 N C
ay =
=
=
= −2.17 × 10 4 m s 2
m
m
6.64 × 10 −27 kg
and
The particle will strike the tube wall when its vertical displacement is ∆y = −0.500 cm, and from
∆y = v0 y t + ay t 2 2 with v0 y = 0, the elapsed time when this occurs is
t=
2 ( ∆y )
=
ay
2 ( −0.500 × 10 −2 m
−2.17 × 10 4 m s 2
) = 6.79 × 10
−4
s
During this time, the distance the alpha particle travels parallel to the axis of the tube is
∆x = v0 x t +
15.26
1 2
ax t = (1 250 m s ) ( 6.79 × 10 −4 s + 0 = 0.849 m
2
)
If the resultant field is to be zero, the contributions
of the two charges must be equal in magnitude
and must have opposite directions. This is only
possible at a point on the line between the two
negative charges.
Assume the point of interest is located on the
y-axis at − 4.0 m < y < 6.0 m . Then, for equal
magnitudes,
ke q1 ke q2
9.0 µC
8.0 µC
= 2 or
2 =
2
r1
r2
( 6.0 m − y ) ( y + 4.0 m )2
Solving for y gives y + 4.0 m =
15.27
8
( 6.0 − y ), or y = +0.85 m .
9
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 m C 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 )
ke q1 ke q2
2 .5 µC
6.0 µC
=
= 2 or
r12
r2
d2
(1.0 m + d )2
2.5
=d
6.0
Solving for d yields
d = 1.8 m,
56157_15_ch15_p001-038.indd 23
or
1.8 m to the left of the − 2.5 µC charge
3/15/08 11:20:22 AM
24
15.28
Chapter 15
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
E1 =
)
9
2
2
−9
ke q1 ( 8.99 × 10 N ⋅ m C ( 3.00 × 10 C
=
h2
( 0.433 m )2
)
= 144 N C
E2 =
and
)
)
2
2
−9
9
ke q2 ( 8.99 × 10 N ⋅ m C ( 8.00 × 10 C
=
= 1.15 × 10 3 N C
2
r22
( 0.250 m )
E3 =
)
)
2
2
−9
9
ke q3 ( 8.99 × 10 N ⋅ m C ( 5.00 × 10 C
=
= 719 N C
r32
( 0.250 m )2
Thus, ΣEx = E2 + E3 = 1.87 × 10 3 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
15.29
From the symmetry of the charge distribution, students
should recognize that the resultant electric field at the
center is
!
ER = 0
If one does not recognize this intuitively, consider:
!
!
!
!
E R = E1 + E2 + E 3, so
Ex = E1x − E2 x =
ke q
k q
cos 30° − e 2 cos 30° = 0
2
r
r
and
Ey = E1y + E2 y − E3 =
ke q
k q
k q
sin 30° + e 2 sin 30° − e 2 = 0
2
r
r
r
Thus, ER = Ex2 + Ey2 = 0
56157_15_ch15_p001-038.indd 24
3/15/08 11:20:23 AM
Electric Forces and Electric Fields
15.30
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)
(b)
Then,
q1 q2 = − 1 3
q2 > 0 because lines emerge from it, and q1 < 0 because lines terminate on it.
15.31
Note in the sketches at the right that
electric field lines originate on
positive charges and terminate on
negative charges. The density of
lines is twice as great for the −2 q
charge in (b) as it is for the 1q
charge in (a).
15.32
Rough sketches for these charge configurations are shown below.
15.33
(a)
The sketch for (a) is shown at
the right. Note that four times
as many lines should leave q1
as emerge from q2 although, for
clarity, this is not shown in this
sketch.
(b)
The field pattern looks the same
here as that shown for (a) with
the exception that the arrows
are reversed on the field lines.
56157_15_ch15_p001-038.indd 25
25
3/15/08 11:20:23 AM
26
15.34
15.35
15.36
Chapter 15
(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.
(a)
Zero net charge on each surface of the sphere.
(b)
The negative charge lowered into the sphere repels − 5µC on the outside surface, and
leaves + 5 µC on the inside surface of the sphere.
(c)
The negative charge lowered inside the sphere neutralizes the inner surface, leaving
zero charge on the inside . This leaves − 5µC on the outside surface of the sphere.
(d)
When the object is removed, the sphere is left with − 5.00 µC on the outside surface
and zero charge on the inside .
(a)
The dome is a closed conducting surface. Therefore, the electric field is zero everywhere
inside it.
At the surface and outside of this spherically symmetric charge distribution, the field is as if
all the charge were concentrated at the center of the sphere.
(b)
At the surface,
E=
(c)
Outside the spherical dome, E =
E=
15.37
)
( 8.99 × 10
9
ke q
. Thus, at r = 4.0 m,
r2
)
N ⋅ m 2 C2 ( 2 .0 × 10 −4 C
( 4.0 m )
2
)=
1.1 × 10 5 N C
For a uniformly charged sphere, the field is strongest at the surface.
Thus, Emax =
ke qmax
,
R2
qmax =
)
6
R 2 Emax ( 2.0 m ) ( 3.0 × 10 N C
= 1.3 × 10 −3 C
=
ke
8.99 × 10 9 N ⋅ m 2 C2
2
or
56157_15_ch15_p001-038.indd 26
)
9
2
2
−4
ke q ( 8.99 × 10 N ⋅ m C ( 2 .0 × 10 C
=
= 1.8 × 10 6 N C
R2
(1.0 m )2
3/15/08 11:20:24 AM
Electric Forces and Electric Fields
15.38
27
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 s 2
13
 3m 
4

