ANSWERS TO QUESTIONS
1.
Usually the number of positively charged particles (protons) is equal to the number of
negatively charged particles (electrons). The total charge is zero.
2. Rub the plastic with the fur. The plastic should become negatively charged. If the charged
solid is attracted to the plastic then the charge on it is positive. If the solid is repelled it is
negatively charged.
3. An atom with fewer electrons than protons is a positive ion since it has a net positive charge.
An atom with more electrons than protons is a negative ion since it has a net negative
charge. (See pages 260-261.)
4. Only the nucleus, which is usually just a proton.
5. A gravitational force acts between any two objects and is always attractive, whereas there is
an electrostatic force acting between two objects only if they have net electric charges and
then the force can be attractive or repulsive. The force laws for the two interactions are very
similar: for example, both forces are inversely proportional to the square of the distance
between two objects. (See page 262.)
6. The net force on each would be zero, since the attractive gravitational force would offset the
equal-sized repulsive electrostatic force. Since both forces vary inversely with the square of
the distance between the objects, the forces would exactly cancel each other no matter how
far apart they were.
7. a) The electrons in the penny are repelled by the negatively charged ball and some of these
electrons move to the far side of the penny away from the ball. The result is positive charge
on the side of the penny closest to the ball and a negative charge on the side farthest away.
Because of the inverse-square distance dependence of the Coulomb force, the attractive
force on the positively charged side is larger than the repulsive force on the negatively
charged side so the net force is attractive.
b) Once the penny touches the iron ball some of the electrons on the ball will flow into the
penny since they are repelled by the extra electrons on the ball. This makes the penny
negatively charged as well so it is then repelled by the ball.
8. The electric field is the agent of the electric force. It exists in the space surrounding charged
objects and causes a force to act on any other nearby charged object. The electric field
around a proton looks like the field in Figure 7.9a. (See page 264.)
9. The electron would feel a force toward the West because electrons are negatively charged.
Positive ions would feel a force toward the East.
10. An electrostatic precipitator removes soot, ash, and dust particles from the gaseous
emissions at power plants and factories. (See page 265.) You can also purchase them for
your home.
11. Since equal amounts of positive and negative charges are flowing in the same direction it is
like two currents flowing simultaneously, but the currents are flowing in opposite directions,
so the total current is zero.
12. There are conductors, in which charge can flow readily; superconductors, which are perfect
conductors since they have zero resistance; insulators, in which charge cannot flow easily;
and semiconductors, which are in between these two extremes with properties that can be
adjusted via “doping.”
95
13. The resistance will be greater since the resistance is proportional to length and inversely
proportional to cross sectional area. Increasing the length and shrinking the radius both act
to increase the resistance.
14. The resistance increases because the student’s hand raises the temperature of the wire.
15. Filaments of light bulbs, heating elements in toasters and electric stoves, resistors in
electrical devices.
16. Current is the rate of flow of charge, resistance is a measure of the opposition to the flow of
charge, and voltage is the energy per unit charge supplied by a power supply. (See Section
7.3.)
17. See Figure 17.21a or Figure 17.17a.
18. The current in a conductor is equal to the voltage applied across it divided by its resistance,
where the resistance is constant for any applied voltage. Hence, the voltage and current are
directly proportional.
19.
20. Series circuits and parallel circuits. In a series circuit there is only one path for the charges
so that one device can control the current flow in the rest of the circuit. In a parallel circuit
the current branches into a separate part for each circuit element. If a device in one branch
fails, the others in different branches are not affected. (It is also possible to create
connections that are neither series nor parallel.)
21.
22. Each 1.5-volt battery gives each coulomb of charge 1.5 joules of energy. When in series,
each coulomb flows through both batteries and is given 1.5 joules of energy twice, the same
as the energy it would get from one 3-volt battery. Two 1.5-volt batteries in parallel are
equivalent to one 1.5-volt battery since each coulomb will flow throw only one of the
batteries. Two batteries in parallel will, however, last twice as long as a single battery, or
they can supply twice as much current for the same length of time.
23. The lower voltage supply can deliver more current. The 12 V supply will be able to deliver
100W/12V=8.33 A but the 6V supply we be able to deliver twice as much current.
24. Reducing the resistance causes the current to increase, by Ohm’s Law. The power output
increases also since power equals voltage times current.
25. A fuse or circuit breaker is designed to stop an electric current if for some reason the current
is so large that it could cause overheating of the wires in the circuit. The fuse or circuit
breaker has to be placed in series with the rest of the circuit.
26. A current large enough to overheat the wires in the circuit could flow, possibly causing a
fire.
