Matter and Interactions, Volume 2: Electric and Magnetic Interactions, 5th Edition
by Ruth W. Chabay, Bruce A. Sherwood - Complete Solutions Included
Chapter 13
Electric Field
13.2-Q-01
The relationship is that the force by an electric field on a particle of charge 𝑞 is F⃗ ⃗ The unit of force
= 𝑞 E.
by E-field on q
is newtons (N) and the unit of electric field is newtons per coulomb (N/C).
13.2-Q-02
(a) At location P, hold the charged object and release it from rest at one end of the meterstick. Watch the direction
of motion and then align the meterstick along this direction. Again, release the object from rest from the end of
the meterstick. Measure the time elapsed (on the stopwatch) as it travels a small distance along the meterstick.
What is small? It depends on the average acceleration of the object. If its acceleration is small, then 1 m will be
fine. If its acceleration is large, then perhaps 0.1 m should be used. The astronaut should measure the distance
travelled 𝑑 and the time interval Δ𝑡. A good experimentalist will make the measurement for multiple trials and
calculate the average time elapsed.
(b) The force on the particle may not be constant (because the electric field at the location of the object may be
different as the object moves). However, for a small time interval, we will assume constant force. Define the +x
direction to be the direction of motion of the object. Use the following procedure to calculate the electric field:
(1) Use 𝑣𝑎𝑣𝑔,𝑥 = 𝑑∕Δ𝑡 to calculate the average x-velocity of the object.
(2) Use 𝑣𝑎𝑣𝑔,𝑥 = (𝑣𝑖 + 𝑣𝑓 )∕2 to calculate the final x-velocity of the object.
(3) Use the Momentum Principle to calculate the net force on the object. 𝐹𝑥 = 𝑀Δ𝑣𝑥 ∕Δ𝑡.
(4) Use 𝐹⃗ = 𝑞 𝐸⃗ to calculate the electric field.
Note that this procedure only works for small time interval because we are assuming a constant net force on the
object.
13.2-Q-03
A particle cannot exert a force on itself. In the equation 𝐹⃗ = 𝑞 𝐸,
⃗ the electric field at the location of the particle 𝑞
is due to other charges.
1
, CHAPTER 13. ELECTRIC FIELD
13.2-Q-04
(a) h. For a proton, the force on the proton is in the same direction as the electric field.
(b) d. For an electron, the force on the electron is in the same direction as the electric field.
13.2-Q-05
(a) The electric field at B due to the proton at A is the same (because the source of the field did not change). So, the
field is E
⃗ .
1
(b) The force on the lithium nucleus is 3F⃗ since 𝐹⃗ = 𝑞 𝐸.
⃗
1
(c) It’ll change the force, but not the E-field.
(d) The electric field at B due to the proton at A is the same (because the source of the field did not change). So, the
field is E
⃗ .
1
| |
(e) The force on the electron is |F⃗ | since the electron and (original) proton have the same magnitude charge.
| 1|
(f) f, because the electron is negatively charged so 𝐹⃗ is opposite 𝐸.
⃗
13.4-Q-01
The simplest case is two equally charged particles in opposite directions and equidistant from the point where the
net electric field is zero. See the figure below..
�
E �
E
2 1
X
proton 1 proton 2
13.5-Q-01
Q is the charge of the distant object. 𝑞 is the magnitude of the charge of each particle of the dipole. 𝑠 is the distance
of separation for the particles that make up the dipole. 𝑟 is the distance from the center of the dipole to the distant
object.
13.5-Q-02
13-2
,CHAPTER 13. ELECTRIC FIELD
The formula given in the problem is an approximation for 𝑟 >> 𝑠. However, in this case 𝑟 = 1.5𝑠 and therefore the
approximation is not valid.
13.5-Q-03
The E-field vectors shown are for an electric dipole. The longest field vectors shown (on the +y axis, relative to
the center of the circle) are for the electric field along the axis of a dipole. They point away from the positive charge
and toward the negative charge. Thus, the dipole is oriented “vertically” with the +𝑞 along the +𝑦 axis and −𝑞 along
the −𝑦 axis, if the origin is defined to be the center of the dipole.
13.5-Q-04
For all of these answers, I will use the convention 𝑥̂ to the right, 𝑦̂ toward the top of the page, and 𝑧̂ outward
perpendicular to the page.
(a) 𝐸̂ =< 0, −1, 0 >. The dipole moment vector 𝑝⃗ points “upward” in the +𝑦 direction. Since point A is on the
perpendicular bisector of the dipole, the electric field at point A points opposite 𝑝⃗ which is in the −𝑦 direction.
(b) 𝐸̂ =< 0, 1, 0 >. Since point B is along the axis of the dipole, the electric field at point B points in the same
direction as the dipole moment, in the +𝑦 direction.
(c) Because the electron starts from rest, it will move in the direction of the electric force on the electron which is
opposite the electric field. Thus, 𝐹̂ =< 0, 1, 0 >.