But, m = ρV = ρ  π r 3  and the radius of the drop is r = 


3
 4πρ 
)
 3 ( 5 × 10 −16 kg 
r=

3
 4π ( 858 kg m 
15.39
(a)
(b)
15.41
)
= 5.2 × 10 −7 m
r ∼ 1 µm
or
)
)
Fe = ma = (1.67 × 10 −27 kg (1.52 × 1012 m s 2 = 2.54 × 10 −15 N in the direction of the
acceleration, or radially outward.
The direction of the field is the direction of the force on a positive charge (such as the
proton). Thus, the field is directed radially outward . The magnitude of the field is
E=
15.40
13
Fe 2.54 × 10 −15 N
= 1.59 × 10 4 N C
=
q 1.60 × 10 −19 C
The flux through an area is Φ E = EA cos θ , where q 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 cos 0° = 2.0 × 10 6 N ⋅ m 2 C
(b)
In this case, θ = 90° and Φ E = 0 .
The area of the rectangular plane is A = ( 0.350 m ) ( 0.700 m ) = 0.245 m 2 .
(a)
When the plane is parallel to the yz-plane, θ = 0°, and the flux is
)
)
Φ E = EA cos θ = ( 3.50 × 10 3 N C ( 0.245 m 2 cos 0° = 858 N ⋅ m 2 C
15.42
(b)
When the plane is parallel to the x-axis, θ = 90° and Φ E = 0 .
(c)
Φ E = EA cos θ = ( 3.50 × 10 3 N C ( 0.245 m 2 cos 40° = 657 N ⋅ m 2 C
(a)
)
)
Gauss’s law states that the electric flux through any closed surface equals the net charge
enclosed divided by ∈0 . We choose to consider a closed surface in the form of a sphere,
centered on the center of the charged sphere and having a radius infinitesimally larger than
that of the charged sphere. The electric field then has a uniform magnitude and is perpendicular to our surface at all points on that surface. The flux through the chosen closed
surface is therefore Φ E = EA = E ( 4π r 2 , and Gauss’s law gives
)
Qinside = ∈0 Φ E = 4π ∈0 Er 2
)
= 4π ( 8.85 × 10 −12 C2 N ⋅ m 2 ( 575 N C ) ( 0.230 m ) = 3.38 × 10 −9 C = 3.38 nC
(b)
56157_15_ch15_p001-038.indd 27
2
Since the electric field displays spherical symmetry, you can conclude that the charge distribution generating that field is spherically symmetric . Also, since the electric field lines are
directed outward away from the sphere, the sphere must contain a net positive charge .
3/15/08 3:23:35 PM
28
15.43
15.44
Chapter 15
From Gauss’s law, the electric flux through any closed surface
is equal to the net charge enclosed divided by ∈0 . Thus, the
flux through each surface (with a positive flux coming outward
from the enclosed interior and a negative flux going inward
toward that interior) is
For S1:
Φ E = Qnet ∈0 = ( +Q − 2Q ) ∈0 = −Q ∈0
For S2:
Φ E = Qnet ∈0 = ( +Q − Q ) ∈0 = 0
For S3:
Φ E = Qnet ∈0 = ( −2Q + Q − Q ) ∈0 = −2Q ∈0
For S4:
Φ E = Qnet ∈0 = ( 0 ) ∈0 = 0
The electric field has constant magnitude and is everywhere perpendicular to the bottom of the
car. Thus, the electric flux through the bottom of the car is
)
Φ E = EA = (1.80 × 10 4 N C ( 5.50 m ) ( 2.00 m )  = 1.98 × 10 5 N ⋅ m 2 C
15.45
We choose a spherical Gaussian surface, concentric with the charged spherical shell and of radius r.
Then, ΣEA cos θ = E ( 4π r 2 cos 0° = 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 .
(b)
Inside the shell, r < a , and the enclosed charge is Q = + q.
Therefore, from Gauss’s law, 4π r 2 E =
kq
q
q
, or E =
= e2
2
r
∈0
4π ∈0 r
! kq
The field for r < a is E = e2 directed radially outward .
r
15.46
(a)
The surface of the cube is a closed surface which surrounds a total charge of
