Physics 11
Chapter 20 HW Solutions
Chapter 20
Conceptual Questions: 5, 11, 13, 20
Problems: 4, 8, 11, 13, 15, 24, 28, 32, 42, 50
Q20.5. Reason: The clothes are charged by rubbing in the drier. Your body is neutral but the clothes can polarize you
a little and attract the opposite charge in your body.
Assess: This is the same way a balloon sticks to the wall after you rub it on your hair.
Q20.11. Reason: If the bees are charged and the grains of pollen are neutral, then the charge on the bees can polarize
the grains of pollen so that they are attracted to the charged bee with an attractive electric force. That is, the charge on the
pollen grain will be somewhat separated, with one side of the grain of pollen becoming slightly negative (and therefore
attracted to the positive bee) and the other side becoming slightly positive (repelled by the bee).
Assess: Bees also attract particles of dust and soil to their bodies the same way. In doing so they provide a chemical
survey of the area around their hive. This enables scientists to use bees to locate landmines and explosives.
Q20.13. Reason: (a) E1 < E2 , because the contributions from the two charges cancel at point 1 ( E1 = 0) , but the field
contribution from each charge points to the right at point 2.
(b) E1 > E2 , because the contributions from each charge point in the same direction (to the right) at point 1 but point in
opposite directions and somewhat cancel at point 2 (the cancellation is not complete since point 2 is closer to the negative
charge than the positive one).
(c) E1 = E2 . The contributions from the different charges are in opposite directions at each of the two points, but in each
case the nearest charge to the point makes the stronger field to the left. The contribution from the other charge reduces the
field a bit, but there is still a net field to the left in each case. The magnitudes of the field strength are equal.
(d) E1 < E2 , because the contributions from the two charges cancel at point 1 ( E1 = 0) , but at point 2 both field contributions
have an upward component that doesn’t cancel.
(e) E1 < E2 . At point 1 the contributions are in opposite directions and partially cancel. While point 2 is farther from the
charge on the left, both contributions are in the same direction, so the field is stronger there.
(f ) E1 > E2 . At point 1 the contributions are in the same direction (to the right), whereas at point 2 they partially cancel
because they are in opposite directions (not to mention that point 2 is farther away from the positive charge).
Assess: It is worth spending a few minutes to get comfortable with all these cases. There are various physics software
packages that allow you to map the fields around various charge distributions; they would be good to play with also.
Q20.20. Reason: When lightning strikes, there is a tremendous transfer of charge from the cloud to the object
struck.
When lightning strikes the metal plane, this charge is distributed over the surface of the plane.
There will initially be movement of charges over the plane (over a very short time interval) but a situation of static
equilibrium will quickly be established.
Since we have established that there is no electric field inside a conductor, the passengers will experience no change.
Assess: In like manner, if you are inside a car struck by lighting, you will be safe. The tires will most likely be damaged
as this tremendous amount of charge moves through them to ground.
P20.4. Prepare: We will use the charge model and the model of a conductor as material through which electrons
move. An electron has a negative charge of magnitude 1.60 × 10−19 C.
Solve: (a) The charge of a plastic rod decreases from −15.0 nC to −10.0 nC. That is, −5.0 nC charge has been removed
from the plastic. Because it is the negatively charged electrons that are transferred, −5.0 nC has been added to the metal
sphere.
(b) Because each electron has a charge of 1.60 × 10−19 C and a charge of 5.0 nC was transferred, the number of electrons
transferred from the plastic rod to the metal sphere is
5.0 × 10−9 C
= 3.1 × 1010
−19
1.60 × 10 C
Assess: A modest charge of 5 nC contains over 30 billion electrons!
P20.8. Prepare: When two identical conducting spheres are in contact the charge is evenly distributed between them.
Solve: (a) The following grid shows the initial charge on each of the spheres and the charge after each event.
(b) The following grid shows the initial charge on each of the spheres and the charge after each event.
Assess:
The end result depends on the order in which the various events occur.
P20.11. Prepare: We need to solve Coulomb’s law (Equation 20.1) for r :
F=K
| q1|| q2 |
r2
where F = 8. 2 × 10−4 N and q1 = − 5 .0 nC, q2 = − 12 nC, and K = 9 .0 × 10 N ⋅ m 2 /C2.
