Halliday Vol 8
Pesquisas Acadêmicas: Halliday Vol 8. Pesquise 860.000+ trabalhos acadêmicosPor: • 27/9/2014 • 9.757 Palavras (40 Páginas) • 634 Visualizações
1. (a) With aunderstood to mean the magnitude of acceleration, Newton’s second and
third laws lead to
ma ma m 22 11 2
7
7
63 10 70
90
4910 = =
×
=×
−
−
..
.
.
kg m s
ms
kg.
2
2
chch
(b) The magnitude of the (only) force on particle 1 is
Fma kqq
r
q
== = × 11
12
2
9
2
2
899 10
0 0032
.
.
. ch
Inserting the values for m1and a1(see part (a)) we obtain |q| = 7.1 ×10
–11
C.
2. The magnitude of the mutual force of attraction at r= 0.120 m is
Fkqq
r
==×××=
−− 12
2
9
66
2
899 10
3 00 10 150 10
0120
281 .
..
.
. chchchN.
3. Eq. 21-1 gives Coulomb’s Law, Fkqq
r
=
12
2
, which we solve for the distance:
()()() 922 6 6
12 8.99 10 N m C 26.0 10 C 47.0 10 C ||| |
1.39m.
5.70N
kq q
r
F
−− ×⋅ × × == =
4. The fact that the spheres are identical allows us to conclude that when two spheres are
in contact, they share equal charge. Therefore, when a charged sphere (q) touches an
uncharged one, they will (fairly quickly) each attain half that charge (q/2). We start with
spheres 1 and 2 each having charge q and experiencing a mutual repulsive force
22/ Fkqr = . When the neutral sphere 3 touches sphere 1, sphere 1’s charge decreases to
q/2. Then sphere 3 (now carrying charge q/2) is brought into contact with sphere 2, a total
amount of q/2 + qbecomes shared equally between them. Therefore, the charge of sphere
3 is 3q/4 in the final situation. The repulsive force between spheres 1 and 2 is finally
2
22 ( / 2)(3 / 4) 3 3 ' 3
0.375.
88 8 qq q F Fk k F rr F ′=== ==
5. The magnitude of the force of either of the charges on the other is given by
F
qQ q
r
=
− 1
4 0
2
πε
bg
where ris the distance between the charges. We want the value of qthat maximizes the
function f(q) = q(Q– q). Setting the derivative df/dqequal to zero leads to Q– 2q= 0, or
q= Q/2. Thus, q/Q= 0.500.
6. For ease of presentation (of the computations below) we assume Q > 0 and q< 0
(although the final result does not depend on this particular choice).
(a) The x-component of the force experienced by q1 = Qis
()()
() ()() 1 2 22 00|| 1||/||cos 45 1
4422 2
x
QQ q Q Qq Q q
F
aa a
§·§· ¨¸ =− °+ = −+ ¨¸ ¨¸©¹ ¨¸ ©¹ πε ε π
which (upon requiring F1x= 0) leads to /| | 2 2 Qq= , or / 2 2 2.83. Qq=− =−
(b) The y-component of the net force on q2= qis
() ()() 22 2 2 22 00|| 1|| || 1 sin 45
44||22 2
y
qQ qqQ F
aaq a
§·§· ¨¸ =°−=−¨¸ ¨¸©¹ ¨¸ ©¹ πε πε
which (if we demand F2y= 0) leads to /1/22 Qq=− . The result is inconsistent with
that obtained in part (a). Thus, we are unable to construct an equilibrium configuration
with this geometry, where the only forces present are given by Eq. 21-1.
7. The force experienced by q3is
31 32 34 3313234 222
0
|||| |||| |||| 1
ˆˆˆˆ j(cos45isin45j)i 4 (2)
qq qq qq FF F F aaa πε
§· =++= − + °+ °+ ¨¸ ©¹ GG G G
(a) Therefore, the x-component of the resultant force on q3is
()()2
7
9 3 2
3422 0
21.0 10 || || 1
| | 8.99 10 2 0.17N.
4 (0.050) 22 22
x
q q
Fqa πε
−
× §· §· =+=× += ¨¸ ¨¸ ©¹ ©¹
(b) Similarly, the y-component of the net force on q3 is
()()2
7
9 3 2
3122 0
21.0 10 || || 1
| | 8.99 10 1 0.046N.
4 (0.050) 22 22
y
q q
Fqa πε
−
× §· §· = −+=× −+=−
¨¸ ¨¸ ©¹ ©¹
8. (a) The individual force magnitudes (acting on Q) are, by Eq. 21-1,
k
a
k
a
aa 1
2
2
2
2
2
−−
=
− bgbg
which leads to |q1| = 9.0 |q2|. Since Qis located between q1and q2, we conclude q1and q2
are like-sign. Consequently, q1/q2= 9.0.
(b) Now we have
k
a
k
a
aa 1
3
2
2
2
3
2
2
−−
=
− bgbg
which yields |q1| = 25 |q2|. Now, Qis not located between q1and q2, one of them must
push and the other must pull. Thus, they are unlike-sign, so q1/q2 = –25.
9. We assume the spheres are far apart. Then the charge distribution on each of them is
spherically symmetric and Coulomb’s law can be used. Let q1 and q2 be the original
charges. We choose the coordinate system so the force on q2is positive if it is repelled by
q1. Then, the force on q2is
F
r
k
r
a =− =−
1
4 0
12
2
12
2
πε
where r= 0.500 m. The negative sign indicates that the spheres attract each other. After
the wire is connected, the spheres, being identical, acquire the same charge. Since charge
is conserved, the total charge is the same as it was originally. This means the charge on
each sphere is (q1+ q2)/2. The force is now one of repulsion and is given by
F
r
k
r
b
qq qq
==+
++
1
440
22
2
122
2
12 12
πε
didibg.
We solve the two force equations simultaneously for q1and q2. The first gives the product
rF
k
a
12
2
2
9
12
0 500 0108
899 10
300 10 =− =−
×⋅ =− ×
−
..
.
., mN
Nm C
C 22
2
bgbg
and the second gives the sum
qq rF
k
b
12 6
2 2 0 500
0 0360
200 10 += = ×⋅ =×−
.
.
. m
N
8.99 10 N m C
C 922 bg
where we have taken the positive root (which amounts to assuming q1+ q2 ≥0). Thus, the
product result provides the relation
()12 2
2
1
3.00 10 C
q
q
−
−× =
which we substitute into the sum result, producing
q
q
1
12
1
6 300 10
200 10 −
×
=× −
−
.
.
C
C.
2
Multiplying by q1and rearranging, we obtain a quadratic equation
qq 1
261
12
2 00 10 3 00 10 0 −× −× = −− ... CC2
c h
The solutions are
q1
662
12
2 00 10 2 00 10 4 3 00 10
2
=
×±−× −−× −− − .. ..
CC C2
c h c h
If the positive sign is used, q1 = 3.00 ×10
–6
C, and if the negative sign is used,
6
1 1.00 10 C q
−
=− × .