Q = 1.70 × 10 2 µC . Thus, by Gauss’s law, the electric flux through the whole surface of the
cube is
ΦE =
(b)
Qinside
1.70 × 10 −4 C
=
= 1.92 × 10 7 N ⋅ m 2 C
∈0
8.85 × 10 −12 C2 N ⋅ m 2
Since the charge is located at the center of the cube, the six faces of the cube are symmetrically positioned around the location of the charge. Thus, one-sixth of the flux passes
through each of the faces, or
Φ face =
(c)
56157_15_ch15_p001-038.indd 28
Φ E 1.92 × 10 7 N ⋅ m 2 C
=
= 3.20 × 10 6 N ⋅ m 2 C
6
6
The answer to part (b) would change because the charge could now be at different distances
from each face of the cube, but the answer to part (a) would be unchanged because the flux
through the entire closed surface depends only on the total charge inside the surface.
3/15/08 11:20:25 AM
Electric Forces and Electric Fields
15.47
29
Note that with the point charge −2.00 nC positioned at the
center of the spherical shell, we have complete spherical
symmetry in this situation. Thus, we can expect the distribution of charge on the shell, as well as the electric fields both
inside and outside of the shell, to also be spherically
symmetric.
(a)
We choose a spherical Gaussian surface, centered
on the center of the conducting shell, with radius
r = 1.50 m < a as shown at the right. Gauss’s law gives
)
Φ E = EA = E ( 4π r 2 =
so
( 8.99 × 10
E=
9
Qinside
Qinside
kQ
or E =
= e center
2
∈0
4π ∈0 r
r2
)
N ⋅ m 2 C2 ( −2.00 × 10 −9 C
(1.50 m )
2
)
and E = −7.99 N C . The negative sign means that the field is radial inward .
(b)
All points at r = 2.20 m are in the range a < r < b, and hence are located within the
conducting material making up the shell. Under conditions of electrostatic equilibrium, the
field is E = 0 at all points inside a conducting material.
(c)
If the radius of our Gaussian surface is r = 2.50 m > b , Gauss’s law (with total spheriQinside
kQ
cal symmetry) leads to E =
= e inside
just as in part (a). However, now
2
4π ∈0 r
r2
Qinside = Qshell + Qcenter = +3.00 nC − 2.00 nC = +1.0
00 nC. Thus, we have
E=
( 8.99 × 10
9
)
N ⋅ m 2 C2 ( +1.00 × 10 −9 C
( 2.50 m )2
)=
+1.44 N C
with the positive sign telling us that the field is radial outward at this location.
(d)
Under conditions of electrostatic equilibrium, all excess charge on a conductor resides
entirely on its surface. Thus, the sum of the charge on the inner surface of the shell and
that on the outer surface of the shell is Qshell = +3.00 nC . To see how much of this is on the
inner surface, consider our Gaussian surface to have a radius r that is infinitesimally larger
than a. Then, all points on the Gaussian surface lie within the conducting material, meaning
that E = 0 at all points and the total flux through the surface is Φ E = 0. Gauss’s law then
states that Qinside = Qinner + Qcenter = 0,
surface
or
Qinner = −Qcenter = − ( −2.00 nC ) = +2.00 nC
surface
The charge on the outer surface must be
Qouter = Qshell − Qinner = 3.00 nC − 2.00 nC = +1.00 nC
surface
56157_15_ch15_p001-038.indd 29
surface
3/15/08 11:20:26 AM
30
15.48
Chapter 15
Please review Example 15.8 in your textbook. There it is shown that the electric field due to a
nonconducting plane sheet of charge parallel to the xy-plane has a constant magnitude given by
