Name: ________________________ Class: ___________________ Date: __________
ID: A
Quiz8--Ch 17 & 18 Electric Fields & Circuits PRACTICE
Multiple Choice
Identify the choice that best completes the statement or answers the question.
1.
Three resistors with values of 3.0 Ω, 6.0 Ω, and 12 Ω are connected in series. What is the equivalent resistance of this combination?
a. 7.0 Ω b. 1.7 Ω c. 0.58 Ω d. 21 Ω
2.
A flashlight bulb with a potential difference of 4.5 V across it has a resistance of 8.0 Ω. How much current is in the bulb filament?
a. 1.8 A b. 0.56 A c. 36 A d. 9.4 A
3.
Electric charge is
a. found only in insulators. b. conserved. c. not conserved. d. found only in a conductor.
4.
A lightbulb has a resistance of 240 Ω when operating at 120 V. What is the current in the lightbulb?
a. 1.0 A b. 0.50 A c. 0.20 A d. 2.0 A
5.
If the batteries in a portable CD player provide a terminal voltage of 12 V, what is the potential difference across the entire CD player?
a. 4.0 V b. 6.0 V c. 3.0 V d. 12 V
6.
What is the equivalent resistance of the resistors in the figure shown above?
a. 18 Ω b. 16 Ω c. 10 Ω d. 7.5 Ω
7.
Three resistors connected in series carry currents labeled I 1 , I 2 , and I 3 , respectively. Which of the following expresses the total current,
I t , in the system made up of the three resistors in series?
−1
ÁÊ 1
1
1 ˜ˆ˜
˜˜
d. It = ÁÁÁ
+
+
ÁÁ I
I2
I 3 ˜˜¯
Ë 1
ÁÊ 1
1
1 ˜ˆ˜
˜˜
a. It = I1 + I2 + I3 b. It = I1 = I2 = I3 c. It = ÁÁÁ
+
+
ÁÁ I
I2
I 3 ˜˜¯
Ë 1
8.
When a positive charge moves in the direction of the electric field, what happens to the electrical potential energy associated with the
charge?
a. It remains the same. b. It decreases. c. It increases. d. It sharply increases, and then decreases.
9.
A repelling force occurs between two charged objects when the charges are of
a. like signs. b. unequal magnitude. c. equal magnitude. d. unlike signs.
10.
Charge is most easily transferred in
a. conductors. b. insulators. c. nonconductors. d. semiconductors.
11.
A circuit has a continuous path through which charge can flow from a voltage source to a device that uses electrical energy. What is the
name of this type of circuit?
a. a circuit schematic b. a closed circuit c. a short circuit d. an open circuit
12.
How does the potential difference across the bulb in a flashlight compare with the terminal voltage of the batteries used to power the
flashlight?
a. It cannot be determined unless the internal resistance of the batteries is known. b. The potential difference is less than the terminal
voltage. c. The potential difference is equal to the terminal voltage. d. The potential difference is greater than the terminal voltage.
1
Name: ________________________
ID: A
13.
Three resistors with values of 4.0 Ω, 6.0 Ω, and 8.0 Ω, respectively, are connected in series. What is their equivalent resistance?
a. 6.0 Ω b. 1.8 Ω c. 8.0 Ω d. 18 Ω
14.
The equivalent resistance of a complex circuit is usually determined by
a. adding and subtracting individual resistances. b. simplifying the circuit into groups of series and parallel circuits. c. dividing the
sum of the individual resistances by the number of resistances. d. inspection.
15.
When you flip a switch to turn on a light, the delay time before the light turns on is determined by
a. the resistance of the wire. b. the drift speed of the electrons in the wire. c. the speed of the electric field moving in the wire.
d. the number of electron collisions per second in the wire.
16.
If the potential difference across a pair of batteries used to power a flashlight is 6.0 V, what is the potential difference across the flashlight
bulb?
a. 3.0 V b. 6.0 V c. 12 V d. 9.0 V
17.
If a 75 W lightbulb operates at a voltage of 120 V, what is the current in the bulb?
a. 1.6 A b. 9.0 × 10 3 A c. 1.95 × 10 2 A d. 0.62 A
18.
