Electricity Review Sheet Solutions
1. answer: D
8. answer: C
The positively charged rod near the electroscope
will attract some electrons into the knob (top) of
the electroscope, leaving the electroscope
polarized, but still neutral. The correct
representation has negative signs in the knob,
and positive signs in the leaves (bottom).
ΔV =
9. answer: C
2. answer: B
10. answer: D
By definition, electric field lines point towards
negative charges and away from positive
charges (because a test charge is always
assumed to be positive). So sphere A must be
negative and sphere B must be positive to create
the field lines shown in the drawing.
3. answer: C
kq1q2 9 × 10 9 (1.6 × 10 −19 )2
Fe = 2 =
r
(1.00 × 10 −8 )2
Fe = 2.30 × 10 −12 N
4. answer: A
number of electrons =
charge
fundamental charge
q
−8 × 10 −6
n= =
= 5 × 1013
e −1.6 × 10 −19
5. answer: B
Electrical resistance depends directly on
temperature, length, and resistivity. It depends
inversely on cross-sectional area. That is, the
wider the diameter of resistor the less the
resistance (much like a drinking straw has less
resistance than a narrow stirring straw.)
6. answer: D
Using the definition of resistance as the ratio of
voltage divided by current:
ΔV
ΔV 0.40 volt
R=
⇒ I=
=
= 0.004 A
I
R
100 Ω
Converting amps to milliamps:
1 mA
0.004 A × −3 = 4.0 mA
10 A
7. answer: A
Ordinarily only electrons can be moved from
one body to another (movement of protons
requires high-energy collisions in accelerators!).
So a glass rod acquires a positive charge, when
rubbed with silk, by losing electrons.
E=
PE work
=
q
q
10 =
work
2 × 10 −4
work = 2 × 10 −3 J
kq
(9 × 10 9 )(q)
3600
=
q = 4 × 10 −9 C = 4 nC
r2
0.12
The electric field inside this parallel plate
capacitor is constant (except at the left and right
ends). The field points down because a positive
test charge is used to define the direction of an
electric field, and the positive test charge would
be repelled by the positive charges above, and
attracted by the negative charges below it.
11. answer: A
An electron placed in an electric field gets
pushed opposite to the direction of the field
lines (because of the positive test charge
convention), so in the previous problem the
electron at point P will move up.
12. answer: C
ΔV = I(R1 + R2 )
12 = 2(3 + R2 )
R2 = 3 Ω
13. answer: D
PE = P(t) = I ΔV (t) = 10(120)(30) = 3.6 × 10 4 J
14. answer: B
1
1
1
1
1
1
=
+
+
=
+
Req R1 R2 R3 20 30
Req = 12 Ω
15. answer: B
I=
ΔV 12
=
= 0.60 A
R
20
16. answer: A
P=
ΔV 2 12 2
=
= 4.8 W
R
30
17. answer: B
k(3 nC)(2 nC) k(1 nC)(2 nC)
=
⇒ r = 1.37 m
(1+ r)2
(r)2
18. answer: D
circuit A: Req = 2 + 2 = 4 Ω
circuit B: Req = (2 −1 + 2 −1 )−1 = 1 Ω
circuit C: Req = 1 + 1 = 2 Ω
circuit D: Req = (1−1 + 1−1 )−1 =
smallest resistance is circuit D
1
2
Ω
19a.
Fe =
21a.
kq1q2 (9 × 10 )(8 × 1.6 × 10
=
r2
(5.9 × 10 −15 )2
9
−19 2
ΔV1 = I1 R1
)
ΔV2 = I 2 R2
b.
−19
kq1q2 (9 × 10 )(−1.6 × 10 )(8 × 1.6 × 10
=
r2
(2.4 × 10 −14 )2
Fe = −3.2 N (to the right)
9
−19
120 = I 2 (240)
I 2 = 0.5 A
c.
)
kq q
(9 × 10 9 )(−1.6 × 10 −19 )(8 × 1.6 × 10 −19 )
Fe = 12 3 =
r
(2.4 × 10 −14 + 5.9 × 10 −15 )2
Fe = −2.06 N (to the right)
P3 = I 3 ΔV3
I 3 = 0.67 A
P = I1ΔV1 = 0.6(120) = 72 W
c.
Fe
5.26
=
= 3.29 × 1019 N/C, to the left
−19
q0 (1.6 × 10 )
80 = I 3 (120)
d.
ΣF = 3.2 + 2.06 = 5.26 N (to the right)
E=
R1 = 200 Ω
b.