(a) Using q2= (–3.00 ×10
–12
)/q1with q1= 3.00 ×10
–6
C, we get
6
2 1.00 10 C q
−
=− × .
(b) If we instead work with the q1= –1.00 ×10
–6
C root, then we find
6
2 3.00 10 C q
−
=× .
Note that since the spheres are identical, the solutions are essentially the same: one sphere
originally had charge –1.00 ×10
–6
C and the other had charge +3.00 ×10
–6
C.
What if we had not made the assumption, above, that q1+ q2 ≥ 0? If the signs of the
charges were reversed (so q1+ q2< 0), then the forces remain the same, so a charge of
+1.00 ×10
–6
C on one sphere and a charge of –3.00 ×10
–6
C on the other also satisfies
the conditions of the problem.
10. With rightwards positive, the net force on q3is
()13 23
31323 2 2
23 12 23
.
qq q q
FF F k kL LL =+= + +
We note that each term exhibits the proper sign (positive for rightward, negative for
leftward) for all possible signs of the charges. For example, the first term (the force
exerted on q3by q1) is negative if they are unlike charges, indicating that q3 is being
pulled toward q1, and it is positive if they are like charges (so q3would be repelled from
q1). Setting the net force equal to zero L23=L12 and canceling k, q3and L12leads to
112
2
0 4.00.
4.00
qqq
q
+= =−
11. (a) Eq. 21-1 gives
()()2
6
922 12
12 2 2
20.0 10 C
8.99 10 N m C 1.60N.
1.50m
Fkd
−
×
==×⋅ =
(b) A force diagram is shown as well as our choice of yaxis (the dashed line).
The y axis is meant to bisect the line between q2 and q3 in order to make use of the
symmetry in the problem (equilateral triangle of side length d, equal-magnitude charges
q1= q2= q3= q). We see that the resultant force is along this symmetry axis, and we
obtain
Fkq
d
y =
F
H
G
I
K
J °= 230277 2
2
cos . . N
12. (a) According to the graph, when q3 is very close to q1 (at which point we can
consider the force exerted by particle 1 on 3 to dominate) there is a (large) force in the
positive xdirection. This is a repulsive force, then, so we conclude q1has the same sign
as q3. Thus, q3is a positive-valued charge.
(b) Since the graph crosses zero and particle 3 is betweenthe others, q1must have the
same sign as q2, which means it is also positive-valued. We note that it crosses zero at r
= 0.020 m (which is a distance d= 0.060 m from q2). Using Coulomb’s law at that point,
we have
q1q3
4πεo r
2=
q3q2
4πεo d
2 q2=
©
¨
§
¹
¸
· d
2
r
2 q1 = 9.0q1,
or q2/q1= 9.0.
3
being pulled by one and pushed by the other (since q1and q2have different signs); in this
region this means the two force arrows on q3are in the same direction and cannot cancel.
It should also be clear that off-axis (with the axis defined as that which passes through the
two fixed charges) there are no equilibrium positions. On the semi-infinite region of the
axis which is nearest q2 and furthest from q1 an equilibrium position for q3 cannot be
found because |q1| < |q2| and the magnitude of force exerted by q2is everywhere (in that
region) stronger than that exerted by q1on q3. Thus, we must look in the semi-infinite
region of the axis which is nearest q1and furthest from q2, where the net force on q3has
magnitude
k
x
k
dx
13
2
23
2
−
+ b g
with d = 10 cm and x assumed positive. We set this equal to zero, as required by the
problem, and cancel kand q3. Thus, we obtain
q
x
q
dx
dx
x
q
q
1
2
2
2
2
2
1
03 −
+
=
+ F
H
G
I
K
J == bg
which yields (after taking the square root)
dx
x
x
d +
= =
−
≈ 3
31
14 cm
for the distance between q3and q1.
(b) As stated above, y= 0.
13. (a) There is no equilibrium position for q betweenthe two fixed charges, because it is
14. Since the forces involved are proportional to q, we see that the essential difference
between the two situations is Fa ∝ qB+ qC(when those two charges are on the same side)
versus Fb ∝ −qB+ qC(when they are on opposite sides). Setting up ratios, we have
Fa
Fb
=
qB+ qC
-qB+ qC
20.14
-2.877
=
1 + r
-1 + r
where in the last step we have canceled (on the left hand side) 10
−24
N from the numerator
and the denominator, and (on the right hand side) introduced the symbol r= qC/qB.
After noting that the ratio on the left hand side is very close to – 7, then, after a couple of
algebra steps, we are led to
71 81.333.
71 6
r
+
=== −
1 2
()( )()() 22 2 2 12 2 1 2 1
0.020 0.035 0.015 0.005 0.056 m. rxxyy =−+−=− − + − =
The magnitude of the force exerted by q1on q2is
()()() 966 12
21 22 12
8.99 10 3.0 10 4.0 10 || 35 N.
(0.056)
Fkr
−− ×× × == =
(b) The vector
G
F21
is directed towards q1and makes an angle θwith the +xaxis, where
1121
21
1.5 0.5
tan tan 10.3 10 .
2.0 3.5
yy
xx θ
−−§·− − §· == =−°≈−° ¨¸¨¸ −−−©¹ ©¹
(c) Let the third charge be located at (x3, y3), a distance rfrom q2. We note that q1, q2and
q3must be collinear; otherwise, an equilibrium position for any one of them would be
impossible to find. Furthermore, we cannot place q3on the same side of q2where we also
find q1, since in that region both forces (exerted on q2by q3and q1) would be in the same
direction (since q2is attracted to both of them). Thus, in terms of the angle found in part
(a), we have x3= x2– rcosθand y3= y2– rsinθ(which means y3> y2since θis negative).
The magnitude of force exerted on q2by q3is
2
23 2 3
|| Fkqqr = , which must equal that of
the force exerted on it by q1(found in part (a)). Therefore,
k
r
k
r
rrq
q
23
2
12
12
2 12
3
1
0 0645 = ==.cm .
Consequently, x3= x2– rcosθ= –2.0 cm – (6.45 cm) cos(–10°) = –8.4 cm,
(d) and y3= y2– rsinθ= 1.5 cm – (6.45 cm) sin(–10°) = 2.7 cm.
15. (a) The distance between qand qis
16. (a) For the net force to be in the +x direction, the y components of the individual
forces must cancel. The angle of the force exerted by the q1= 40 µC charge on
3
20 qCµ = is 45°, and the angle of force exerted on q3by Qis at –θwhere
θ=
F
H
G
I
K
J=° −
tan
.
.
.. 1 20
30
33 7
Therefore, cancellation of ycomponents requires
() ( )
13 3
22||
sin 45 sin
0.02 2 (0.030) (0.020)
qq Q q
kk θ °=
+
from which we obtain |Q| = 83 µC. Charge Qis “pulling” on q3, so (since q3> 0) we
conclude Q= –83 µC.
(b) Now, we require that the xcomponents cancel, and we note that in this case, the angle
of force on q3exerted by Qis +θ(it is repulsive, and Qis positive-valued). Therefore,
() ( )
13 3
22cos 45 cos
0.02 2 (0.030) (0.020)
qq Qq
kk θ °=
+
from which we obtain Q= 55.2 µC 55 C µ ≈ .