If the current through a 5.00 × 10 2 W heater is 4.00 A, what is the potential difference across the ends of the heating element?
a. 2.50 × 10 1 V b. 8.00 × 10 −3 V c. 1.25 × 10 2 V d. 2.00 × 10 3 V
19.
The energy gained by electrons as they are accelerated by an electric field is
a. equal to the average loss in energy due to collisions. b. greater than the average loss in energy due to collisions. c. less than the
average loss in energy due to collisions. d. not affected by the gain in energy due to collisions.
20.
An attracting force occurs between two charged objects when the charges are of
a. equal magnitude. b. unequal magnitude. c. like signs. d. unlike signs.
21.
A circuit is composed of resistors wired in series. What is the relationship between the equivalent resistance of the circuit and the
resistance of the individual resistors?
a. The equivalent resistance is less than the smallest resistance in the circuit. b. The equivalent resistance is greater than the sum of all
the resistances in the circuit. c. The equivalent resistance is equal to the largest resistance in the circuit. d. The equivalent resistance
is equal to the sum of the individual resistances.
22.
Which of the following wires would have the least resistance?
a. an aluminum wire 40 cm in diameter at 40°C b. an aluminum wire 20 cm in diameter at 40°C c. an aluminum wire 40 cm in
diameter at 60°C d. an aluminum wire 20 cm in diameter at 60°C
23.
How is current affected if the number of charge carriers decreases?
a. The current is not affected. b. The current increases. c. The current initially decreases and then is gradually restored. d. The
current decreases.
24.
Which sentence best describes electrical conductors?
a. Electrical conductors have high tensile strength. b. Electrical conductors have low mass density. c. Electrical conductors are poor
heat conductors. d. Electrical conductors have electric charges that move freely.
25.
Which of the following is not true for both gravitational and electric forces?
a. Potential energy is a function of distance of separation. b. The inverse square distance law applies. c. Forces are either attractive
or repulsive. d. Forces are proportional to physical properties.
26.
Three resistors with values of 3.0 Ω, 6.0 Ω, and 12 Ω are connected in parallel. What is the equivalent resistance of this combination?
a. 9.0 Ω b. 0.26 Ω c. 1.7 Ω d. 21 Ω
27.
Two point charges, initially 2 cm apart, are moved to a distance of 10 cm apart. By what factor does the resulting electric force between
them change?
1
1
a. 5 b.
c. 25 d.
25
5
2
Name: ________________________
ID: A
28.
Increasing the potential difference between the plates of a capacitor will produce what effect on the capacitor?
a. It will decrease the charge on each plate. b. It will increase the capacitance. c. It will decrease the capacitance. d. It will
increase the charge on each plate.
29.
Resultant force on a charge is the ____ sum of the individual forces on that charge.
a. individual b. negative c. scalar d. vector
30.
What is the equivalent resistance for the resistors in the figure shown above?
a. 3.0 Ω b. 2.2 Ω c. 1.3 Ω d. 12.0 Ω
31.
How much does it cost to operate a 695 W heater for exactly 30.0 min if electrical energy costs $0.060 per kW•h?
a. $0.36 b. $0.02 c. $0.18 d. $0.90
32.
If the potential difference across the bulb in a camping lantern is 9.0 V, what is the potential difference across the battery used to power
it?
a. 1.0 V b. 3.0 V c. 9.0 V d. 18 V
33.
Three resistors with values of 4.0 Ω, 6.0 Ω, and 10.0 Ω are connected in parallel. What is their equivalent resistance?
a. 1.9 Ω b. 6.0 Ω c. 20.0 Ω d. 7.3 Ω
34.
Three resistors with values of R 1 , R 2 , and R 3 are connected in parallel. Which of the following expresses the total resistance, R eq , of the
three resistors?
ÊÁ 1
1
1 ˆ˜˜
Á
˜˜ b. Req = R1 = R2 = R3 c. Req = R1 + R2 + R3 d. Req =
a. Req = ÁÁ
+
+
ÁÁ R
˜˜
R
R
1
2
3
Ë
¯
−1
ÊÁ 1
1
1 ˆ˜˜
ÁÁ
˜˜
+
+
ÁÁ
R2
R 3 ˜˜¯
ÁË R 1
35.