Solve:
| q || q |
r2 = K 1 2
F
9
r= K
| q1|| q2 |
| − 0.5 × 10−9 C|| −12 × 10−9 C|
9
= (9 .0 × 10 N ⋅ m 2 /C 2 )
= 0 . 026 m = 2 . 6 cm
F
8.2 × 10−4 N
Assess: Notice the N and C cancel out leaving units of m. Comparing with Problem 20.10, the answer of 2.6 cm seems
to be in the right ballpark.
P20.13. Prepare: We will model the glass bead and the ball bearing as point charges. A visual overview of the forces
G
and the coordinate system is shown. The ball bearing experiences a downward electric force F1 on 2 . By Newton’s third
law, F2 on 1 = F1 on 2.
Solve:
Using Coulomb’s law,
F1 on 2 = K
| q1|| q2|
(9.0 × 109 N ⋅ m2 /C 2 )(20 × 10−9 C) | q2|
⇒
0.018
N
=
⇒ | q2| = 1.0 × 10−8 C
r122
(1.0 × 10−2 m) 2
Because the force F1 on 2 is attractive and q 1 is a positive charge, the charge q 2 is a negative charge. Thus,
q2 = − 1.0 × 10−8 C = −10 nC .
P20.15. Prepare: Please refer to Figure P20.15. The charged particles are point charges. The charge q2 is in static
equilibrium, so the net force on q2 is zero. If q2 is positive, q1 will have to be positive to make the net force zero on q2.
And, if q2 is negative, q1 will still have to be positive for q2 to be in equilibrium. We will assume that the charge q2 is
positive. For this situation, the force on q2 by the –2 nC charge is to the left and by q1 is to the right.
Solve: We have
G
G
G
⎛ 1
⎞ ⎛ 1 (2 × 10−9 C)| q2|
⎞
| q1|| q2|
Fnet on q2 = Fq1 on q2 + F−2 nC on q2 = ⎜
, + x-direction ⎟ + ⎜
, − x-direction ⎟ = 0 N/C
2
2
4
(0.2
m)
4
(0.10
m)
πε
πε
0
0
⎝
⎠ ⎝
⎠
Thus,
q1
2 × 10−9 C
−
= 0 N/C ⇒ q1 = 8.0 nC
2
(0.2 m)
(0.10 m) 2
Assess: If the charge q2 is assumed negative, the force on q2 by the –2 nC charge is to the right
and by q1 is to the left. The magnitude of q1 remains unchanged.
P20.24. Prepare: The electric field is that of the two charges placed on the y-axis. Please refer to Figure P20.24. We
denote the upper charge by q1 and the lower charge by q2. Because both the charges are positive, their electric fields at P
are directed away from the charges.
Solve: The electric field strength of q1 is
| q | (9.0 × 10 9 N ⋅ m2 /C2 )(1 × 10−9 C)
E1 = K 21 =
= 1800 N/C
r1
(0.050 m) 2 + (0.050 m) 2
Similarly, the electric field strength of q2 is
| q | (9.0 × 10 9 N ⋅ m2 /C2 )(1 × 10−9 C)
E2 = K 22 =
= 1800 N/C
r2
(0.050 m) 2 + (0.050 m) 2
We will now calculate the components of these electric fields. The electric field due to q1 is away from q1 in the fourth
quadrant and that due to q2 is away from q2 in the first quadrant. Their components are
E1 x = E1 cos 45 °
E1 y = − E1 sin 45 °
E2 x = E2 cos 45 °
E2 y = E2 sin 45 °
The x and y components of the net electric field are:
( Enet ) x = E1x + E2 x = E1 cos 45 ° + E2 cos 45 ° = 2500 N/C
( Enet ) y = E1 y + E2 y = −E1 sin 45 ° + E2 sin 45 ° = 0 N/C
G
⇒ Enet at dot = (2500 N/C, along + x axis)
Thus, the strength of the electric field is 2500 N/C and its direction is horizontal.
Assess: Because the charges are located symmetrically on either side of the y-axis and are of equal value, the ycomponents of their fields will cancel when added.
P20.28. Prepare: The charged plastic bead is a point charge. The bead hangs suspended in the air when the net force
acting on the bead is zero. The two forces that act on the bead are the electric force and the weight. Because the bead is
negatively charged, the electric field must be pointed downward to cause an upward force, which will balance the weight.