17. (a) If the system of three charges is to be in equilibrium, the force on each charge
must be zero. The third charge q3must lie between the other two or else the forces acting
on it due to the other charges would be in the same direction and q3 could not be in
equilibrium. Suppose q3is at a distance xfrom q, and L– xfrom 4.00q. The force acting
on it is then given by
() 33 3 2 2
0
4 1
4
qq qq
F
x Lx πε
§· =−¨¸−
©¹
where the positive direction is rightward. We require F3= 0 and solve for x. Canceling
common factors yields 1/x
2
= 4/(L– x)
2
and taking the square root yields 1/x= 2/(L– x).
The solution is x= L/3.
The force on qis
2
3
22 0
14.00.
4
q
qq q
F
xL ε
§· −
=+¨¸ ©¹ π
The signs are chosen so that a negative force value would cause qto move leftward. We
require Fq= 0 and solve for q3:
2
3
3 2
44 4 0.444
99q qx
qqLq =− =− =− =−
where x= L/3 is used. We may easily verify that the force on 4.00qalso vanishes:
()
()2 22 22 0
4 2 22222 00 0449 4 14 14 14 4 0
44494 q
q qq qq qq F
LLLLL Lx
§·§·− §· =+=+ =−= ¨¸¨¸¨¸ ¨¸ ¨¸− ©¹ ©¹ ©¹ ππ π εε ε.
(b) As seen above, q3is located at x= L/3. With L= 9.00 cm, we have x= 3.00 cm.
(c) Similarly, the ycoordinate of q3is y= 0.
18. (a) We note that cos(30º) =
1
2
3 , so that the dashed line distance in the figure is
2/3 rd= . We net force on q1due to the two charges q3and q4(with |q3| = |q4| = 1.60 ×
10
−19
C) on the yaxis has magnitude
13 13
22 00 || 33|| 2cos(30) 416 qq qq
rd πε πε
°= .
This must be set equal to the magnitude of the force exerted on q1by q2= 8.00 ×10
−19
C
= 5.00 |q3| in order that its net force be zero:
13 12
22 00
33|| ||
16 4 ( )
qq qq
dDd πε πε
=
+
D = d
©
¨
§
¹
¸
·
2
5
3 3
−1 = 0.9245 d.
Given d= 2.00 cm, then this leads to D= 1.92 cm.
(b) As the angle decreases, its cosine increases, resulting in a larger contribution from the
charges on the y axis. To offset this, the force exerted by q2must be made stronger, so
that it must be brought closer to q1 (keep in mind that Coulomb’s law is inversely
proportional to distance-squared). Thus, Dmust be decreased.
2
ρ= b/r, we have
qdq brdr brr r
r
== = − z z
421
2
2
2
1
2
π π c h.
With b= 3.0 µC/m
2
, r2= 0.06 m and r1= 0.04 m, we obtain q= 0.038 µC = 3.8 ×10
−8
C.
19. The charge dqwithin a thin shell of thickness dris ρ A drwhere A= 4πr. Thus, with
20. If θis the angle between the force and the x-axis, then
cosθ=
x
x
2
+ d
2
.
We note that, due to the symmetry in the problem, there is no ycomponent to the net
force on the third particle. Thus, Frepresents the magnitude of force exerted by q1or q2
on q3. Let e= +1.60 ×10
−19
C, then q1= q2= +2eand q3= 4.0eand we have
Fnet = 2F cosθ=
2(2e)(4e)
4πεo (x
2
+ d
2
)
x
x
2
+ d
2
=
4e
2
x
πεo (x
2
+ d
2
)
3/2.
(a) To find where the force is at an extremum, we can set the derivative of this expression
equal to zero and solve for x, but it is good in any case to graph the function for a fuller
understanding of its behavior – and as a quick way to see whether an extremum point is a
maximum or a miminum. In this way, we find that the value coming from the derivative
procedure is a maximum (and will be presented in part (b)) and that the minimum is
found at the lower limit of the interval. Thus, the net force is found to be zero at x= 0,
which is the smallest value of the net force in the interval 5.0 m ≥ x ≥0.
(b) The maximum is found to be at x= d/ 2 or roughly 12 cm.
(c) The value of the net force at x= 0 is Fnet = 0.
(d) The value of the net force at x= d/ 2 is Fnet = 4.9 × 10
−26
N.
F
r
k
q
r
==bgbg
4 0
2
2
2
πε
where qis the charge on either of them and ris the distance between them. We solve for
the charge:
qrF
k
==× ×
×⋅ =× −
−
−
50 10
37 10
899 10
32 10
10
9
9
19
.
.
.
. m
N
Nm C
C. 22 c h
(b) Let Nbe the number of electrons missing from each ion. Then, Ne= q, or
N
q
e
==×
×
=
−
−
32 10
16 10
2
9
19
.
.
.
C
C
21. (a) The magnitude of the force between the (positive) ions is given by
22. The magnitude of the force is
Fke
r
== ×⋅ F
H
G
I
K
J
×
×
=× −
−
−
2
2
9
19
2
10
2
9
899 10
160 10
282 10
289 10 .
.
.
.
Nm
C
C
m
N.
2
2
c h
ch
23. Eq. 21-11 (in absolute value) gives
n
q
e
==×
×
=×
−
−
10 10
16 10
63 10
7
19
11
.
.
.. C
C
24. (a) Eq. 21-1 gives
F=
×⋅ ×
×
=× −
−
−
8 99 10 100 10
100 10
899 10
9162
2
2
19
...
.
Nm C C
m
N.
22 c hc h
ch (b) If nis the number of excess electrons (of charge –eeach) on each drop then
n
q
e
=− =−
−×
×
=
−
−
100 10
160 10
625
16
19
.
.
.
C
C
25. The unit Ampere is discussed in §21-4. The proton flux is given as 1500 protons per
square meter per second, where each proton provides a charge of q= +e. The current
through the spherical area 4π R
2
= 4π(6.37 ×10
6
m)
2
= 5.1 ×10
14
m
2
would be
i=× ⋅
F
H
G
I
K
J ×=−
51 10 1500 16 10 0122
14 2
2
19
.... m
protons
sm
Cproton A c h c h
26. The volume of 250 cm
3
corresponds to a mass of 250 g since the density of water is
1.0 g/cm
3
. This mass corresponds to 250/18 = 14 moles since the molar mass of water is
18. There are ten protons (each with charge q= +e) in each molecule of H2O, so
()()() 23 19 7
14 14 6.02 10 10 1.60 10 C 1.3 10 C.
A
QNq −
== × × =×
27. Since the graph crosses zero, q1must be positive-valued: q1 = +8.00e. We note that it
crosses zero at r = 0.40 m. Now the asymptotic value of the force yields the magnitude
and sign of q2:
q1q2
4πεo r
2= F q2=
©
§
¹
·
1.5x 10
-25
kq1
r
2
= 2.086x 10
−18
C = 13e .