To find the current in a complex circuit, it is necessary to know the
a. potential difference in each device in the circuit. b. current in each device in the circuit. c. number of branches in the circuit.
d. equivalent resistance of the circuit.
36.
What is the equivalent resistance for the resistors in the figure shown above?
a. 10.0 Ω b. 25 Ω c. 5.0 Ω d. 7.5 Ω
37.
When a glass rod is rubbed with silk and becomes positively charged,
a. electrons are removed from the rod. b. the silk remains neutral. c. protons are added to the silk. d. protons are removed from the
silk.
38.
A surface charge can be produced on insulators by
a. grounding. b. contact. c. polarization. d. induction.
3
Name: ________________________
ID: A
39.
Two point charges, initially 1 cm apart, are moved to a distance of 3 cm apart. By what factor do the resulting electric and gravitational
forces between them change?
1
1
b. 3 c.
d. 9
a.
9
3
40.
Three resistors connected in parallel have individual values of 4.0 Ω, 6.0 Ω, and 10.0 Ω, as shown above. If this combination is connected
in series with a 12.0 V battery and a 2.0 Ω resistor, what is the current in the 10.0 Ω resistor?
a. 11 A b. 16 A c. 0.58 A d. 1.0 A
41.
Which of the following is the best description of a schematic diagram?
a. shows the parts of a circuit and how the parts connect to each other b. shows some of the parts that make up a circuit
c. determines the location of the parts of a circuit d. uses pictures to represent the parts of a circuit
42.
Which of the following does not affect a material’s resistance?
a. Ohm’s law b. the temperature of the material c. the length of the material d. the type of material
43.
Which is the most correct statement regarding the drawing of electric field lines?
a. Electric field lines can start on a charge of either polarity. b. Electric field lines always form closed loops. c. Electric field lines
never cross each other. d. Electric field lines always connect from one charge to another.
44.
A microwave draws 5.0 A when it is connected to a 120 V outlet. If electrical energy costs $0.090 per kW•h, what is the cost of running
the microwave for exactly 6 h?
a. $2.70 b. $0.32 c. $0.72 d. $1.60
45.
Three resistors with values of R 1 , R 2 , and R 3 are connected in series. Which of the following expresses the total resistance, R eq , of the
three resistors?
−1
ÊÁ 1
1
1 ˆ˜˜
ÁÁ
˜
a. Req = Á
+
+
b. Req =
˜
ÁÁ R
R2
R 3 ˜˜¯
Ë 1
ÊÁ 1
1
1 ˆ˜˜
ÁÁ
˜˜ c. Req = R1 + R2 + R3 d. Req = R1 = R2 = R3
+
+
ÁÁ
R2
R 3 ˜˜¯
ÁË R1
Problem
46.
A 13.2 Ω resistor has 0.049 A of current in it. What is the potential difference across the resistor?
47.
Two point charges are 10.0 cm apart and have charges of 2.0 µC and –2.0 µC, respectively. What is the magnitude of the electric field at
the midpoint between the two charges?
48.
Two equal charges are separated by 3.7 × 10 −10 m. The force between the charges has a magnitude of 2.37 × 10 −3 N. What is the
magnitude of q on the charges? (k C = 8.99 × 10 9 N•m 2/C 2 )
49.
What amount of charge moves through an electric fan in 15.1 s if the current through the fan is 1.27 A?
50.
Two charges are located on the positive x-axis of a coordinate system. Charge q 1 = 2.00 × 10 −9 C, and it is 0.020 m from the origin.
Charge q 2 = –3.00 × 10 −9 C, and it is 0.040 m from the origin. What is the electric force exerted by these two charges on a third charge,
q 3 = 5.00 × 10 −9 , located at the origin? (k C = 8.99 × 10 9 N•m 2/C 2 )
51.
A 2.2 kΩ resistor has 0.042 A of current in it. What is the potential difference across the resistor?
4
Name: ________________________
52.