Ez = σ sheet 2 ∈0 , where σ sheet is the uniform charge per unit area on the sheet. This field is everywhere perpendicular to the xy-plane, is directed away from the sheet if it has a positive charge
density, and is directed toward the sheet if it has a negative charge density.
In this problem, we have two plane sheets of charge, both parallel to the xy-plane and separated
by a distance of 2.00 cm. The upper sheet has charge density σ sheet = −2σ , while the lower sheet
has σ sheet = +σ . Taking upward as the positive z-direction, the fields due to each of the sheets in
the three regions of interest are:
Region
Lower sheet (at z = 0)
Upper sheet (at z = 2.00 cm)
Electric Field
Electric Field
Ez = −
+σ
σ
=−
2 ∈0
2 ∈0
Ez = +
−2σ
σ
=+
2 ∈0
∈0
0 > z > 2.00 cm
Ez = +
+σ
σ
=+
2 ∈0
2 ∈0
Ez = +
−2σ
σ
=+
2 ∈0
∈0
z > 2.00 cm
Ez = +
+σ
σ
=+
2 ∈0
2 ∈0
Ez = −
−2σ
σ
=−
2 ∈0
∈0
z<0
When both plane sheets of charge are present, the resultant electric field in each region is the vector sum of the fields due to the individual sheets for that region.
56157_15_ch15_p001-038.indd 30
(a)
For z < 0 :
Ez = Ez ,lower + Ez ,upper = −
σ
σ
σ
+
= +
2 ∈0 ∈0
2 ∈0
(b)
For 0 < z < 2.00 cm:
Ez = Ez ,lower + Ez ,upper = +
σ
σ
3σ
+
= +
2 ∈0 ∈0
2 ∈0
(c)
For z > 2.00 cm:
Ez = Ez ,lower + Ez ,upper = +
σ
σ
σ
−
= −
2 ∈0 ∈0
2 ∈0
3/15/08 11:20:27 AM
Electric Forces and Electric Fields
15.49
31
The radius of each sphere is small in comparison to the distance to the nearest neighboring charge
(the other sphere). Thus, we shall assume that the charge is uniformly distributed over the surface
of each sphere and, in its interaction with the other charge, treat it as though it were a point
charge. In this model, we then have two identical point charges, of magnitude 35.0 mC, separated
by a total distance of 310 m (the length of the cord plus the radius of each sphere). Each of these
charges repels the other with a force of magnitude
Q2
Fe = ke 2 = ( 8.99 × 10 9 N ⋅ m 2 C2
r
( 35.0 × 10
)
( 3.10 × 10
−3
2
)
m)
C
2
2
= 115 N
Thus, to counterbalance this repulsion and hold each sphere in equilibrium, the cord must have a
tension of 115 N so it will exert a 115 N on that sphere, directed toward the other sphere.
15.50
(a)
As shown in Example 15.8 in the textbook, the electric field due to a nonconducting plane
sheet of charge has a constant magnitude of E = σ 2 ∈0 , where σ is the uniform charge per
unit area on the sheet. The direction of the field at all locations is perpendicular to the plane
sheet and directed away from the sheet if σ is positive, and toward the sheet if σ is negative. Thus, if σ = +5.20 µC m 2 , the magnitude of the electric field at all distances greater
than zero from the plane (including the distance of 8.70 cm) is
E=
(b)
15.51
σ
+5.20 × 10 −6 C m 2
=
= 2.94 × 10 5 N C
2 ∈0 2 ( 8.85 × 10 −12 C2 N ⋅ m 2 )
The field does not vary with distance as long as the distance is small compared with the
dimensions of the plate.
The three contributions to the resultant
electric field at the point of interest are
shown in the sketch at the right.
The magnitude of the resultant field is
ER = − E1 + E2 + E3
ER = −
 q
ke q1 ke q2 ke q3
q
q 
+ 2 + 2 = ke  − 21 + 22 + 23 
2
r1
r2
r3
r2
r3 
 r1