3 4
+y axis, where the symbol q is assumed to be a positive value. Similarly, d is the
(positive) distance from the origin q4= −on the −yaxis. If we take each angle θin the
figure to be positive, then we have tanθ= d/Rand cosθ= R/r(where ris the dashed line
distance shown in the figure). The problem asks us to consider θto be a variable in the
sense that, once the charges on the xaxis are fixed in place (which determines R), dcan
then be arranged to some multiple of R, since d = Rtanθ. The aim of this exploration is
to show that if qis bounded then θ (and thus d) is also bounded.
From symmetry, we see that there is no net force in the vertical direction on q2= –e
sitting at a distance Rto the left of the coordinate origin. We note that the net xforce
caused by q3and q4on the yaxis will have a magnitude equal to
2
q e
4πεo r
2cos(θ) =
2 q e cos(θ)
4πεo (R/cos(θ))
2=
2 q e cos
3
(θ)
4πεo R
2 .
Consequently, to achieve a zero net force along the x axis, the above expression must
equal the magnitude of the repulsive force exerted on q2by q1= –e. Thus,
2 q e cos
3
(θ)
4πεo R
2 =
e
2
4πεo R
2 q =
e
2 cos
3
(θ)
.
Below we plot q/eas a function of the angle (in degrees):
The graph suggests that q/e< 5 for θ< 60º, roughly. We can be more precise by solving
the above equation. The requirement that q ≤5e leads to
e
2 cos
3
(θ)
≤5e
1
(10)
1/3 ≤ cosθ
28. Let dbe the vertical distance from the coordinate origin to q= −qand q= −qon the
which yields θ ≤ 62.34º. The problem asks for “physically possible values,” and it is
reasonable to suppose that only positive-integer-multiple values of eare allowed for q. If
we let q= Ne, for N= 1 … 5, then θNwill be found by taking the inverse cosine of the
cube root of (1/2N).
(a) The smallest value of angle is θ1= 37.5º (or 0.654 rad).
(b) The second smallest value of angle is θ2= 50.95º (or 0.889 rad).
(c) The third smallest value of angle is θ3= 56.6º (or 0.988 rad).
29. (a) Every cesium ion at a corner of the cube exerts a force of the same magnitude on
the chlorine ion at the cube center. Each force is a force of attraction and is directed
toward the cesium ion that exerts it, along the body diagonal of the cube. We can pair
every cesium ion with another, diametrically positioned at the opposite corner of the cube.
Since the two ions in such a pair exert forces that have the same magnitude but are
oppositely directed, the two forces sum to zero and, since every cesium ion can be paired
in this way, the total force on the chlorine ion is zero.
(b) Rather than remove a cesium ion, we superpose charge –e at the position of one
cesium ion. This neutralizes the ion, and as far as the electrical force on the chlorine ion
is concerned, it is equivalent to removing the ion. The forces of the eight cesium ions at
the cube corners sum to zero, so the only force on the chlorine ion is the force of the
added charge.
The length of a body diagonal of a cube is 3a, where ais the length of a cube edge.
Thus, the distance from the center of the cube to a corner isda= 32 di. The force has
magnitude
Fke
d
ke
a
== =×⋅ ×
×
=×
−
−
−
2
2
2
2
922 192
9
2
9
34
8 99 10 160 10
34 040 10
19 10
b g
c hc h
bgch ...
.. Nm C C
m
N
Since both the added charge and the chlorine ion are negative, the force is one of
repulsion. The chlorine ion is pushed away from the site of the missing cesium ion.
30. (a) Since the proton is positively charged, the emitted particle must be a positron
(as opposed to the negatively charged electron) in accordance with the law of charge
conservation.
(b) In this case, the initial state had zero charge (the neutron is neutral), so the sum of
charges in the final state must be zero. Since there is a proton in the final state, there
should also be an electron (as opposed to a positron) so that Σq= 0.
number of neutrons, and the number of electrons are each conserved. Atomic numbers
(numbers of protons and numbers of electrons) and molar masses (combined numbers of
protons and neutrons) can be found in Appendix F of the text.
(a)
1
H has 1 proton, 1 electron, and 0 neutrons and
9
Be has 4 protons, 4 electrons, and 9 –
4 = 5 neutrons, so X has 1 + 4 = 5 protons, 1 + 4 = 5 electrons, and 0 + 5 – 1 = 4 neutrons.
One of the neutrons is freed in the reaction. X must be boron with a molar mass of 5 + 4
= 9 g/mol:
9
B.
(b)
12
C has 6 protons, 6 electrons, and 12 – 6 = 6 neutrons and
1
H has 1 proton, 1 electron,
and 0 neutrons, so X has 6 + 1 = 7 protons, 6 + 1 = 7 electrons, and 6 + 0 = 6 neutrons. It
must be nitrogen with a molar mass of 7 + 6 = 13 g/mol:
13
N.
(c)
15
N has 7 protons, 7 electrons, and 15 – 7 = 8 neutrons;
1
H has 1 proton, 1 electron,
and 0 neutrons; and
4
He has 2 protons, 2 electrons, and 4 – 2 = 2 neutrons; so X has 7 +
1 – 2 = 6 protons, 6 electrons, and 8 + 0 – 2 = 6 neutrons. It must be carbon with a molar
mass of 6 + 6 = 12:
12
C.
31. None of the reactions given include a beta decay, so the number of protons, the
32. We note that the problem is examining the force oncharge A, so that the respective
distances (involved in the Coulomb force expressions) between Band A, and between C
and A, do not change as particle B is moved along its circular path. We focus on the
endpoints (θ= 0º and 180º) of each graph, since they represent cases where the forces (on
A) due to Band Care either parallel or antiparallel (yielding maximum or minimum force
magnitudes, respectively). We note, too, that since Coulomb’s law is inversely
proportional to r² then the (if, say, the charges were all the same) force due to Cwould be
one-fourth as big as that due to B(since Cis twice as far away from A). The charges, it
turns out, are not the same, so there is also a factor of the charge ratio ξ(the charge of C
divided by the charge of B), as well as the aforementioned ¼ factor.That is, the force
exerted by Cis, by Coulomb’s law equal to ±¼ξmultiplied by the force exerted by B.
(a) The maximum force is 2F0and occurs when θ= 180º (Bis to the left of A, while Cis
the right of A). We choose the minus sign and write
2F0= (1 −¼ξ)F0 ξ= – 4 .
One way to think of the minus sign choice is cos(180º) = –1. This is certainly consistent
with the minimum force ratio (zero) at θ= 0º since that would also imply
0 = 1 + ¼ξ ξ= – 4 .
(b) The ratio of maximum to minimum forces is 1.25/0.75 = 5/3 in this case, which
implies
5
3
=
1 + ¼ξ
1 −¼ξ
ξ= 16 .
Of course, this could also be figured as illustrated in part (a), looking at the maximum
force ratio by itself and solving, or looking at the minimum force ratio (¾) at θ= 180º
and solving for ξ.