ID: A
In 9.85 × 10 −2 s, 2.43 × 10 −1 C of charge moves through the filament in the magnetron tube of a microwave oven. What is the current in
the magnetron filament?
5
ID: A
Quiz8--Ch 17 & 18 Electric Fields & Circuits PRACTICE
Answer Section
MULTIPLE CHOICE
1.
ANS: D
Given
R1 = 3.0 Ω
R2 = 6.0 Ω
R3 = 12 Ω
Solution
Req = R1 + R2 + R3 = 3.0 Ω + 6.0 Ω + 12 Ω = 21 Ω
2.
PTS: 1
ANS: B
Given
∆V = 4.5 V
R = 8.0 Ω
DIF:
I
OBJ:
18-2.1
DIF:
PTS:
IIIA
1
OBJ:
DIF:
17-3.3
I
OBJ:
16-1.1
IIIA
1
OBJ:
DIF:
17-3.3
II
OBJ:
18-1.3
Solution
∆V = IR
Rearrange to solve for I.
I=
3.
4.
∆V
4.5 V
=
= 0.56 A
8.0 Ω
R
PTS: 1
ANS: B
ANS: B
Given
∆V = 120 V
R = 240 Ω
Solution
∆V = IR
Rearrange to solve for I.
I=
5.
∆V
120 V
=
= 0.50 A
R
240 Ω
PTS:
ANS:
1
D
DIF:
PTS:
1
ID: A
6.
ANS: C
Given
R1 = 10.0 Ω
R2 = 10.0 Ω
R3 = 16 Ω
R4 = 8.0 Ω
R5 = 8.0 Ω
∆V = 60 V
Solution
R1,2 = R1 + R2 = 10.0 Ω + 10.0 Ω = 20.0 Ω
1
1
1
1
1
=
+
=
+
R 4, 5
R4
R5
8.0 Ω
8.0 Ω
1
=
R 4, 5
R 4, 5 =
R 3, 4, 5
1
R eq
1
R eq
R eq =
7.
8.
9.
10.
11.
12.
13.
2.0
8.0 Ω
8.0 Ω
= 4.0 Ω
2.0
= R 3 + R 4, 5 = 16 Ω + 4.0 Ω = 20 Ω
=
=
1
R 1, 2
+
1
R 3, 4, 5
=
1
20.0 Ω
+
1
20 Ω
2
20.0 Ω
20.0 Ω
2
PTS: 1
ANS: B
ANS: B
ANS: A
ANS: A
ANS: B
ANS: C
ANS: D
Given
R1 = 4.0 Ω
R2 = 6.0 Ω
R3 = 8.0 Ω
= 10 Ω
DIF:
PTS:
PTS:
PTS:
PTS:
PTS:
PTS:
IIIB
1
1
1
1
1
1
OBJ:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
18-3.1
II
I
I
I
II
II
OBJ:
OBJ:
OBJ:
OBJ:
OBJ:
OBJ:
18-2.1
17-1.1
16-1.1
16-1.2
18-1.2
18-1.3
OBJ:
DIF:
DIF:
DIF:
18-2.1
I
II
II
OBJ:
OBJ:
OBJ:
18-3.1
17-3.2
18-1.3
Solution
Req = R1 + R2 + R3 = 4.0 Ω + 6.0 Ω + 8.0 Ω = 18 Ω
14.
15.
16.
PTS:
ANS:
ANS:
ANS:
1
B
C
B
DIF:
PTS:
PTS:
PTS:
I
1
1
1
2
ID: A
17.
ANS: D
Given
P = 75 W
∆V = 120 V
Solution
P = I∆V
Rearrange to solve for I.
I=
18.
P
75 W
=
= 0.62 A
120 V
∆V
PTS:
ANS:
Given
1
C
DIF:
IIIA
OBJ:
17-4.3
OBJ:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
17-4.3
I
I
II
I
II
I
I
OBJ:
OBJ:
OBJ:
OBJ:
OBJ:
OBJ:
OBJ:
17-3.2
16-1.1
18-2.1
17-3.4
17-3.1
16-1.2
16-2.2
OBJ:
DIF:
DIF:
DIF:
18-2.2
II
I
I
OBJ:
OBJ:
OBJ:
16-2.2
17-2.1
16-2.3
P = 5.00 × 10 2 W
I = 4.00 A
Solution
P = I∆V
Rearrange to solve for ∆V.
