??
ED
OV
PR
AP
A?_
UVI
ST
, TABLE OF CONTENTS
Solutions Manual: Fundamentals of Physics, Extended, 12th Edition
(Volume 2)
ST
Authors: David Halliday, Robert Resnick, Jearl Walker
U
Chapter 21. Coulomb's Law
Chapter 22. Electric Fields
VI
Chapter 23. Gauss' Law
Chapter 24. Electric Potential
A
Chapter 25. Capacitance
Chapter 26. Current and Resistance
?_
Chapter 27. Circuits
Chapter 28. Magnetic Fields
Chapter 29. Magnetic Fields Due to Currents
AP
Chapter 30. Induction and Inductance
Chapter 31. Electromagnetic Oscillations and Alternating Current
Chapter 32. Maxwell's Equations; Magnetism of Matter
PR
Chapter 33. Electromagnetic Waves
Chapter 34. Images
Chapter 35. Interference
Chapter 36. Diffraction
OV
Chapter 37. Relativity
Chapter 38. Photons and Matter Waves
Chapter 39. More About Matter Waves
ED
Chapter 40. All About Atoms
Chapter 41. Conduction of Electricity in Solids
Chapter 42. Nuclear Physics
Chapter 43. Energy from the Nucleus
?
Chapter 44. Quarks, Leptons, and the Big Bang
?
, Chapter 21
ST
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
U
q1 and q2 and separated by distance r is given by Coulomb’s law (see Eq. 21.1.1):
VI
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
A
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).
AP
ANALYZE Setting the derivative df/dq equal to zero leads to Q – 2q = 0, or q = Q/2.
Thus, q/Q = 0.500.
PR
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)
OV
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
ED
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
ST
which can be used to solve for the distance:
r=
U
F
ANALYZE Substituting our values of q1 = 2.60 10−6 C, q2 = −47.0 10−6 C, and
VI
k = 8.99 109 N m 2 /C2 , we find
A
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.
AP
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.
PR
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
OV
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.
ED
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.
ED
OV
PR
AP
A?_
UVI
ST
, TABLE OF CONTENTS
Solutions Manual: Fundamentals of Physics, Extended, 12th Edition
(Volume 2)
ST
Authors: David Halliday, Robert Resnick, Jearl Walker
U
Chapter 21. Coulomb's Law
Chapter 22. Electric Fields
VI
Chapter 23. Gauss' Law
Chapter 24. Electric Potential
A
Chapter 25. Capacitance
Chapter 26. Current and Resistance
?_
Chapter 27. Circuits
Chapter 28. Magnetic Fields
Chapter 29. Magnetic Fields Due to Currents
AP
Chapter 30. Induction and Inductance
Chapter 31. Electromagnetic Oscillations and Alternating Current
Chapter 32. Maxwell's Equations; Magnetism of Matter
PR
Chapter 33. Electromagnetic Waves
Chapter 34. Images
Chapter 35. Interference
Chapter 36. Diffraction
OV
Chapter 37. Relativity
Chapter 38. Photons and Matter Waves
Chapter 39. More About Matter Waves
ED
Chapter 40. All About Atoms
Chapter 41. Conduction of Electricity in Solids
Chapter 42. Nuclear Physics
Chapter 43. Energy from the Nucleus
?
Chapter 44. Quarks, Leptons, and the Big Bang
?
, Chapter 21
ST
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
U
q1 and q2 and separated by distance r is given by Coulomb’s law (see Eq. 21.1.1):
VI
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
A
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).
AP
ANALYZE Setting the derivative df/dq equal to zero leads to Q – 2q = 0, or q = Q/2.
Thus, q/Q = 0.500.
PR
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)
OV
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
ED
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
ST
which can be used to solve for the distance:
r=
U
F
ANALYZE Substituting our values of q1 = 2.60 10−6 C, q2 = −47.0 10−6 C, and
VI
k = 8.99 109 N m 2 /C2 , we find
A
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.
AP
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.
PR
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
OV
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.
ED
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.