(d) Because it starts from rest, it will move in the direction of the electric force on the proton which is in the same
direction as the electric field. Thus, 𝐹̂ =< 0, 1, 0 >.
(e) The force on the dipole by the electron is opposite the force on the electron by the dipole. Thus the direction of the
force on the dipole by the electron is 𝐹̂ =< 0, −1, 0 >. Because the dipole starts from rest, it will begin moving in
the direction < 0, −1, 0 >.
13.5-Q-05
1. true
2. true
3. false
4. false
5. true
13.5-Q-06
Since 𝑑 >> 𝑠, then the electric field varies as 1∕𝑑 3 . If you triple the distance 𝑑, then the electric field changes by
a factor 1∕33 = 1∕27.
13-3
, CHAPTER 13. ELECTRIC FIELD
13.5-Q-07
(a) The electric field at the location of Q is due to the dipole, and it remains the same. Since the force on Q due to
the field is 𝐹 = |𝑞| 𝐸, then increasing |𝑞| by 9 will increase the magnitude of the force by a factor of 9; therefore,
𝐹 = 9𝐹0 .
(b) Yes, the force on −9𝑄 is opposite the force on 𝑄 and in this case will be in the +x direction.
13.5-Q-08
Since 𝑑 >> 𝑠, then the electric field varies as 1∕𝑑 3 . If you change the distance 𝑑 by 1/2, then the electric field
changes by a factor 1∕(1∕2)3 = 8. Thus 𝐹 = 8𝐹0 .
13.5-Q-09
A sketch of all vectors is shown in the figure below..
�
E �F
�F �
E due to ball
on ball due to dipole on dipole
(a) The electric field due to the dipole at the location of the ball is in the +x direction, in the same direction as the
dipole moment.
(b) Since 𝑄 is negative, the force on the ball by the electric field is in the -x direction, opposite the electric field at
this location.
(c) The electric field due to the ball at the location of the dipole is in the −𝑥 direction, toward the ball, since it is
negatively charged.
(d) The force by the ball on the dipole is to the right because the force on −𝑞 (which is to the right) is greater than the
force on +𝑞 (which is to the left). This is consistent with Newton’s second law, that in an interaction the objects
exert forces on each other that are equal in magnitude and opposite in direction.
13.5-Q-10
The force on the ball by the dipole is directly proportional to the electric field created by the dipole. The electric
field by the dipole depends on 1∕𝑟3 , for 𝑟 >> 𝑠. If the distance 𝑟 is doubled, then the electric field changes by a factor
1∕23 = 1∕8. Therefore the force also changes by 1∕8.
13-4
by Ruth W. Chabay, Bruce A. Sherwood - Complete Solutions Included
Chapter 13
Electric Field
13.2-Q-01
The relationship is that the force by an electric field on a particle of charge 𝑞 is F⃗ ⃗ The unit of force
= 𝑞 E.
by E-field on q
is newtons (N) and the unit of electric field is newtons per coulomb (N/C).
13.2-Q-02
(a) At location P, hold the charged object and release it from rest at one end of the meterstick. Watch the direction
of motion and then align the meterstick along this direction. Again, release the object from rest from the end of
the meterstick. Measure the time elapsed (on the stopwatch) as it travels a small distance along the meterstick.
What is small? It depends on the average acceleration of the object. If its acceleration is small, then 1 m will be
fine. If its acceleration is large, then perhaps 0.1 m should be used. The astronaut should measure the distance
travelled 𝑑 and the time interval Δ𝑡. A good experimentalist will make the measurement for multiple trials and
calculate the average time elapsed.
(b) The force on the particle may not be constant (because the electric field at the location of the object may be
different as the object moves). However, for a small time interval, we will assume constant force. Define the +x
direction to be the direction of motion of the object. Use the following procedure to calculate the electric field:
(1) Use 𝑣𝑎𝑣𝑔,𝑥 = 𝑑∕Δ𝑡 to calculate the average x-velocity of the object.
(2) Use 𝑣𝑎𝑣𝑔,𝑥 = (𝑣𝑖 + 𝑣𝑓 )∕2 to calculate the final x-velocity of the object.
(3) Use the Momentum Principle to calculate the net force on the object. 𝐹𝑥 = 𝑀Δ𝑣𝑥 ∕Δ𝑡.
(4) Use 𝐹⃗ = 𝑞 𝐸⃗ to calculate the electric field.
Note that this procedure only works for small time interval because we are assuming a constant net force on the
object.
13.2-Q-03
A particle cannot exert a force on itself. In the equation 𝐹⃗ = 𝑞 𝐸,
⃗ the electric field at the location of the particle 𝑞
is due to other charges.
1
, CHAPTER 13. ELECTRIC FIELD
13.2-Q-04
(a) h. For a proton, the force on the proton is in the same direction as the electric field.
(b) d. For an electron, the force on the electron is in the same direction as the electric field.
13.2-Q-05
(a) The electric field at B due to the proton at A is the same (because the source of the field did not change). So, the
field is E
⃗ .