∆V =
19.
20.
21.
22.
23.
24.
25.
26.
P
5.00 × 10 2 W
=
= 1.25 × 10 2 V
I
4.00 A
PTS: 1
ANS: B
ANS: D
ANS: D
ANS: A
ANS: D
ANS: D
ANS: C
ANS: C
Given
R1 = 3.0 Ω
R2 = 6.0 Ω
R3 = 12 Ω
DIF:
PTS:
PTS:
PTS:
PTS:
PTS:
PTS:
PTS:
IIIA
1
1
1
1
1
1
1
Solution
1
1
1
1
1
1
1
=
+
+
=
+
+
R eq
R1
R2
R3
3.0 Ω
6.0 Ω
12 Ω
1
R eq
R eq =
27.
28.
29.
PTS:
ANS:
ANS:
ANS:
=
4.0
12 Ω
12 Ω
7.0
1
B
D
D
+
2.0
12 Ω
+
1.0
12 Ω
=
7.0
12 Ω
= 1.7 Ω
DIF:
PTS:
PTS:
PTS:
IIIA
1
1
1
3
ID: A
30.
ANS: B
Given
R1 = 3.0 Ω
R2 = 3.0 Ω
R3 = 3.0 Ω
R4 = 3.0 Ω
Solution
R1,2,3 = R1 + R2 + R3 = 3.0 Ω + 3.0 Ω + 3.0 Ω = 9.0 Ω
1
1
1
1
1
=
+
=
+
R eq
R 1, 2, 3
R4
9.0 Ω
3.0 Ω
1
R eq
R eq =
31.
1.0
=
9.0 Ω
9.0 Ω
4.0
+
3.0
9.0 Ω
=
4.0
9.0 Ω
= 2.2 Ω
PTS: 1
DIF:
ANS: B
Given
P = 695 W
∆t = 30.0 min
Energy cost = $0.060 per kW•h
IIIA
OBJ:
18-3.1
Solution
ÊÁ 1 h
∆t = ( 30.0 min ) ÁÁÁ
ÁË 60.0 min
ˆ˜
˜˜ = 0.500 h
˜˜
¯
Energy = P∆t = (695 W)(0.500 h) = 348 Wh
ÊÁ
Á 1 kW
Energy = (348 Wh) ÁÁ
ÁË 10 3 W
ˆ˜
˜˜
˜˜ = 0.348 kWh
¯
ÁÊÁ $0.060 ˜ˆ˜
˜˜ = $0.02
$ = (Energy)(Energy cost) = (0.348 kWh) ÁÁ
Á 1 kWh ˜˜
Ë
¯
32.
PTS:
ANS:
1
C
DIF:
PTS:
IIIA
1
OBJ:
DIF:
17-4.3
I
4
OBJ:
18-1.3
ID: A
33.
ANS: A
Given
R1 = 4.0 Ω
R2 = 6.0 Ω
R3 = 10.0 Ω
Solution
1
1
1
1
1
1
1
=
+
+
=
+
+
R eq
R1
R2
R3
4.0 Ω
6.0 Ω
10.0 Ω
1
R eq
R eq =
34.
35.
36.
=
0.25
1Ω
1Ω
0.52
+
0.17
1Ω
+
0.100
1Ω
=
0.52
1Ω
= 1.9 Ω
PTS: 1
ANS: D
ANS: D
ANS: A
Given
R1 = 8.0 Ω
R2 = 2.0 Ω
R3 = 10.0 Ω
R4 = 5.0 Ω
DIF:
PTS:
PTS:
IIIA
1
1
OBJ:
DIF:
DIF:
18-2.2
II
I
OBJ:
OBJ:
18-2.2
18-3.1
OBJ:
DIF:
DIF:
DIF:
18-3.1
I
I
II
OBJ:
OBJ:
OBJ:
16-1.1
16-1.3
16-2.2
Solution
R1,2 = R1 + R2 = 8.0 Ω + 2.0 Ω = 10.0 Ω
1
1
1
1
1
2
=
+
=
+
=
R 1, 2, 3
R 1, 2
R3
10.0 Ω
10.0 Ω
10.0 Ω
10.0 Ω
= 5.00 Ω
2
Req = R1,2,3 + R4 = 5.00 Ω + 5.0 Ω = 10.0 Ω
R 1, 2, 3 =
37.