N ⋅ m 2   4.0 × 10 −9 C 5.0 × 10 −9 C 3.0 × 10 −9 C 
ER =  8.99 × 10 9
+
+
−

C2   ( 2 .5 m )2

(1.2 m )2 
( 2 .0 m )2
!
ER = + 24 N C, or E R = 24 N C in the +x direction
15.52
Consider the free-body diagram shown at the right.
ΣFy = 0 ⇒ T cos θ = mg or T =
mg
cos θ
ΣFx = 0 ⇒ Fe = T sin θ = mg tan θ
Since Fe = qE , we have
qE = mg tan θ , or q =
q=
56157_15_ch15_p001-038.indd 31
( 2 .00 × 10
−3
)
mg tan θ
E
)
kg ( 9.80 m s 2 tan 15.0°
1.00 × 10 N C
3
= 5.25 × 10 −6 C = 5.25 µC
3/15/08 11:20:28 AM
32
15.53
Chapter 15
(a)
At a point on the x-axis, the contributions by the two
charges to the resultant field have equal magnitudes
kq
given by E 1 = E2 = e2 .
r
The components of the resultant field are
 k q
 k q
Ey = E1y − E2 y =  e2  sin θ −  e2  sin θ = 0
 r 
 r 
and
 k ( 2q ) 
 k q
 k q
Ex = E1x + E2 x =  e2  cos θ +  e2  cos θ =  e 2  cos θ
 r 
 r 
 r

b
Since cos θ = b r = b =
2
2
3
2
r
r
r
( a + b2
!
ER =
(b)
ke ( 2 q ) b
(a
2
)
+ b2
32
)
32
, the resultant field is
in the +x direction
Note that the result of part (a) may be written as ER =
ke ( Q ) b
(a
2
+ b2
)
32
where Q = 2 q is the
total charge in the charge distribution generating the field.
In the case of a uniformly charged circular ring, consider the ring to consist of a very large
number of pairs of charges uniformly spaced around the ring. Each pair consists of two
identical charges located diametrically opposite each other on the ring. The total charge
of pair number i is Qi . At a point on the axis of the ring, this pair of charges generates an
ke b Qi
electric field contribution that is parallel to the axis and has magnitude Ei =
32 .
2
(a + b2
The resultant electric field of the ring is the summation of the contributions by all pairs of
charges, or
)

ke b
ER = ΣEi = 
2
 (a + b2

)

