Ch. 21: 7, 11, 13, 16, 32, 34, 40, 47, 48, 49, 58, 59, 62
7.
Since the magnitude of the force is inversely proportional to the square of the separation
1
distance, F ∝ 2 , if the force is tripled, the distance has been reduced by a factor of 3 .
r
r
8.45 cm
r= 0 =
= 4.88 cm
3
3
Let one of the charges be q , and then the other charge is QT − q. The force
11.
(a)
between the
charges is FE = k
q ( QT − q )
2
=
k
2
( qQ
T
)
− q2 . To find the maximum and minimum
r
r
force, set the first derivative equal to 0. Use the second derivative test as well.
k
dFE k
2
= ( QT − 2q ) = 0 → q = 12 QT
FE = 2 ( qQT − q ) ;
dq r 2
r
d 2 FE
dq
2
=−
2k
r2
< 0 → q = 12 QT gives ( FE ) max
So q1 = q2 = 12 QT gives the maximum force.
( b)
If one of the charges has all of the charge, and the other has no charge, then the
force between
them will be 0, which is the minimum possible force. So q1 = 0, q2 = QT gives the
minimum force.
13. The forces on each charge lie along a line connecting the charges. Let the
variable d represent the length of a side of the triangle. Since the triangle
is equilateral, each angle is 60o. First calculate the magnitude of each
individual force.
F12 = k
Q1Q2
d2
(
= 8.988 × 10 N ⋅ m C
9
2
2
)
( 7.0 × 10 C)(8.0 × 10 C)
−6
F13 = k
d2
d
−6
(1.20 m) 2
r
F23
(
= 8.988 × 10 N ⋅ m C
9
2
2
)
( 7.0 × 10 C)( 6.0 × 10 C )
−6
−6
(1.20 m ) 2
= 0.2622 N
F23 = k
Q2Q3
d2
(
= 8.988 ×10 N ⋅ m C
9
2
2
)
(8.0 ×10 C)( 6.0 ×10 C) = 0.2996 N = F
−6
(1.20 m )
−6
2
r
F12
Q1
d
Q2
r
F21
= 0.3495 N
Q1Q3
r
F13
32
Now calculate the net force on each charge and the direction of that net force, using
components.
Q3
d
r
F32
r
F31
F1 x = F12 x + F13 x = − ( 0.3495 N ) cos 60o + ( 0.2622 N ) cos 60o = −4.365 × 10−2 N
F1 y = F12 y + F13 y = − ( 0.3495 N ) sin 60o − ( 0.2622 N ) sin 60o = −5.297 × 10−1 N
θ1 = tan −1
F1 = F12x + F12y = 0.53N
F1 y
= tan −1
F1 x
−5.297 × 10−1 N
−4.365 × 10−2 N
= 265°
F2 x = F21 x + F23 x = ( 0.3495 N ) cos 60o − ( 0.2996 N ) = −1.249 × 10−1 N
F2 y = F21 y + F23 y = ( 0.3495 N ) sin 60o + 0 = 3.027 × 10−1 N
F2 y
θ 2 = tan −1
F2 = F22x + F22y = 0.33N
F2 x
= tan −1
3.027 × 10−1 N
−1.249 × 10−1 N
= 112°
16. Determine the force on the upper right charge, and then use the symmetry of the
configuration to determine the force on the other charges.
The force at the upper right corner of the square is the vector sum of the
forces due to the other three charges. Let the variable d represent the
0.100 m length of a side of the square, and let the variable Q represent