38.
39.
PTS:
ANS:
ANS:
ANS:
1
A
C
A
DIF:
PTS:
PTS:
PTS:
IIIA
1
1
1
5
ID: A
40.
ANS: C
Given
R1 = 2.0 Ω
R2 = 4.0 Ω
R3 = 6.0 Ω
R4 = 10.0 Ω
∆Vbatt = 12 V
Solution
1
R 2, 3, 4
1
=
=
R 2, 3, 4
R 2, 3, 4 =
1
+
R2
0.25
1Ω
1Ω
0.52
1
R3
+
+
0.17
1Ω
1
R4
+
=
1
4.0 Ω
0.100
1Ω
=
+
1
6.0 Ω
+
1
10.0 Ω
0.52
1Ω
= 1.9 Ω
R 1, 2, 3, 4 = R 1 + R 2, 3, 4 = 2.0 Ω + 1.9 Ω = 3.9 Ω
I total =
∆V batt
R 1, 2, 3, 4
=
12 V
3.9 Ω
= 3.1 A
∆VR1 = R1 × Itotal = 2.0 Ω × 3.1 A = 6.2 V
∆VR4 = ∆Vbatt – ∆VR1 = 12 V – 6.2 V = 5.8 V
∆V R4
5.8 V
I R4 =
=
= 0.58 A
R4
10.0 Ω
41.
42.
43.
44.
PTS: 1
DIF:
ANS: A
PTS:
ANS: A
PTS:
ANS: C
PTS:
ANS: B
Given
I = 5.0 A
∆V = 120 V
Energy cost = $0.090 per kW•h
∆t = 6 h
IIIA
1
1
1
OBJ:
DIF:
DIF:
DIF:
18-3.2
I
I
II
OBJ:
OBJ:
OBJ:
18-1.1
17-3.4
16-3.2
17-4.3
II
OBJ:
18-2.1
Solution
P = I∆V = (5.0 A)(120 V) = 6.0 × 10 2 W
Energy = P∆t = (6.0 × 10 2 W)(6 h) = 3.6 × 10 3 Wh
ÁÊÁ 1 kW ˜ˆ˜
˜ = 3.6 kWh
Energy = (3.6 × 10 3 Wh) ÁÁ
ÁË 10 3 W ˜˜¯
ÊÁÁ $0.090 ˆ˜˜
˜ = $0.32
$ = (Energy)(Energy cost) = (3.6 kWh) ÁÁ
Á 1 kWh ˜˜
¯
Ë
45.
PTS:
ANS:
1
C
DIF:
PTS:
IIIA
1
OBJ:
DIF:
6
ID: A
PROBLEM
46.
ANS:
0.65 V
Given
R = 13.2 Ω
I = 0.049 A
Solution
∆V = I × R = (0.049 A)(13.2 Ω) = 0.65 V
47.
PTS:
ANS:
1
DIF:
IIIA
OBJ:
17-3.3
1.4 × 10 7 N/C
Given
q 1 = 2.0 × 10 −6 C
q 2 = –2.0 × 10 −6 C
θ 1 = 0°
θ 2 = 180°
r total = 10.0 cm = 0.100 m
r 1 = r 2 = 5.00 cm = 0.0500 m
k C = 8.99 × 10 9 Nm2 /C 2
Solution
E 1 = E x,1 = k C
E 2 = E x,2 = k C
q1
r 12
q2
r 22
ÊÁÁ
ˆ˜
ÁÁ 2.0 × 10 −6 C ˜˜˜
ˆ˜ Ë
¯
ÁÊ
= ÁÁ 8.99 × 10 9 Nm2 /C 2 ˜˜
= 7.2 × 10 6 N/C
Ë
¯ ( 0.0500 m ) 2
ÊÁÁ
ˆ˜
ÁÁ −2.0 × 10 −6 C ˜˜˜
ˆ˜ Ë
¯
ÁÊ
= ÁÁ 8.99 × 10 9 Nm2 /C 2 ˜˜
= −7.2 × 10 6 N/C
Ë
¯ ( 0.0500 m ) 2
E x,total = (cos 0°)E x,1 + (cos 180°)E x,2 = (1)(7.2 × 10 6 N/C) + (–1)(−7.2 × 10 6 N/C) = 1.4 × 10 7 N/C
PTS:
1
DIF:
IIIB
OBJ:
16-3.1
7
ID: A
48.