33. We note that, as result of the fact that the Coulomb force is inversely proportional to
r
2
, a particle of charge Qwhich is distance dfrom the origin will exert a force on some
charge qoat the origin of equal strength as a particle of charge 4Qat distance 2dwould
exert on qo. Therefore, q6= +8eon the –yaxis could be replaced with a +2ecloser to the
origin (at half the distance); this would add to the q5= +2ealready there and produce +4e
below the origin which exactly cancels the force due to q2= +4eabove the origin.
Similarly, q4= +4eto the far right could be replaced by a +eat half the distance, which
would add to q3= +e already there to produce a +2e at distance d to the right of the
central charge q7. The horizontal force due to this +2eis cancelled exactly by that of q1=
+2eon the –xaxis, so that the net force on q7is zero.
34. For the Coulomb force to be sufficient for circular motion at that distance (where r=
0.200 m and the acceleration needed for circular motion is a = v
2
/r)the following
equality is required:
Q q
4πεo r
2= −
m v
2
r
.
With q= 4.00 × 10
−6
C, m= 0.000800 kg, v= 50.0 m/s, this leads to Q= −1.11 × 10
−5
C.
12
G
denotes the force on q1exerted by q2and
12 F be its magnitude.
(a) We consider the net force on q1.
12 F
G
points in the +xdirection since q1is attracted to
q2.
13 F
G
and
14 F
G
both point in the –xdirection sinceq1is repelled by q3and q4. Thus, using
d= 0.0200 m, the net force is
F1= F12– F13– F14=
2
0
2| |
4
ee
d πε
−
−
(2e)(e)
4πεo (2d)
2 −
(2e)(4e)
4πεo(3d)
2= + 3.52 × 10
−25
N ,
or
25
1
ˆ
(3.52 10 N)i. F
−
=× G
(b) We now consider the net force on q2. We note that
21 12 FF=−
GGpoints in the –x
direction, and
23 F
G
and
24 F
G
both point in the +xdirection. The net force is
23 24 21 222 000
4| | | | 2| |
0
4(2)4 4
ee ee ee FFF ddd πε πε πε
−−− +−= + − =
35. Let F
1
2
A , where
qA is the initial charge of sphere A. As a result of the second action, sphere Whas charge
1
2
(
1
2
qA −32e ).
As a result of the final action, sphere Wnow has charge equal to
1
2
[
1
2
(
1
2
qA −32e ) + 48e] .
Setting this final expression equal to +18e as required by the problem leads (after a
couple of algebra steps) to the answer: qA = +16e.
36. As a result of the first action, both sphere Wand sphere Apossess charge q
cosθ=
d2
d1
2
+ d2
2
.
Thus, using Coulomb’s law for F, we have
Fx= F cosθ=
q1q2
4πεo (d1
2
+ d2
2
)
d2
d1
2
+ d2
2= 1.31 × 10
−22
N .
37. If θis the angle between the force and the x axis, then
3 . In the initial (highly symmetrical) configuration,
the net force on the central bead is in the –ydirection and has magnitude 3Fwhere Fis
the Coulomb’s law force of one bead on another at distance d= 10 cm. This is due to the
fact that the forces exerted on the central bead (in the initial situation) by the beads on the
x axis cancel each other; also, the force exerted “downward” by bead 4 on the central
bead is four times larger than the “upward” force exerted by bead 2. This net force along
the yaxis does not change as bead 1 is now moved, though there is now a nonzero xcomponent Fx . The components are now related by
tan(30°) =
Fx
Fy
1
3
=
Fx
3F
which implies Fx= 3 F. Now, bead 3 exerts a “leftward” force of magnitude Fon the
central bead, while bead 1 exerts a “rightward” force of magnitude F′. Therefore,
F′ − F= 3 F. F′= ( 3 + 1) F.
The fact that Coulomb’s law depends inversely on distance-squared then implies
r
2
=
d
2
3 + 1
r=
d
3 + 1
where ris the distance between bead 1 and the central bead. Thus r= 6.05 cm.
(b) To regain the condition of high symmetry (in particular, the cancellation of xcomponents) bead 3 must be moved closer to the central bead so that it, too, is the
distance r(as calculated in part(a)) away from it.
38. (a) We note that tan(30°) = 1/
1
–9
9
2 80 10 C Q
−
=+ × is on the yaxis at y= –0.003 m. The force on particle 3 (which has a
charge of q= +18 ×10
–9
C) is due to the vector sum of the repulsive forces from Q1and
Q2. In symbols,
31 3 2 3
, FF F += GG Gwhere
|| ||
|| .
G G
Fkqq
r
Fkqq
r
31
31
31
2 32
32
32
2
==and
Using the Pythagorean theorem, we have r31= r32 = 0.005 m. In magnitude-angle
notation (particularly convenient if one uses a vector-capable calculator in polar mode),
the indicated vector addition becomes
()()() 3
0.518 37 0.518 37 0.829 0 . F=∠−°+∠°=∠° G
Therefore, the net force is 3
ˆ
(0.829 N)i F=
G
.
(b) Switching the sign of Q2 amounts to reversing the direction of its force on q.
Consequently, we have
()( )() 3
0.518 37 0.518 143 0.621 90 . F= ∠−°+ ∠− °= ∠−°
G
Therefore, the net force is 3
ˆ
(0.621 N)j F=−
G
.
39. (a) Charge Q = +80 ×10 C is on the y axis at y = 0.003 m, and charge
40. (a) Let xbe the distance between particle 1 and particle 3. Thus, the distance between
particle 3 and particle 2 is L – x. Both particles exert leftward forces on q3(so long as it is
on the line betweenthem), so the magnitude of the net force on q3is
Fnet= |F1 3
→
|+ |F2 3
→
|=
|q1q3|
4πεo x
2+
|q2q3|
4πεo (L− x)
2=
e
2
πεo ©
¨
§
¹
¸
· 1
x
2+
27
(L −x)
2
with the values of the charges (stated in the problem) plugged in. Finding the value of x
which minimizes this expression leads to x = ¼ L. Thus, x= 2.00 cm.
(b) Substituting x = ¼ Lback into the expression for the net force magnitude and using
the standard value for eleads to Fnet= 9.21 × 10
−24
N.
41. The individual force magnitudes are found using Eq. 21-1, with SI units (so
0.02 m a= ) and kas in Eq. 21-5. We use magnitude-angle notation (convenient if one
uses a vector-capable calculator in polar mode), listing the forces due to +4.00q, +2.00q,
and –2.00qcharges:
4 6010 180 2 3010 90 102 10 145 616 10 152
24 24 24 24
.. . . ×∠°+×∠−°+×∠−°=×∠−° −− − − c h c h c h c h
(a) Therefore, the net force has magnitude 6.16 ×10
–24
N.
(b) The direction of the net force is at an angle of –152° (or 208° measured
counterclockwise from the +xaxis).
42. The charge dq within a thin section of the rod (of thickness dx) is ρ A dx where
42 4.00 10 m A
−
=× and ρis the charge per unit volume. The number of (excess) electrons
in the rod (of length L= 2.00 m) is N= q/(–e) where eis given in Eq. 21-14.