ke b Q

3 2 ΣQi =
2

( a + b2

)
32
where Q = ΣQi is the total charge on the ring.
!
ER =
15.54
(a
keQ b
2
+ b2
)
32
in the +x direction
It is desired that the electric field exert a retarding force on the electrons, slowing them down
and bringing them to rest. For the retarding force to have maximum effect, it should be
anti-parallel to the direction of the electrons motion. Since the force an electric field exerts
on negatively charged particles (such as electrons) is in the direction opposite to the field,
the electric field should be in the directioon of the electron’s motion .
The work a retarding force of magnitude Fe = q E = eE does on the electrons as they move
distance d is W = Fe d cos180° = − Fe d = − eEd . The work-energy theorem (W = ∆KE ) then gives
− eEd = KE f − KEi = 0 − K
and the magnitude of the electric field required to stop the electrons in distance d is
E=
56157_15_ch15_p001-038.indd 32
−K
− ed
or
E = K ed
3/15/08 11:20:29 AM
Electric Forces and Electric Fields
15.55
(a)
Without the electric field present, the ball comes to equilibrium when the upward force
exerted on the ball by the spring equals the downward gravitational force acting on it, or
when kx = mg. The distance the ball stretches the spring before reaching the equilibrium
position in this case is
x=
(b)
(a)
)
2
mg ( 4.00 kg ) ( 9.80 m s
= 4.64 × 10 −2 m = 4.64 cm
=
845 N m
k
With the electric field present, the positively charged ball experiences an upward electric
force in addition to the upward spring force and downward gravitational force. When the
ball comes to equilibrium, with the spring stretched a distance x ′ , the total upward force
equals the magnitude of the downward force, or qE + kx ′ = mg . The distance between the
unstretched position and the new equilibrium position of the ball is
x′ =
15.56
33
)
2
mg − qE ( 4.00 kg ) ( 9.80 m s − ( 0.050 0 C ) ( 355 N C )
=
= 2.54 × 10 −2 m = 2.54 cm
845 N m
k
The downward electrical force acting on the ball is
)
)
Fe = qE = ( 2 .00 × 10 −6 C (1.00 × 10 5 N C = 0.200 N
The total downward force acting on the ball is then
)
)
F = Fe + mg = 0.200 N + (1.00 × 10 −3 kg ( 9.80 m s 2 = 0.210 N
Thus, the ball will behave as if it was in a modified gravitational field where the effective
free-fall acceleration is
“g” =
0.210 N
F
=
= 210 m s 2
m 1.00 × 10 −3 kg
The period of the pendulum will be
T = 2π
(b)
L
0.500 m
= 2π
= 0.307 s
“g”
210 m s 2
Yes . The force of gravity is a significant portion of the total downward force acting on
the ball. Without gravity, the effective acceleration would be
“g” =
0.200 N
Fe
=
= 200 m s 2
m 1.00 × 10 −3 kg
giving T = 2π
0.500 m
= 0.314 s
200 m s 2
a 2.28% difference from the correct value with gravity included.
56157_15_ch15_p001-038.indd 33
3/15/08 11:20:30 AM
34
15.57
Chapter 15
The sketch at the right gives a free-body diagram of the
2
positively charged sphere. Here, F1 = ke q r 2 is the attractive