ANS:
1.9 × 10 −16 C
Given
q 1 = q2
F electric = 2.37×10 −3 N
r = 3.7 × 10 −10 m
k C = 8.99 × 10 9 N•m 2 /C 2
Solution
F electric = k C
q=
q=
q1 q2
r2
F electric r 2
=
kC
=
kC q2
r2
(2.37 × 10 − 3 N)(3.7 × 10 −10 m) 2
8.99 × 10 9 Nm2 C 2
(2.37 × 10 −3 N)(1.4 × 10 −19 m2 )
8.99 × 10 9 Nm2 /C 2
q = 1.9×10 −16 C
49.
PTS: 1
ANS:
19.2 C
DIF:
IIIB
OBJ:
16-2.1
OBJ:
17-3.1
Given
∆t = 15.1 s
Ι = 1.27 A
Solution
∆Q
I=
∆t
Rearrange to solve for ∆Q.
∆Q = I∆t = (1.27 A)(15.1 s) = 19.2 C
PTS:
1
DIF:
IIIB
8
ID: A
50.
ANS:
1.4 × 10 −4 N
Given
q 1 = 2.00 ×10 −9 C
q 2 = –3.00 ×10 −9 C
q 3 = 5.00 ×10 −9 C
r 3,1 = 0.020 m = 2.0 ×10 −2 m
r 3,2 = 0.040 m = 4.0 ×10 −2 m
k C = 8.99 ×10 9 Nm2 /C 2
Solution
F 3,1
F 3,2
ÊÁ
ˆÊ
ˆ
ÁÁ 5.00 × 10 −9 C ˜˜˜ ÁÁÁ 2.00 × 10 −9 C ˜˜˜
q 3q 1
ˆ˜ Ë
¯Ë
¯
ÁÊ
= ÁÁ 8.99 × 10 9 Nm2 /C 2 ˜˜
= 2.2 × 10 −4 N
= kC
2
ÊÁ r ˆ˜ 2 Ë
¯
ÊÁ
ˆ
˜
−2
ÁË 3,1 ¯
ÁÁ 2.0 × 10 m ˜˜
Ë
¯
ÊÁ
ˆ
Ê
ˆ
ÁÁ 5.00 × 10 −9 C ˜˜˜ ÁÁÁ −3.00 × 10 −9 C ˜˜˜
q 3q 2
¯Ë
¯
ÁÊ
˜ˆ Ë
= ÁÁ 8.99 × 10 9 Nm2 /C 2 ˜˜
= −8.4 × 10 −5 N
= kC
2
¯
ÊÁ r ˆ˜ 2 Ë
ÊÁ
ˆ
ÁË 3,2 ˜¯
Á 4.0 × 10 −2 m ˜˜
ÁË
˜¯
F 3 = 1.4 × 10 −4 N
51.
PTS:
ANS:
92 V
1
DIF:
IIIC
OBJ:
16-2.3
IIIA
OBJ:
17-3.3
IIIB
OBJ:
17-3.1
Given
R = 2.2 kΩ = 2.2 × 10 3 Ω
I = 0.042 A
Solution
∆V = I × R = (0.042 A)(2.2 × 10 3 Ω) = 92 V
52.
PTS: 1
ANS:
2.47 A
DIF:
Given
∆Q = 2.43 × 10 −1 C
∆t = 9.85 × 10 −2 s
Solution
I=
∆Q
2.43 × 10 −2 C
=
= 2.47 A
∆t
9.85 × 10 −2 s
PTS:
1
DIF:
9