the 4.15 mC charge at each corner.
r
F41
Q1
r
F42
Q4
r
F43
d
F41 = k
F42 = k
F43 = k
Q
d
2
→ F41x = −k
2
Q
2
2d
2
Q2
→ F42 x = k
Q
2
, F41 y = 0
d2
Q
2d
2
2Q
cos45o = k
→ F43 x = 0 , F43 y = −k
Q3
Q2
2
4d
2
2
, F42 y = k
2Q
2
4d 2
Q2
d2
d2
Add the x and y components together to find the total force, noting that F4 x = F4 y .
F4 x = F41x + F42 x + F43 x = −k
F4 = F + F = k
2
4x
2
4y
Q2
d2
d2
+k
( 0.64645)
2Q 2
4d 2
+0 = k
2 =k
Q2
d2
(
9
2
Q2 
2
Q2
d2 
4 
d2
 −1 +
 = −0.64645k
= F4 y
( 0.9142 )
( 4.15 × 10 C) ( 0.9142) = 1.42 × 10 N
C )
2
−3
= 8.988 × 10 N ⋅ m
θ = tan −1
Q2
2
7
( 0.100 m )
2
F4 y
= 225o from the x-direction, or exactly towards the center of the square.
F4 x
For each charge, there are two forces that point towards the adjacent corners, and one force
that points away from the center of the square. Thus for each charge, the net force will be
the magnitude of 1.42 × 107 N and will lie along the line from the charge inwards towards
the center of the square.
32.
The field due to the negative charge will point towards
the negative charge, and the field due to the positive charge will point towards the negative
charge. Thus the magnitudes of the two fields can be added together to find the charges.
Enet = 2 EQ = 2k
Q
( l / 2)
2
=
8kQ
l
→ Q=
2
El 2
=
8k
( 586 N C)( 0.160 m)2
(
8 8.988 × 10 N ⋅ m C
9
2
34. The field at the center due to the two −27.0µC negative charges
on opposite corners (lower right and upper left in the diagram)
will cancel each other, and so only the other two charges need to
be considered. The field due to each of the other charges will
point directly toward the charge. Accordingly, the two fields are
in opposite directions and can be combined algebraically.
Q
Q
Q −Q
E = E1 − E2 = k 2 1 − k 2 2 = k 1 2 2
l 2
l 2
l 2
(
= 8.988 × 109 N ⋅ m2 C2
)(
38.6 − 27.0) × 10 C
2
)
= 2.09 × 10−10 C
Q2 = −27.0 µ C
Q2
r
E1
r
E2
l
Q2
Q1 = −38.6 µ C
−6
( 0.525m )
2
2
= 7.57 × 106 N C, towards the − 38.6µC charge
40. Consider Example 21-9. We use the result from this example, but
shift the center of the ring to be at x = 12 l for the ring on the right,
and at x = − 12 l for the ring on the left. The fact that the original
expression has a factor of x results in the interpretation that the sign
of the field expression will give the direction of the field. No special
consideration needs to be given to the location of the point at which
the field is to be calculated.
y
R
R
1
2
l
O
1
2
x
l
r r
r
E = Eright + Eleft
=
Q ( x − 12 l )
1
4πε 0 ( x − 1 l ) 2 + R 2 
2