force exerted by the negatively charged sphere and F2 = qE
is exerted by the electric field.
ΣFy = 0 ⇒ T cos 10° = mg or T =
mg
cos 10°
2
ΣFx = 0 ⇒ F2 = F1 + T sin 10°
or qE =
ke q
+ mg tann 10°
r2
At equilibrium, the distance between the two spheres is r = 2 ( L sin 10°). Thus,
E=
=
ke q
4 ( L sin 10°)
( 8.99 × 10
9
2
+
mg tan 10°
q
)(
N ⋅ m 2 C2 5.0 × 10 −8 C
4 ( 0.100 m ) sin 10° 
2
or the needed electric field strength is
15.58
−3
(
)(
)
kg 9.80 m s 2 tan 10°
5.0 × 10 −8 C
)
E = 4.4 × 10 5 N C
(a)
At any point on the x-axis in the range 0 < x < 1.00 m, the contributions made to the
resultant electric field by the two charges are both in the positive x direction. Thus, it is not
possible for these contributions to cancel each other and yield a zero field.
(b)
Any point on the x-axis in the range x < 0 is located closer to the larger magnitude charge
( q = 5.00 µC) than the smaller magnitude charge ( q = 4.00 µC). Thus, the contribution to
the resultant electric field by the larger charge will always have a greater magnitude than the
contribution made by the smaller charge. It is not possible for these contributions to cancel to
give a zero resultant field.
(c)
If a point is on the x-axis in the region x > 1.00 m, the contributions made by the two charges
are in opposite directions. Also, a point in this region is closer to the smaller magnitude
charge than it is to the larger charge. Thus, there is a location in this region where the contributions of these charges to the total field will have equal magnitudes and cancel each other.
(d)
When the contributions by the two charges cancel each other, their magnitudes must be
equal. That is,
ke
( 5.00 µC) = k ( 4.00 µC )
e
x2
( x − 1.00 m )2
Thus, the resultant field is zero at
56157_15_ch15_p001-038.indd 34
) + ( 2 .0 × 10
or
x=
x − 1.00 m = +
4
x
5
1.00 m
= + 9.47 m
1− 4 5
3/15/08 11:20:30 AM
Electric Forces and Electric Fields
15.59
35
We assume that the two spheres have equal charges, so the
repulsive force that one exerts on the other has magnitude
Fe = ke q 2 r 2 .
From Figure P15.59 in the textbook, observe that the distance
separating the two spheres is
r = 3.0 cm + 2 ( 5.0 cm ) sin 10°  = 4.7 cm = 0.047 m
From the free-body diagram of one sphere given above, observe that
ΣFy = 0 ⇒ T cos 10° = mg or T = mg cos10°
and
 mg 
sin 10° = mg tan10°
ΣFx = 0 ⇒ Fe = T sin 10° = 
 cos 10° 
Thus, ke q 2 r 2 = mg tan 10°
or
q=
( 0.015 kg ) ( 9.8
)
m s 2 ( 0.047 m ) tan 10°
2
8.99 × 10 9 N ⋅ m 2 C2
q = 8.0 × 10 −8 C or q ~ 10 −7 C
giving
15.60
mgr 2 tan 10°
=
ke
The charges on the spheres will be equal in magnitude and opposite in sign. From F = ke q 2 r 2,
this charge must be
q=
F ⋅ r2
=
ke
(1.00 × 10 N ) (1.00 m )
4
8.99 × 10 9 N ⋅ m 2 C2
2
= 1.05 × 10 −3 C
The number of electrons transferred is
n=
q 1.05 × 10 −3 C
=
= 6.59 × 1015
e 1.60 × 10 −19 C
The total number of electrons in 100 g of silver is
electrons  