27. High voltages allow the transmission of electrical energy with relatively small currents. This
makes it possible to use relatively small, and consequently less expensive, wires without a
large loss of energy due to ohmic heating.
28. AC, alternating current, consists of electric charges flowing back and forth. It is used by
electric utilities because transformers make it possible to easily step up and step down AC
voltages. DC, direct current, consists of electric charges flowing steadily in one direction.
Flashlights use DC because batteries produce DC.
29. Since the charges in a wire move forward and backward along the wire in response to an AC
voltage, the closer analogy is a longitudinal wave.
30. Electric clocks and timers would run slow. Electric lights would flicker—flash brighter and
dimmer at a rate slow enough see.
ANSWERS TO EVEN NUMBERED PROBLEMS
I = 100 C ÷ 0.05 s = 2000 A
q = 0.0001 A × 5 min × 60 s/min = 0.03 C
R = 6Ω
I = 120 V ÷ 80 Ω = 1.5 A
a) V = IR = 0.3 A × 150 Ω = 45 V
b) I = V/R = 18 V ÷ 150 Ω = 0.12 A
12. Eel’s power = V × I = 400 V × 0.5 A = 200 W
2.
4.
6.
8.
10.
14. a) P = V × I = 120 V × 20 A = 2400 W
b) Two 1200-W appliances will test it to the limit.
16. I = P ÷ V = 40 W ÷ 120 V = 0.33 A
(one-tenth of the current in a 40-W, 12-V headlight)
18. E = P × t = 2 W × 24 h × 60 min/h × 60 s/min = 172,800 J = 0.048 kW-h
20. P = V × I = 200,000,000 V × 1000 A = 200,000,000,000 W
22. a) P = V × I = 120 V × 10 A = 1200 W
b) E = P × t = 1200 W × 4 h × 60 min/h × 60 s/min = 17,280,000 J = 4.8 kW-h
24. a) I = P ÷ V = 60 W ÷ 120 V = 0.5 A
b) R = V ÷ I = 120 V ÷ 0.5 A = 240 Ω
c) I = V ÷ R = 60 V ÷ 240 Ω = 0.25 A
d) P = V × I = 60 V × 0.25 A = 15 W
26. E = P × t = V × I × t = V × V/R × t = 120 V × 120 V/10 Ω × 30 min × 60 s/min
= 2,592,000 J
= 0.72 kW-h
ANSWERS TO CHALLENGES
1.
F = – 8.2 × 10-8 N
(The minus sign just means the force on the electron is inward, toward the proton.)
2.
3.
v2 = 4.78 × 1012 m2/s2
v = 2,186,000 m/s
First, find how much charge is flowing per second:
I = q÷t
q = I t = 0.2 A × 1 s
q = 0.2 C every second.
Now find how many electrons this is per second:
The size of the charge on each electron is
qe = 1.6 × 10-19 C
The number of electrons making up charge q is = q ÷ qe
4.
5.
6.
= (0.2 C) ÷ (1.6 × 10-19 C)
= 1.25 × 1018
1.25 × 1018 electrons flow through the wire every second.
Resistance is due to the collisions of electrons with atoms as the electrons move through the
solid. At higher temperatures, the atoms vibrate with larger amplitude because of their
increased internal energy. This increases the likelihood of each electron striking atoms so
the resistance is higher.
When the bulb is first turned on, its filament is cool so its resistance is low. A fraction of a
second later, the filament is very hot so its resistance is higher. This higher resistance
reduces the current.
Use P = VI
I = P÷V
For V = 12 V
I = (1,200 W) ÷ (12 V ) = 100 A
Similarly, for V = 30 V
I = 40 A
V = 60 V
I = 20 A
V = 120 V I = 10 A
The lower voltage dryers would have to have successively thicker wires to handle the larger
currents.
8. How high would a 150-lb person have to climb to gain 360,000 joules of energy?
E = 360,000 J = 266,000 ft-lb
PE = mgh =W × h
h = (PE) ÷ W = (266,000 ft-lb) ÷ (150 lb)
h = 1,770 ft
9. P = VI and
V = IR
P = (IR)I = I2 R
If I is doubled, P becomes 22 or 4 times as large—the cable would waste 40 kW-h per day.
10. The average current I = q ÷ t = 0.1 C ÷ 2 ms = 0.1 C ÷ 0.002 s = 50 A.
The average power output = P = V × I = 3000 V × 50 A = 150,000 W.
The total energy used = E = P × t = 150,000 W × 2 ms = 150,000 W × 0.002 s = 300 J.
7.