3/ 2

Q ( x + 12 l )
ˆi + 1
ˆi
3/ 2
2
2
4πε 0 ( x + 1 l ) + R 
2




Q 
( x − 12 l )
( x + 12 l )

= ˆi
+


3
/
2
3
/
2
4πε 0  ( x − 1 l ) 2 + R 2 
( x + 12 l )2 + R 2  
2





47. If we consider just one wire, then from the answer to problem 46, we
would have the following. Note that the distance from the wire to the
point in question is x = z 2 + ( l 2) .
2
Ewire =
λ
2πε 0
z + ( l 2)
( 4  z
2
wire
2
2
θ
)
z2 + ( l 2)
1/ 2
+ ( l 2 )  + l
But the total field is not simply four times the above expression,
because the fields due to the four wires are not parallel to each other.
Consider a side view of the problem. The two dots represent two
2
Eleft
wire
l
2
Eright θ
z
l 2
l 2
2
parallel wires, on opposite sides of the square. Note that only the vertical component of the
field due to each wire will actually contribute to the total field. The horizontal components
will cancel.
z
Ewire = 4 ( Ewire ) cos θ = 4 ( Ewire )
2
z2 + ( l 2)
Ewire

λ
= 4
 2πε 0

z2 + ( l 2)
2
(

z

1/ 2
2
 z 2 + ( l 2)2
4  z 2 + ( l 2 )  + l 2

l
)
8λ l z
=
πε 0 ( 4 z 2 + l 2 )( 4 z 2 + 2 l 2 )
1/ 2
The direction is vertical, perpendicular to the loop.
48. From the diagram, we see that the x components of the two fields will cancel each other at
the point P. Thus the net electric field will be in
+Q
the negative y-direction, and will be twice the ycomponent of either electric field vector.
a
kQ
Enet = 2 E sin θ = 2 2
sin θ
2
x
θ
x +a
r
2kQ
a
E− Q
= 2
a
1/ 2
2
x + a x2 + a2
(
=
)
−Q
2kQa
(x
2
+ a2
)
3/ 2
in the negative y direction
49. Select a differential element of the arc which makes an
angle of θ with the x axis. The length of this element
is Rd θ , and the charge on that element is dq = λ Rdθ .
r
The magnitude of the field produced by that element is
dEbottom
1 λ Rdθ
dE =
. From the diagram, considering
4πε 0 R 2
r
pieces of the arc that are symmetric with respect to the x
dE top
axis, we see that the total field will only have an x
component. The vertical components of the field due to
symmetric portions of the arc will cancel each other.
So we have the following.
1 λ Rdθ
dE horizontal =
cos θ
4πε 0 R 2
θ0
Ehorizontal =
r
EQ
∫
−θ0
1
4πε 0
cos θ
λ Rdθ
R2
θ0
Rdθ
R
θ
θ0
λ
λ
2λ sin θ 0
=
cos θ dθ =
[sin θ 0 − sin ( −θ0 )] =
∫
4πε 0 R −θ
4πε 0 R
4πε 0 R
0
x
The field points in the negative x direction, so E = −
2λ sin θ 0 ˆ
i
4πε 0 R
58.
(a)
The electron will experience a force in the opposite direction to the electric field.
Since the
electron is to be brought to rest, the electric field must be in the same direction as the
initial velocity of the electron, and so is to the right .
(b) Since the field is uniform, the electron will experience a constant force, and therefore
have a constant acceleration. Use constant acceleration relationships to find the field
strength.
qE
qE
∆x →
F = qE = ma → a =
v 2 = v02 + 2a∆x = v02 + 2
m
m
E=
(
m v 2 − v02
2q∆x
) = −mv
( 9.109 ×10 kg )( 7.5 × 10 m s)
=−
2q∆x
2 ( −1.602 × 10 C) ( 0.040 m)
−31
2
0
5
−19
2
= 40 N C
( 2 sig. fig.)
59. The angle is determined by the velocity. The x component of the velocity is constant. The
time to pass through the plates can be found from the x motion. Then the y velocity can be
found using constant acceleration relationships.
x = v0t → t =
tan θ =
vy
vx
−
=
x
v0
eE x
m v0
v0
; vy = vy0 + ayt = −
=−
eEx
mv02
eE x
m v0
(1.60 × 10 C)( 5.0 × 10 N C) ( 0.049 m) = −.4303
( 9.11 × 10 kg )(1.00 × 10 m s )
−19
=−
3
−31
7
2
→
θ = tan −1 ( −0.4303) = −23°
62. (a) The dipole moment is given by the product of the positive charge and the separation
distance.
(
)(
)
p = Ql = 1.60 × 10−19 C 0.68 × 10−9 m = 1.088 × 10−28 C m ≈ 1.1 × 10−28 C m
(b) The torque on the dipole is given by Eq. 21-9a.
τ = pE sin θ = (1.088 × 10−28 C m )( 2.2 × 104 N C ) ( sin 90° ) = 2.4 × 10−24 C m
(
)(
)
(c) τ = pE sin θ = 1.088 × 10−28 C m 2.2 × 104 N C ( sin 45° ) = 1.7 × 10−24 N m
(d) The work done by an external force is the change in potential energy. Use Eq. 21-10.
W = ∆U = ( − pE cos θfinal ) − ( − pE cos θinitial ) = pE ( cos θinitial − cos θfinal )
(
)(
)
= 1.088 × 10−28 C m 2.2 × 104 N C [1 − ( −1) ] = 4.8 × 10−24 J