23 atoms   1 mole 
N =  47
(100 g ) = 2 .62 × 10 25

  6.02 × 10

atom  
mole   107.87 g 
Thus, the fraction transferred is
n 6.59 × 1015
=
= 2 .51 × 10 −10 (that is, 2.51 out of every 10 billion).
N 2 .62 × 10 25
56157_15_ch15_p001-038.indd 35
3/15/08 11:20:31 AM
36
15.61
Chapter 15
Because of the spherical symmetry of the
charge distribution, any electric field present will
be radial in direction. If a field does exist at
distance R from the center, it is the same as if the
net charge located within r ≤ R were concentrated as a point charge at the center of the inner
sphere. Charge located at r > R does not contribute to the field at r = R .
(a)
At r = 1.00 cm, E = 0 since static electric
fields cannot exist within conducting materials.
(b)
The net charge located at r ≤ 3.00 cm is
Q = +8.00 µ C.
Thus, at r = 3.00 cm,
E=
=
keQ
r2
( 8.99 × 10
9
)
N ⋅ m 2 C2 ( 8.00 × 10 −6 C
( 3.00 × 10
−2
m
)
2
)=
7.99 × 10 7 N C ( outward )
(c)
At r = 4.50 cm, E = 0 since this is located within conducting materials.
(d)
The net charge located at r ≤ 7.00 cm is Q = + 4.00 µC.
Thus, at r = 7.00 cm,
E=
=
56157_15_ch15_p001-038.indd 36
keQ
r2
( 8.99 × 10
9
)
N ⋅ m 2 C2 ( 4.00 × 10 −6 C
( 7.00 × 10
−2
m
)
2
)=
7.34 × 10 6 N C ( outward )
3/15/08 11:20:31 AM
Electric Forces and Electric Fields
15.62
37
Consider the free-body diagram of the rightmost charge given below.
ΣFy = 0 ⇒ T cos θ = mg
T = mg cos θ
or
and ΣFx = 0 ⇒ Fe = T sin θ = ( mg cos θ ) sin θ = mg tan θ
But, Fe =
ke q 2 ke q 2
ke q 2
ke q 2
5 ke q 2
+ 2 =
2 +
2 =
2
r1
r2
( L sin θ ) ( 2 L sin θ ) 4 L2 sin 2 θ
4 L2 mg sin 2 θ tan θ
5 ke
5 ke q 2
= mg tan θ or q =
4 L2 sin 2 θ
Thus,
If θ = 45°, m = 0.10 kg, and L = 0.300 m then
(
(a)
(
5 8.99 × 10 N ⋅ m C2
9
2
)
q = 2.0 × 10 −6 C = 2.0 µC
or
15.63
)
4 ( 0.300 m ) ( 0.10 kg ) 9.80 m s 2 sin 2 ( 45° ) tan ( 45° )
2
q=
When an electron (negative charge) moves distance ∆x in the direction of an electric field,
the work done on it is
W = Fe ( ∆x ) cos θ = eE ( ∆x ) cos 180° = − eE ( ∆x )
)
(
From the work-energy theorem Wnet = KE f − KEi with KE f = 0, we have
− eE ( ∆x ) = − KEi , or E =
(b)
1.60 × 10 −17 J
KEi
=
= 1.00 × 10 3 N C
e ( ∆x ) (1.60 × 10 −19 C ( 0.100 m )
)
The magnitude of the retarding force acting on the electron is Fe = eE , and Newton’s second
law gives the acceleration as a = − Fe m = − eE m . Thus, the time required to bring the electron to rest is
t=
v − v0 0 − 2 ( KEi ) m
=
=
− eE m
a
2 m ( KEi )
eE
or
t=
(c)
(1.60 × 10
−19
)
C (1.00 × 10
3
) = 3.37 × 10
N C)
−8
s = 33.7 ns
After bringing the electron to rest, the electric force continues to act on it causing the
electron to accelerate in the direction opposite to the field at a rate of
a =
56157_15_ch15_p001-038.indd 37
)
2 ( 9.11 × 10 −31 kg (1.60 × 10 −17 J
)
)
−19
3
eE (1.60 × 10 C (1.00 × 10 N C
=
= 1.76 × 1014 m s 2
m
9.11 × 10 −31 kg
3/15/08 11:20:32 AM
38
15.64
Chapter 15
(a)
The acceleration of the protons is downward (in the direction of the field) and
ay =
)
−19
Fe eE (1.60 × 10 C ( 720 N C )
=
=
= 6.90 × 1010 m s 2
m m
1.67 × 10 −27 kg
The time of flight for the proton is twice the time required to reach the peak of the arc, or
 v0 y  2 v sin θ
0
t = 2t peak = 2 
=
a
ay
 y
The horizontal distance traveled in this time is
 2 v sin θ  v 2 sin 2θ
0
R = v0 x t = ( v0 cos θ )  0
=
a
ay


y
Thus, if R = 1.27 × 10 −3 m, we must have
sin 2θ =
ay R
v
2
0
( 6.90 × 10
10
=
)
m s 2 (1.27 × 10 −3 m
( 9 550
m s)
2
) = 0.961
giving 2θ = 73.9° or 2θ = 180° − 73.9° = 106.1°.
Hence, θ = 37.0° or 53.0° .
(b)
56157_15_ch15_p001-038.indd 38
The time of flight for each possible angle of projection is:
For θ = 37.0°: t =
2 v0 sin θ 2 ( 9 550 m s ) sin 37.0°
=
= 1.66 × 10 −7 s
6.90 × 1010 m s 2
ay
For θ = 53.0°: t =
2 v0 sin θ 2 ( 9 550 m s ) sin 53.0°
=
= 2.21 × 10 −7 s
6.90 × 1010 m s 2
ay
3/15/08 3:25:02 PM