1.
2. A. Yes, there is an electric field inside the box. Electric fields exist as a consequence of a charge body, and electric field is present at a point even if there is no charge at that point of space as long as there are nearby charged bodies.
F=8.988 ×109 N∙m2C22 C×0.01 C3 m2
F=1.9973 ×107 N
C.E=kq1r2
E=8.988 ×109 N∙m2C22 C3 m2
E=1.9973 ×109 NC
3. A. The direction of the force would be downwards, or pointing towards the negatively charged plate.
B.
C.E=FC
E=9×10-6 N-3 ×10-9 C=-3×103 NC
D.Vab=Ed
Vab=-3×103 NC×0.03 m=90 volts
E. The voltage on the battery will be the same as the potential difference calculated on 3.D, which is 90 volts.
4. A.
B.
5. A.
B. negative charge
C. Charge = Current x Time (s)
Q=It
Q=3.5 A ×30s=105 C
D. The switch acts as a bridge from the batteries to the bulb. In this case, the bulb receives no charge if the switch is open. If the switch is closed eternally, both the switch and the bulb receive a certain charge. On the other hand, if the switch is open, no charge is received by the bulb. Therefore, in general, more charge must be received by the switch than the bulb.
6. According to Kirchoff’s rules, the current will be distributed accordingly in a parallel circuit. Since the resistance of the resistors are the same, the current will be distributed evenly in the resistors connected in parallel. On the other hand, at the end joint of the parallel connection, the current goes back to 6 Amperes. Since current is not exhausted, the current will be the same in the last 2 resistors.
7. On the first resistor, the voltage is 6 V, while the current must be similar with the outgoing current, which is 5 A. On the second resistor, the 5-A current is distributed since there is a junction. On the first wire, the current is 3 amperes. Therefore, the current on the second resistor must be 2 amperes.
8. Use V = IR.
On the first resistor:
V=IR
6 V=5 A×R1
R1=1.2 Ω
On the second resistor:
V=IR
2 V=2 A×R2
R2=1 Ω
9.
A.
B. On the voltmeter, the voltage reading must be 6 V. On the ammeter, we will use the equation V = IR to find the current.
V=IR
6 V=I×0.5 Ω
I=12 A
10. A.Req=R1+R2+R3
Req=5 Ω+1 Ω+2 Ω=8 Ω
B.
V=IR
6 V=I×8 Ω
I=0.67 A
C. Since the resistors are in series, the current for each resistor is the same, which is 0.67 amperes.
D.
For the first resistor:
V=IR
V=0.67 A×5 Ω
V=3.33 V
For the second resistor:
V=IR
V=0.67 A×1 Ω
V=0.67 V
For the third resistor:
V=IR
V=0.67 A×2 Ω
V=1.33 V
Adding all the computed voltages:
Veq=V1+V2+V3
Veq=3.33 V+0.67 V+1.33 V
Veq=6 V
11.
A. When the switch is open, only two bulbs receive electrical charge. Moreover, these two bulbs are connected in series. Therefore, the total resistance must be:
Req=R1+R2
Req=4 Ω+4 Ω=8 Ω
24 V=I×8 Ω
I=3 A
And since the bulbs are connected in series, the current throughout the circuit is 3 A. On the other hand, the voltage is distributed among the two bulbs. Note that the resistances of the bulbs are equal, therefore, the voltage must also be equal
V=IR
V=3 A×4 Ω
V=12 V on each bulb
B. First, solve for the total resistance.
Req=11R1+1R2+R3
Req=114 Ω+14 Ω+4 Ω
Req=2 Ω+4 Ω=6 Ω
Then, solve for the total current.
Veq=I×Req
24 V=I×6 Ω
I=4 A
On the first bulb:
V=IR
V=4 A×4 Ω
V=16 V
On the second and third bulb, the current is distributed evenly since the resistance of each bulb is the same. Therefore, the current across the second and third bulb must be 2 amperes. To solve for their voltages: V=IR
V=2 A×4 Ω
V=8 V on the second and third bulbs
C. When the switch is closed, the first bulb and the bulb controlled by the switch get brighter.
12.Req=1 Ω+1.5 Ω+111Ω+1.5Ω+12Ω+13Ω
Req=2.5Ω+13730Ω=3.73 Ω
13.
A. Req=114Ω+116Ω+2Ω=5.2Ω
B. V=IR
24V=I×5.2Ω
I=4.62 A
C.
Let I1 be the current flowing on the 4-Ω resistor, and I2 on the 16-Ω resistor. Using the loop rule:
24V-I216Ω-I1+I22Ω=0 [1]
24V-I14Ω-I1+I22Ω=0 2
I216Ω-I14Ω=0 [3]
Using equation 3, we can say that:
I216Ω-I14Ω=0
4I2=I1 [4]
Note that the total current, which is the sum of I1 and I2, is equal to 4.62 amperes.
I1+I2=4.62 A [5]
Using equation 4 and 5:
I1+I2=4.62 A
4I2++I2=4.62 A
I2=0.923 A
I1=3.697 A
Use these values to solve for the voltage on each resistor.
On the 4-Ω resistor:
V=IR
V=3.697A×4Ω
V=14.79 V
On the 16-Ω resistor:
V=IR
V=0.923×16Ω
V=14.79V
On the 2-Ω resistor:
V=IR
V=4.62A×2Ω
V=9.24 V