Solve:
For the bead to be in static equilibrium,
( Fnet ) y = qE − mg = 0 N ⇒ E =
mg (0.10 × 10−3 kg)(9.8 N/kg)
=
= 6.1 × 105 N/C
q
(1.0 × 1010 )(1.60 × 10−19 C)
G
Thus the required field is E = (6.1 × 105 N/C, down).
G
G
G
G
Assess: F = qE means the sign of the charge q determines the direction of F or E . For positive
G
G
G
G
q, E and F are pointing in the same direction. But E and F point in opposite directions when q is
negative.
P20.32. Prepare: The high part of the bottom conductor is closest to the upper conductor, so the positive charges like to
congregate there.
Solve:
(c) We see from the figures that the field is strongest at the high point of the bottom conductor (the earth); this is where the
air is mostly likely to break down and become conducting and allow the opposite charges to rush toward each other (we
call this lightning).
Assess: This is why we are told to stay away from high and tall things in a lightning storm.
Stay away from tall things and crouch down low and hug your knees if you must remain outside.
Inside metal cars is usually a safe place to be.
P20.42. Prepare: The two charged spheres are point charges. The electric force on one charged sphere due to the other
charged sphere is equal to the sphere’s mass multiplied by its acceleration. Because the spheres are identical and equally
charged, m1 = m2 = m and q1 = q2 = q.
Solve: We have
Kq q
Kq 2
F2 on 1 = F1 on 2 = 12 2 = 2 = ma
r
r
2
(1.0
mar
× 10−3 kg)(225 m/s 2 )(2.0 × 10−2 m) 2
⇒ q2 =
=
= 1.0 × 10−14 C2
9.0 × 109 N ⋅ m2/C 2
K
⇒ q = 1.0 × 10−7 C = 100 nC
P20.50. Prepare: The charges are point charges. Please refer to Figure P20.50.
Solve: Placing the 1 nC charge at the origin and calling it q1, the q2 charge is in the first quadrant, the q3 charge is in the
fourth quadrant, the q4 charge is in the third quadrant, and the q5 charge is in the second quadrant. The electric force on q1 is
the vector sum of the electric forces from the other four charges q2, q3, q4, and q5. The magnitude of these four forces is the
same because all four charges are equal in magnitude and are equidistant from q1. So,
F2 on 1 = F3 on 1 = F4 on 1 = F5 on 1 =
(9.0 × 10 9 N ⋅ m2/C 2 )(2 × 10−9 C)(1 × 10−9 C)
= 3.6 × 10−4 N
(0.5 × 10−2 m) 2 + (0.5 × 10−2 m) 2
G
Thus, Fon 1 = (3.6 × 10−4 N, away from q2) + (3.6 × 10−4 N, away from q3) + (3.6 × 10−4 N, toward q4) + (3.6 ×
10−4 N, toward q5).
G
( F2 on 1 ) x = − (3.6 × 10−4 N)(cos 45 °) = − (2.55 × 10−4 N)
G
( F2 on 1 ) y = − (3.6 × 10−4 N)(sin 45 °) = −(2.55 × 10−4 N)
G
( F3 on 1 ) x = − (3.6 × 10−4 N)(cos 45 °) = − (2.55 × 10−4 N)
G
( F3 on 1 ) y = (3.6 × 10−4 N)(sin 45 °) = (2.55 × 10−4 N)
G
( F4 on 1 ) x = − (3.6 × 10−4 N)(cos 45 °) = − (2.55 × 10−4 N)
G
( F4 on 1 ) y = − (3.6 × 10−4 N)(sin 45 °) = −(2.55 × 10−4 N)
G
( F5 on 1 ) x = − (3.6 × 10−4 N)(cos 45 °) = −(2.55 × 10−4 N)
G
( F5 on 1 ) y = (3.6 × 10−4 N)(sin 45 °) = (2.55 × 10−4 N)
G
G
G
G
G
( Fon 1 ) x = ( F2 on 1 ) x + ( F3 on 1 ) x + ( F4 on 1 ) x + ( F5 on 1 ) x = − 1.02 × 10−3 N
G
G
G
G
G
( Fon 1 ) y = ( F2 on 1 ) y + ( F3 on 1 ) y + ( F4 on 1 ) y + ( F5 on 1 ) y = 0 N
So the force on the 1 nC charge is 1.02 × 10–3 N directed to the left.