(a) In the case where ρ= –4.00 ×10
–6
C/m
3
, we have
10
0
||
= 2.00 10
L qA AL Ndxee e == = × −−³
ρρ .
(b) With ρ= bx
2
(b= –2.00 ×10
–6
C/m
5
) we obtain
3
210 0
||
1.33 10 .
3
L bA b AL
Nxdxee ===× −
³
43. The magnitude of the net force on the q= 42 ×10
–6
C charge is
k
k
qq 1
2
2
2
028 044 .
||
.
+
where q1= 30 ×10
–9
C and |q2| = 40 ×10
–9
C. This yields 0.22 N. Using Newton’s
second law, we obtain
m
F
a
==×
=×− 022
10
22 10
3
6
.
.
N
100 m s
kg.
2
44. Let q1be the charge of one part and q2that of the other part; thus, q1 + q2 = Q= 6.0 µC.
The repulsive force between them is given by Coulomb’s law:
F=
q1q2
4πεo r
2=
q1(Q - q1)
4πεo r
2 .
If we maximize this expression by taking the derivative with respect to q1 and setting
equal to zero, we find q1 = Q/2 , which might have been anticipated (based on symmetry
arguments). This implies q2 = Q/2 also. With r= 0.0030 m and Q= 6.0 × 10
−6
C, we find
F=
(Q/2)(Q/2)
4πεo r
2 ≈9.0 ×10
3
N .
1 2 = qmust
exactly cancel the force of attraction caused by q4= –2Q. Consequently,
2
2 22 0 00|2 |
cos 45
4 4(2) 4 2
Qq Q Q Q
a aa πε πε πε
=°=
or q = Q/ 2 .This implies that / 1/ 2 0.707. qQ==
45. For the net force on q= +Qto vanish, the x force component due to q
46. We are looking for a charge qwhich, when placed at the origin, experiences
G
Fnet =0,
where
G G G G
FFFF net =++ 123.
The magnitude of these individual forces are given by Coulomb’s law, Eq. 21-1, and
without loss of generality we assume q> 0. The charges q1(+6 µC), q2(–4 µC), and q3
(unknown), are located on the +xaxis, so that we know
G
F1points towards –x,
G
F2
points
towards +x, and
G
F3points towards –xif q3> 0 and points towards +xif q3< 0. Therefore,
with r1= 8 m, r2= 16 m and r3= 24 m, we have
0
1
1
2
2
2
2
3
3
2
=− + − k
r
k
r
k
r
||
.
Simplifying, this becomes
0
6
8
4
16 24
22
3
2
=− + −
q
where q3is now understood to be in µC. Thus, we obtain q3= –45 µC.
47. There are two protons (each with charge q= +e) in each molecule, so
QNqA ==× × =× = −
602 10 2 160 10 19 10
2319 5
.. . c hbgc h C C 0.19 MC.
48. (a) Since the rod is in equilibrium, the net force acting on it is zero, and the net torque
about any point is also zero. We write an expression for the net torque about the bearing,
equate it to zero, and solve for x. The charge Q on the left exerts an upward force of
magnitude (1/4πε0) (qQ/h
2
), at a distance L/2 from the bearing. We take the torque to be
negative. The attached weight exerts a downward force of magnitude W, at a distance
/2 x L − from the bearing. This torque is also negative. The charge Qon the right exerts
an upward force of magnitude (1/4πε0) (2qQ/h
2
), at a distance L/2 from the bearing. This
torque is positive. The equation for rotational equilibrium is
22 00 1120.
42 24 2 qQ L L qQ L
Wx hh εε − §· −−+ = ¨¸π ©¹ π
The solution for xis
x
LqQ
hW
=+F
H
G
I
K
J
2
1
1
4
0
2
πε
.
(b) If FN is the magnitude of the upward force exerted by the bearing, then Newton’s
second law (with zero acceleration) gives
22 00
1120.
44 N
qQ qQ
WFhh πε πε
−− −=
We solve for hso that FN = 0. The result is
h
W
=
1
4
3
0 πε
.
49. Charge q1= –80 ×10
–6
C is at the origin, and charge q2= +40 ×10
–6
C is at x= 0.20
m. The force on q3= +20 ×10
–6
C is due to the attractive and repulsive forces from q1
and q2, respectively. In symbols,
G G G
FFF 33132 net =+, where
31 32 31 32 22 31 3 2
||and | | .
qq qq Fk Fk rr == GG
(a) In this case r31 = 0.40 m and r32 = 0.20 m, with
G
F31directed towards –x and
G
F32
directed in the +xdirection. Using the value of kin Eq. 21-5, we obtain
3net
ˆ
(89.9 N)i F =
G
.
(b) In this case r31 = 0.80 m and r32 = 0.60 m, with
G
F31directed towards –x and
G
F32
towards +x. Now we obtain
3net
ˆ
(2.50 N)i F =−
G
.
(c) Between the locations treated in parts (a) and (b), there must be one where
G
F3
0
net = .
Writing r31= x and r32= x – 0.20 m, we equate
G
F31
and
G
F32
, and after canceling
common factors, arrive at
||
.
.
q
x
q
x
1
2
2
2
02
=
− bg
This can be further simplified to
(.)
||
.
x
x
q
q
−
== 02 1
2
2
2
2
1
Taking the (positive) square root and solving, we obtain x= 0.683 m. If one takes the
negative root and ‘solves’, one finds the location where the net force wouldbe zero if q1
and q2were of like sign (which is not the case here).
(d) From the above, we see that y= 0.
50. We are concerned with the charges in the nucleus (not the “orbiting” electrons, if
there are any). The nucleus of Helium has 2 protons and that of Thorium has 90.
(a) Eq. 21-1 gives
()9 22 19 19
2
2
215 2 8.99 10 N m C (2(1.60 10 C))(90(1.60 10 C))
5.1 10 N.
(9.0 10 m)
q
Fkr
−− −
×⋅ × × == =× ×
(b) Estimating the helium nucleus mass as that of 4 protons (actually, that of 2 protons
and 2 neutrons, but the neutrons have approximately the same mass), Newton’s second
law leads to
a
F
m
== ×
×
=× −
51 10
4 167 10
77 10
2
27
28
.
.
.. N
kg
ms2
c h
51. Coulomb’s law gives
()922 192 2
22 152 0
8.99 10 N m C (1.60 10 C) |||| (3)
3.8 N.
49(2.610m) qqke F
rr ε
−
−
×⋅ × ⋅
=== = π×
52. (a) Since qA= –2.00 nC and qC= +8.00 nC Eq. 21-4 leads to
922 9 9 6
22 0
||| (8.99 10 Nm C ) ( 2.00 10 C)(8.00 10 C) |
| | 3.60 10 N.
4 (0.200 m)
AC
AC
F
d πε
−− − ×⋅ −× × == =× G
(b) After making contact with each other, both Aand Bhave a charge of
() 2.00 4.00
nC 3.00 nC.