Fe = 424 N (repulsive)
Fe =
120 = 0.60R1
R1
R
200
I
0.6
V
120
P
72
R2
240
0.5
120
60
R3
180
0.67
120
80
Req
67.9
1.767
120
212
22a.
alternate solution
R12 = R1 + R2 = 6 + 4 = 10 Ω
kq2 (9 × 10 9 )(8 × 1.6 × 10 −19 )2
E= 2 =
r
(2.4 × 10 −14 )2
Req = R123 + R4 + R5 = 6 + 4 + 2 = 12 Ω
R123 = (R12 −1 + R3−1 )−1 = (10 −1 + 15 −1 )−1 = 6 Ω
E = 2.0 × 10 N/C (to the left)
I=
ΔV 36
=
= 3A
Req 12
P=
ΔV 2 36 2
=
= 108 W
Req
12
19
E=
−19 2
kq3 (9 × 10 )(8 × 1.6 × 10 )
=
r2
(2.4 × 10 −14 + 5.9 × 10 −15 )2
9
E = 1.29 × 1019 N/C (to the left)
ΣF = 1.15 × 10 20 + 7.41 × 1019
b.
= 3.29 × 1019 N/C (to the left)
I = I4 = I5 = 3 A
20a.
ΔV = ΔV1 + ΔV2
12 = 10 + ΔV2
ΔV2 = 2 v
b.
I = I1 = I 2 = 0.5 A
ΔV1 = I1 R1
10 = 0.5R1
R1 = 20 Ω
12 = 0.5Req
Req = 24 Ω
P4 = I 4 ΔV4 = 3(12) = 36 W
ΔV3 = ΔV123 = ΔV − ΔV45 = 36 − 18 = 18 v
ΔV3 = I 3 R3
d.
18 = I 3 (15)
I 3 = 1.2 A
P3 = I 3 ΔV3 = 1.2(18) = 21.6 W
P2 = I 2 ΔV2 = 0.5(10) = 5 W
table method:
R
R1
4
P5 = I 5 ΔV5 = 3(6) = 18 W
ΔV4 = I 4 R4 = 3(4) = 12 v
c.
ΔV = IReq
ΔV5 = I 5 R5 = 3(2) = 6 v
I1 = I 2 = I − I 3 = 3 − 1.2 = 1.8 A
I
0.5
V
2
P
1
ΔV2 = I 2 R2 = 1.8(4) = 7.2 v
P2 = I 2 ΔV2 = 1.8(7.2) = 12.96 W
R2
20
0.5
10
5
ΔV1 = I1 R1 = 1.8(6) = 10.8 v
Req
24
0.5
12
6
P1 = I1ΔV1 = 1.8(10.8) = 19.44 W
23.
Series:
R1
R
6
I
1.8
V
10.8
P
19.44
R2
4
1.8
7.2
12.96
R12
10
1.8
18
32.4
Parallel:
R12
R
10
I
1.8
V
18
P
32.4
R3
15
1.2
18
21.6
R123
6
3
18
54
Series:
R123
R
6
I
3
V
18
P
54
R4
4
3
12
36
R5
2
3
6
18
Req
12
3
36
108
Using the table method, work through each resistor column
from the first table, to the second, then to the third.
Then complete the third table for the final series circuit.
Only the known values are shown in bold. All others are
found by multiplication or division.
Then complete the second table. Notice the bottom row is
simply taken from the table below it.
Finally, complete the first table. Again, notice the bottom
row is simply taken from the table below it
assume that current points away from
the positive battery terminals:
I1 in top branch is towards the left
I 2 in middle branch is towards the right
I 3 in bottom branch is towards the right
outer loop, starting at upper, right corner:
−5I1 + 18 − 8I1 − 11I 2 + 12 − 7I 2 = 0
30 − 13I1 − 18I 2 = 0
30 − 18I 2
I1 =
13
lower loop, starting at lower, right corner:
7I 2 − 12 + 11I 2 − 5I 3 + 36 = 0
24 + 18I 2 − 5I 3 = 0
24 + 18I 2
5
junction rule for current:
I1 = I 2 + I 3
combining equations:
30 − 18I 2
24 + 18I 2
= I2 +
13
5
multiply by lcd (65):
5(30 − 18I 2 ) = 65I 2 + 13(24 + 18I 2 )
gather terms:
5(30) − 13(24) = [65 + 5(18) + 13(18)]I 2
−162
I2 =
= −0.416 A
389
direction was assumed wrong, so I 2 = 0.416 A, left
I3 =
30 − 18(−0.416)
= 2.88 A, to the left
13
24 + 18(−0.416)
I3 =
= 3.30 A, to the right
5
note: the current direct, I 2 , was wrong because
the 36 v battery dominates the current direction.
I1 =