Solutions
Manual
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,Chapter 21
1. THINK After the transfer, the charges on the two spheres are Q − q and q.
EXPRESS The magnitude of the electrostatic force between two charges of magnitudes
q1 and q2 and separated by distance r is given by Coulomb’s law (see Eq. 21.1.1):
q1q2
F =k ,
r2
where k = 1/40 = 8.99109 N m 2 /C2. In our case, q1 = Q − q and q2 = q, so the
magnitude of the force of either charge on the other is
1 q (Q − q )
F= .
40 r2
We want the value of q that maximizes the function f(q) = q(Q – q).
ANALYZE Setting the derivative df/dq equal to zero leads to Q – 2q = 0, or q = Q/2.
Thus, q/Q = 0.500.
LEARN The force between the two spheres is maximum when the total charge is
distributed evenly between them.
2. 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 F = kq2 /r2. 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 + q becomes 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
(q/2)(3q/4) 3 q2 3 F 3
F = k = k 2 = F = = 0.375.
r2 8 r 8 F 8
3. THINK The magnitude of the electrostatic force between two charges q1 and q2
separated by distance r is given by Coulomb’s law.
,CHAPTER 21 1025
EXPRESS Equation 21.1.1 gives Coulomb’s law,
q1 q2
F =k ,
r2
which can be used to solve for the distance:
r=
F
ANALYZE Substituting our values of q1 = 2.60 10−6 C, q2 = −47.0 10−6 C, and
k = 8.99 109 N m 2 /C2 , we find
r=
5.70 N
= 1.39 m.
LEARN The electrostatic force between two charges decreases as 1/r2. The same
inverse-square nature is also seen in the gravitational force between two masses.
4. The unit ampere is discussed in Module 21.4. Using i for current, we find that the
charge transferred is
q = it = (2.5 104 A)(20 10−6 s) = 0.50 C.
5. The magnitude of the mutual force of attraction at r = 0.120 m is
q1 q2
= ( 8.99 109 N m 2 /C2 (3.00 10−6 C)(1.50 10−6 C) = 2.81 N.
F =k
r2
) (0.120 m)2
6. (a) With a understood to mean the magnitude of acceleration, Newton’s second
and third laws lead to
m a =ma m =(
6.310−7 kg )( 7.0 m/s2 )
= 4.9 10−7 kg.
2 2 1 1 2 2
9.0 m/s
(b) The magnitude of the (only) force on particle 1 is
q q q
2
F = m1a1 = k
1 2
= ( 8.99 10 N m /C
9 2 2
) (0.0032 m)2 .
r2
Inserting the values for m1 and a1 (see part (a)), we obtain q = 7.110–11 C.
, 1026 CHAPTER 21
7. With rightward positive, the net force on q3 is
q2q3
F =F +F =k q1q3 +k .
3 13 23
( L12 + L23 )
2 2
L 23
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 q3 by 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 q3 would be repelled from
q1). Setting the net force equal to zero L23 = L12 and canceling k, q3, and L12 leads to
q1 q1
+ q2 = 0 = −4.00.
4.00 q2
8. 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 A and sphere C each
acquired charge 3Q/4. Then, sphere C touches B and those spheres split up their
total charge (3Q/4 plus –Q/4), so B ends up with charge equal to Q/4. The force of
repulsion between A and B is therefore
(3Q/4)(Q/4)
F1 = k
d2
at the end of experiment 1. Now, in experiment 2, sphere C first touches B, which
leaves each of them with charge Q/8. When C next touches A, sphere A is left with
charge 9Q/16. Consequently, the force of repulsion between A and B is
(9Q/16)(Q/8)
F2 = k
d2
at the end of experiment 2. The ratio is
F2 (9/16)(1/8)
= = 0.375.
F1 (3/4)(1/4)
9. THINK Since charges with opposite signs attract, the initial charge configurations
must be of opposite signs. Similarly, since charges with the same sign repel, the final
charge configurations must be with the same sign.
EXPRESS We assume that the spheres are far enough 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 that
the force on q2 is positive if it is repelled by q1. Then the force on q2 is
1 q1q2 q1q2
F =− = −k ,
a
4 0 r 2
r 2
Manual
Want to earn $1,500
extra per year?