22 AB qq§·−+− +
==− ¨¸ ©¹
When Bis grounded its charge is zero. After making contact with C, which has a charge
of +8.00 nC, Bacquires a charge of [0 + (–8.00 nC)]/2 = –4.00 nC, which charge Chas as
well. Finally, we have QA= –3.00 nC and QB= QC= –4.00 nC. Therefore,
922 9 9 6
22 0
||| (8.99 10 Nm C ) ( 3.00 10 C)( 4.00 10 C) |
| | 2.70 10 N.
4 (0.200 m)
AC
AC
F
d πε
−− − × ⋅ −× −×
== =× G
(c) We also obtain
922 9 9 6
22 0
||| (8.99 10 Nm C ) ( 4.00 10 C)( 4.00 10 C) |
| | 3.60 10 N.
4 (0.200 m)
BC
BC
F
d πε
−− − × ⋅ −× −×
== =× G
53. Let the two charges be q1and q2. Then q1+ q2= Q= 5.0 ×10
–5
C. We use Eq. 21-1:
()922
12
2
8.99 10 N m C
1.0N .
2.0m
qq ×⋅ =
We substitute q2= Q– q1and solve for q1using the quadratic formula. The two roots
obtained are the values of q1 and q2, since it does not matter which is which. We get
5
1.2 10 C
−
× and 3.8 ×10
–5
C. Thus, the charge on the sphere with the smaller charge is
5
1.2 10 C
−
× .
54. The unit Ampere is discussed in §21-4. Using ifor current, the charge transferred is
()() 46 2.5 10 A 20 10 s 0.50 C. qit −
== × × =
55. (a) Using Coulomb’s law, we obtain
()() ()
2
922 2
9 12
2 22 0
8.99 10 N m C 1.00C
8.99 10 N.
4 1.00m
qq kq
F
rr ε
×⋅ === =× π
(b) If r= 1000 m, then
()()
()
2
922 2
3 12
2 22 3
0
8.99 10 N m C 1.00C
8.99 10 N.
4 1.00 10 m
qq kq
F
rr ε
×⋅ === =× π ×
56. Keeping in mind that an Ampere is a Coulomb per second, and that a minute is 60
seconds, the charge (in absolute value) that passes through the chest is
| q | = ( 0.300
Coulomb
second
) ( 120 seconds) = 36.0 Coulombs.
This charge consists of a number Nof electrons (each of which has an absolute value of
charge equal to e). Thus,
N =
| q |
e
=
36.0 C
1.60x 10
-19
C
= 2.25 × 10
20
.
57. When sphere C touches sphere A, they divide up their total charge (Q/2 plus Q)
equally between them. Thus, sphere Anow has charge 3Q/4, and the magnitude of the
force of attraction between Aand Bbecomes
19
2
(3 /4)( /4)
4.68 10 N.
QQ Fk d
−
==×
58. In experiment 1, sphere C first touches sphere A, and they divided up their total
charge (Q/2 plus Q) equally between them. Thus, sphere Aand sphere Ceach acquired
charge 3Q/4. Then, sphere Ctouches Band those spheres split up their total charge (3Q/4
plus –Q/4) so that Bends up with charge equal to Q/4. The force of repulsion between A
and Bis therefore
1 2
(3 / 4)( / 4) QQ Fk d
=
at the end of experiment 1. Now, in experiment 2, sphere Cfirst touches Bwhich leaves
each of them with charge Q/8. When Cnext touches A, sphere Ais left with charge 9Q/16.
Consequently, the force of repulsion between Aand Bis
2 2
(9 /16)( /8) QQ Fk d
=
at the end of experiment 2. The ratio is
2
1
(9 /16)(1/8)
0.375.
(3 / 4)(1/ 4)
F
F
==
59. If the relative difference between the proton and electron charges (in absolute value)
were
e
pe−
=0 0000010 .
then the actual difference would be qqpe−=×−
16 10
25
..CAmplified by a factor of 29 ×
3 ×10
22
as indicated in the problem, this amounts to a deviation from perfect neutrality of
∆q=×× × = −
29 3 10 16 10 014
22 25
c hc h ..CC
in a copper penny. Two such pennies, at r= 1.0 m, would therefore experience a very
large force. Eq. 21-1 gives
Fkq
r
==× ∆ b g
2
2
8
17 10 ..N
60. With F= meg, Eq. 21-1 leads to
()() ()()2
922 19 2
2
2 31
8.99 10 N m C 1.60 10 C
9.11 10 kg 9.8m s e
ke
y
mg
−
−
×⋅ × == ×
which leads to y= ±5.1 m. We choose 5.1 m y=− since the second electron must be
below the first one, so that the repulsive force (acting on the first) is in the direction
opposite to the pull of Earth’s gravity.
61. Letting kq
2
/r
2
= mg, we get
()()() 922 19
27 2
8.99 10 N m C
1.60 10 C 0.119m.
1.67 10 kg 9.8 m s
k
rq
mg
−
−
×⋅ ==× = ×
62. The net charge carried by John whose mass is mis roughly
()
() 23 19
5
0.0001
(90kg)(6.02 10 molecules mol)(18 electron proton pairs molecule) (1.6 10 C)
0.0001
0.018 kg mol
8.7 10 C,
A
mN Ze
q
M
−
=
×× =
=×
and the net charge carried by Mary is half of that. So the electrostatic force between them
is estimated to be
()()()
52 922 18
2 2
2 (8.7 10 C)
8.99 10 N m C 4 10 N.
230m
Fkd
×
≈=×⋅ ≈×
Thus, the order of magnitude of the electrostatic force is
18
10 N .
63. (a) Eq. 21-11 (in absolute value) gives
n
q
e
== ×
×
=× −
−
200 10
160 10
125 10
6
19
13
.
.
.. C
C
electrons
(b) Since you have the excess electrons (and electrons are lighter and more mobile than
protons) then the electrons “leap” from you to the faucet instead of protons moving from
the faucet to you (in the process of neutralizing your body).
(c) Unlike charges attract, and the faucet (which is grounded and is able to gain or lose
any number of electrons due to its contact with Earth’s large reservoir of mobile charges)
becomes positively charged, especially in the region closest to your (negatively charged)
hand, just before the spark.
(d) The cat is positively charged (before the spark), and by the reasoning given in part (b)
the flow of charge (electrons) is from the faucet to the cat.
(e) If we think of the nose as a conducting sphere, then the side of the sphere closest to
the fur is of one sign (of charge) and the side furthest from the fur is of the opposite sign
(which, additionally, is oppositely charged from your bare hand which had stroked the
cat’s fur). The charges in your hand and those of the furthest side of the “sphere”
therefore attract each other, and when close enough, manage to neutralize (due to the
“jump” made by the electrons) in a painful spark.
64. The two charges are q= αQ(where αis a pure number presumably less than 1 and
greater than zero) and Q– q= (1 – α)Q. Thus, Eq. 21-4 gives
F
d
Q
d
=
−
=
− 1
4
1 1
4
0
2
2
0
2
ππ ε
αα αα
ε
b g b g c h b g
.