1
(b) The force on the lithium nucleus is 3F⃗ since 𝐹⃗ = 𝑞 𝐸.
⃗
1
(c) It’ll change the force, but not the E-field.
(d) The electric field at B due to the proton at A is the same (because the source of the field did not change). So, the
field is E
⃗ .
1
| |
(e) The force on the electron is |F⃗ | since the electron and (original) proton have the same magnitude charge.
| 1|
(f) f, because the electron is negatively charged so 𝐹⃗ is opposite 𝐸.
⃗
13.4-Q-01
The simplest case is two equally charged particles in opposite directions and equidistant from the point where the
net electric field is zero. See the figure below..
�
E �
E
2 1
X
proton 1 proton 2
13.5-Q-01
Q is the charge of the distant object. 𝑞 is the magnitude of the charge of each particle of the dipole. 𝑠 is the distance
of separation for the particles that make up the dipole. 𝑟 is the distance from the center of the dipole to the distant
object.
13.5-Q-02
13-2
,CHAPTER 13. ELECTRIC FIELD
The formula given in the problem is an approximation for 𝑟 >> 𝑠. However, in this case 𝑟 = 1.5𝑠 and therefore the
approximation is not valid.
13.5-Q-03
The E-field vectors shown are for an electric dipole. The longest field vectors shown (on the +y axis, relative to
the center of the circle) are for the electric field along the axis of a dipole. They point away from the positive charge
and toward the negative charge. Thus, the dipole is oriented “vertically” with the +𝑞 along the +𝑦 axis and −𝑞 along
the −𝑦 axis, if the origin is defined to be the center of the dipole.
13.5-Q-04
For all of these answers, I will use the convention 𝑥̂ to the right, 𝑦̂ toward the top of the page, and 𝑧̂ outward
perpendicular to the page.
(a) 𝐸̂ =< 0, −1, 0 >. The dipole moment vector 𝑝⃗ points “upward” in the +𝑦 direction. Since point A is on the
perpendicular bisector of the dipole, the electric field at point A points opposite 𝑝⃗ which is in the −𝑦 direction.
(b) 𝐸̂ =< 0, 1, 0 >. Since point B is along the axis of the dipole, the electric field at point B points in the same
direction as the dipole moment, in the +𝑦 direction.
(c) Because the electron starts from rest, it will move in the direction of the electric force on the electron which is
opposite the electric field. Thus, 𝐹̂ =< 0, 1, 0 >.
(d) Because it starts from rest, it will move in the direction of the electric force on the proton which is in the same
direction as the electric field. Thus, 𝐹̂ =< 0, 1, 0 >.
(e) The force on the dipole by the electron is opposite the force on the electron by the dipole. Thus the direction of the
force on the dipole by the electron is 𝐹̂ =< 0, −1, 0 >. Because the dipole starts from rest, it will begin moving in
the direction < 0, −1, 0 >.
13.5-Q-05
1. true
2. true
3. false
4. false
5. true
13.5-Q-06
Since 𝑑 >> 𝑠, then the electric field varies as 1∕𝑑 3 . If you triple the distance 𝑑, then the electric field changes by
a factor 1∕33 = 1∕27.
13-3
, CHAPTER 13. ELECTRIC FIELD
13.5-Q-07
(a) The electric field at the location of Q is due to the dipole, and it remains the same. Since the force on Q due to
the field is 𝐹 = |𝑞| 𝐸, then increasing |𝑞| by 9 will increase the magnitude of the force by a factor of 9; therefore,
𝐹 = 9𝐹0 .
(b) Yes, the force on −9𝑄 is opposite the force on 𝑄 and in this case will be in the +x direction.
13.5-Q-08
Since 𝑑 >> 𝑠, then the electric field varies as 1∕𝑑 3 . If you change the distance 𝑑 by 1/2, then the electric field
changes by a factor 1∕(1∕2)3 = 8. Thus 𝐹 = 8𝐹0 .
13.5-Q-09
A sketch of all vectors is shown in the figure below..
�
E �F
�F �
E due to ball
on ball due to dipole on dipole
(a) The electric field due to the dipole at the location of the ball is in the +x direction, in the same direction as the
dipole moment.
(b) Since 𝑄 is negative, the force on the ball by the electric field is in the -x direction, opposite the electric field at
this location.
(c) The electric field due to the ball at the location of the dipole is in the −𝑥 direction, toward the ball, since it is
negatively charged.
(d) The force by the ball on the dipole is to the right because the force on −𝑞 (which is to the right) is greater than the
force on +𝑞 (which is to the left). This is consistent with Newton’s second law, that in an interaction the objects
exert forces on each other that are equal in magnitude and opposite in direction.
13.5-Q-10
The force on the ball by the dipole is directly proportional to the electric field created by the dipole. The electric
field by the dipole depends on 1∕𝑟3 , for 𝑟 >> 𝑠. If the distance 𝑟 is doubled, then the electric field changes by a factor
1∕23 = 1∕8. Therefore the force also changes by 1∕8.
13-4