,Chapter 21
1. THINK After the transfer, the charges on the two spheres are Q − q and q.
EXPRESS The magnitude of the electrostatic force between two charges of magnitudes
q1 and q2 and separated by distance r is given by Coulomb’s law (see Eq. 21.1.1):
q1q2
F =k ,
r2
where k = 1/40 = 8.99109 N m 2 /C2. In our case, q1 = Q − q and q2 = q, so the
magnitude of the force of either charge on the other is
1 q (Q − q )
F= .
40 r2
We want the value of q that maximizes the function f(q) = q(Q – q).
ANALYZE Setting the derivative df/dq equal to zero leads to Q – 2q = 0, or q = Q/2.
Thus, q/Q = 0.500.
LEARN The force between the two spheres is maximum when the total charge is
distributed evenly between them.
2. 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 F = kq2 /r2. 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 + q becomes 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
(q/2)(3q/4) 3 q2 3 F 3
F = k = k 2 = F = = 0.375.
r2 8 r 8 F 8
3. THINK The magnitude of the electrostatic force between two charges q1 and q2
separated by distance r is given by Coulomb’s law.
,CHAPTER 21 1025
EXPRESS Equation 21.1.1 gives Coulomb’s law,
q1 q2
F =k ,
r2
which can be used to solve for the distance:
r=
F
ANALYZE Substituting our values of q1 = 2.60 10−6 C, q2 = −47.0 10−6 C, and
k = 8.99 109 N m 2 /C2 , we find
r=
5.70 N
= 1.39 m.
LEARN The electrostatic force between two charges decreases as 1/r2. The same
inverse-square nature is also seen in the gravitational force between two masses.
4. The unit ampere is discussed in Module 21.4. Using i for current, we find that the
charge transferred is
q = it = (2.5 104 A)(20 10−6 s) = 0.50 C.
5. The magnitude of the mutual force of attraction at r = 0.120 m is
q1 q2
= ( 8.99 109 N m 2 /C2 (3.00 10−6 C)(1.50 10−6 C) = 2.81 N.
F =k
r2
) (0.120 m)2
6. (a) With a understood to mean the magnitude of acceleration, Newton’s second
and third laws lead to
m a =ma m =(
6.310−7 kg )( 7.0 m/s2 )
= 4.9 10−7 kg.
2 2 1 1 2 2
9.0 m/s
(b) The magnitude of the (only) force on particle 1 is
q q q
2
F = m1a1 = k
1 2
= ( 8.99 10 N m /C
9 2 2
) (0.0032 m)2 .
r2
Inserting the values for m1 and a1 (see part (a)), we obtain q = 7.110–11 C.
, 1026 CHAPTER 21
7. With rightward positive, the net force on q3 is
q2q3
F =F +F =k q1q3 +k .
3 13 23
( L12 + L23 )
2 2
L 23
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 q3 by 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 q3 would be repelled from
q1). Setting the net force equal to zero L23 = L12 and canceling k, q3, and L12 leads to
q1 q1
+ q2 = 0 = −4.00.
4.00 q2
8. 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 A and sphere C each
acquired charge 3Q/4. Then, sphere C touches B and those spheres split up their
total charge (3Q/4 plus –Q/4), so B ends up with charge equal to Q/4. The force of
repulsion between A and B is therefore
(3Q/4)(Q/4)
F1 = k
d2
at the end of experiment 1. Now, in experiment 2, sphere C first touches B, which
leaves each of them with charge Q/8. When C next touches A, sphere A is left with
charge 9Q/16. Consequently, the force of repulsion between A and B is
(9Q/16)(Q/8)
F2 = k
d2
at the end of experiment 2. The ratio is
F2 (9/16)(1/8)
= = 0.375.
F1 (3/4)(1/4)
9. THINK Since charges with opposite signs attract, the initial charge configurations
must be of opposite signs. Similarly, since charges with the same sign repel, the final
charge configurations must be with the same sign.
EXPRESS We assume that the spheres are far enough 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 that
the force on q2 is positive if it is repelled by q1. Then the force on q2 is
1 q1q2 q1q2
F =− = −k ,
a
4 0 r 2
r 2