The graph below, of Fversus α, has been scaled so that the maximum is 1. In actuality,
the maximum value of the force is Fmax= Q
2
/16πε0 d
2
.
(a) It is clear that α=
1
2
= 0.5 gives the maximum value of F.
(b) Seeking the half-height points on the graph is difficult without grid lines or some of
the special tracing features found in a variety of modern calculators. It is not difficult to
algebraically solve for the half-height points (this involves the use of the quadratic
formula). The results are
1211 11 1 0.15 and 1 0.85.
2222 αα§· §· =− ≈ =+ ≈ ¨¸ ¨¸ ©¹ ©¹
Thus, the smaller value of αis
1
0.15 α= ,
(c) and the larger value of αis
2
0.85 α = .
65. (a) The magnitudes of the gravitational and electrical forces must be the same:
1
4 0
2
22 πε
q
r
G
mM
r
=
where q is the charge on either body, r is the center-to-center separation of Earth and
Moon, Gis the universal gravitational constant, Mis the mass of Earth, and mis the mass
of the Moon. We solve for q:
q GmM = 4
0 πε .
According to Appendix C of the text, M= 5.98 ×10
24
kg, and m= 7.36 ×10
22
kg, so
(using 4πε0= 1/k) the charge is
q=
×⋅ × × ×⋅ =×
−
6 67 10 7 36 10 5 98 10
899 10
57 10
11 22 24
9
13
....
.
N m kg kg kg
Nm C
C.
22
22
c hc hc h
(b) The distance r cancels because both the electric and gravitational forces are
proportional to 1/r
2
.
(c) The charge on a hydrogen ion is e= 1.60 ×10
–19
C, so there must be
q
e
=
×
×
=× −
57 10
16 10
36 10
13
19
32
.
.
.
C
C
ions.
Each ion has a mass of 1.67 ×10
–27
kg, so the total mass needed is
36 10 167 10 60 10
32 27 5
.. . ××=× −
c hc h kg kg.
66. (a) A force diagram for one of the balls is shown below. The force of gravity mg
G
acts
downward, the electrical force
G
Fe
of the other ball acts to the left, and the tension in the
thread acts along the thread, at the angle θto the vertical. The ball is in equilibrium, so its
acceleration is zero. The ycomponent of Newton’s second law yields Tcosθ– mg= 0
and the xcomponent yields Tsinθ– Fe= 0. We solve the first equation for Tand obtain T
= mg/cosθ. We substitute the result into the second to obtain mgtanθ– Fe= 0.
Examination of the geometry of Figure 21-43 leads to
tan . θ=
−
x
Lx
2
2
2
2
bg
If Lis much larger than x(which is the case if θis very small), we may neglect x/2 in the
denominator and write tanθ ≈ x/2L. This is equivalent to approximating tanθby sinθ. The
magnitude of the electrical force of one ball on the other is
F
q
x
e =
2
0
2
4πε
by Eq. 21-4. When these two expressions are used in the equation mg tanθ= Fe, we
obtain
mgx
L
q
x
x
qL
mg 2
1
420
2
2
2
0
13
≈ ≈
F
H
G
I
K
J
ππ εε
/
.
(b) We solve x
3
= 2kq
2
L/mg) for the charge (using Eq. 21-5):
()()() ()()
3
2
3
8
922
0.010 kg 9.8 m s 0.050 m
2.4 10 C.
2 2 8.99 10 N m C 1.20 m
mgx
q
kL
−
== =±×
×⋅
Thus, the magnitude is
8
||2.410C. q
−
=×
67. (a) If one of them is discharged, there would no electrostatic repulsion between the
two balls and they would both come to the position θ= 0, making contact with each other.
(b) A redistribution of the remaining charge would then occur, with each of the balls
getting q/2. Then they would again be separated due to electrostatic repulsion, which
results in the new equilibrium separation
() () 1/3
2 1/3 1/3
0
2 115.0 cm 3.1 cm.
244 qL xxmg ε
ªº§· §· ′=== = «»¨¸ ¨¸ π ©¹ ©¹ «» ¬¼
68. Regarding the forces on q3exerted by q1and q2, one must “push” and the other must
“pull” in order that the net force is zero; hence, q1 and q2 have opposite signs. For
individual forces to cancel, their magnitudes must be equal:
()()
13 23 22 12 23 23
|||| |||| qq q q kkLL L =
+
.
With 23 12
2.00 LL= , the above expression simplifies to
|| ||
.
qq 12
94= Therefore,
129/4 qq=− , or
12/ 2.25. qq=−
69. (a) The charge qplaced at the origin is a distance r from Q(which is the positive
charge on which the forces are being evaluated), and the charge qplaced at x = dis a
distance r´ from Q. Depending on what region Qis located in, the relation between r, r´
and dwill be either
r´ = r + d if Qis along the –xaxis (region A)
r´ = d – r if Qis between the charges (region B)
r´ = r – d if Qis at x> d(region C).
Since all charges in this problem are taken to be positive, then the net force in region A
will in the –x direction; its magnitude will consist of the individual force magnitudes
added together. In region C the net force will be in the +x direction and will consist
again of the individual force magnitudes added together. It is in region B where the
individual force magnitudes must be subtracted, and in order for the result to exhibit the
correct sign (positive when the net force F
→
should point in the +x direction, and so
forth), we must write
FB
→
=
q Q
4πεo r
2 −
q Q
4πεo r´
2=
q Q
4πεo r
2 −
q Q
4πεo (d – r)
2.
If we further adopt the notation suggested in the problem, then r= αdin regions Band C,
and r= −αd in region A.(since r must by definition be a positive number, yet α is
negative-valued in region A). Using this notation, too, it is clear that we can factor out a
common qQ/4πεod² from our expressions. For brevity we will use the notation
J=
q Q
4πεo d
2.
Then, using the observations noted above, we are able to write down the expressions for
the force in each region:
FA
→
= −J
©
¨
§
¹
¸
· 1
α
2+
1
(1− α)
2
FB
→
= J
©
¨
§
¹
¸
· 1
α
2 −
1
(1− α)
2
FC
→
= J
©
¨
§
¹
¸
· 1
α
2+
1
(α − 1)
2
(b) We set J=1 in our plot of the force, below.
70. The mass of an electron is m = 9.11 ×10
–31
kg, so the number of electrons in a
collection with total mass M= 75.0 kg is
N
M
m
==×
=× −
75 0
10
823 10 31
31
.
.
kg
9.11 kg
electrons.
The total charge of the collection is
qNe =− =− × × =− ×
−
8 23 10 160 10 132 10
31 19 13
.. . c hc h CC.
71. (a) If a (negative) charged particle is placed a distance x to the right of the +2q
particle, then its attraction to the +2qparticle will be exactly balanced by its repulsion
from the –5qparticle is we require
22 52 ()Lx x
=
+
which is obtained by equating the Coulomb force magnitudes and then canceling
common factors. Cross-multiplying and taking the square root, we obtain
2
5
x
Lx
=
+
which can be rearranged to produce
1.72
2
1
5
L
x L =≈−
(b) The y coordinate of particle 3 is y = 0.
...