PHYS 202 (DDE)
Solved problems in electricity
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1. Common static electricity involves charges ranging from nanocoulombs to microcoulombs.
(a) How many electrons are needed to form a charge of -2.00 nC? (b) How many electrons must be removed from a neutral object to leave a net charge of 0.500 µC?
(a)
2.00×10−9 C
= 1.25×1010 electrons
1.6×10−19 C/e−
(b)
0.500×10−6 C
= 3.13×1012 electrons
1.6×10−19 C/e−
2. An amoeba has 1.00×1016 protons and a net charge of 0.300 pC. (a) How many fewer
electrons are there than protons? (b) If you paired them up, what fraction of the
protons would have no electrons?
(a)
0.300×10−12 C
= 1.88×106 fewer electrons
1.6×10−19 C/e−
(b)
1.88×106
= 1.88×10−10
1.00×1016
3. What is the repulsive force between two pith balls that are 8.00 cm apart and have
equal charges of -30.0 nC?
F = k
q1 q2
r2
= (9.0×109 N-m2 /C2 )
(-30×10−9 C)2
(0.080 m)2
= 1.27×10−3 N
4. How far apart must two point charges of 75.0 nC (typical of static electricity) be to
have a force of 1.00 N between them?
q1 q2
F =k 2
r
so
r =
s
s
kq1 q2
F
9.0×109 N-m2 /C2
(7.5×10−8 C)
1.0 N
= 7.1 mm
=
PHYS 202 (DDE)
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5. By what factor must you change the distance between two point charges to change the
force between them by a factor of 10?
Force is proportional to 1/r 2. Therefore to change the force by a factor of 10, you
√ must
2
change the quantity r by a factor of 10. So you must change r by a factor of 10 or
3.16.
6. A simple and common technique for accelerating electrons is shown in the figure, where
a uniform electric field is created between two plates. Electrons are released, usually
from a hot filament, near the negative plate, and there is a small hole in the positive
plate that allows the electrons to continue moving. (a) Calculate the acceleration of
the electron if the field strength is 2.50×104 N/C. (b) Explain why the electron will
not be pulled back to the positive plate once it moves through the hole.
(a) F = ma so a = F/m, and F = Eq.
(2.50×104 N/C)(1.6×10−19 C)
Eq
=
m
9.1×10−31 kg
= 4.4×1015 m/s2
a =
e
(b) The electric field outside the plates is small. Since
each plate is charged with equal numbers of charges (but
opposite signs), all of the field lines starting at the positive plate will terminate at the negative charges on the
other plate. They do not pass through the plates.
7. The earth has a net charge that produces an electric field of approximately 150 N/C
downward at its surface. (a) What is the magnitude and sign of the excess charge, noting the electric field of a conducting sphere is equivalent to a point charge at its center?
(b) What acceleration will the field produce on a free electron near the earth’s surface?
(c) What mass object with a single extra electron will have its weight supported by
this field?
(a) E = kq/r 2 so q = Er 2 /k.
(150 N/C)(6.4×106 m)2
9.0×109 N-m2 /C2
= 6.83×105 C, negative charge
q =
(b) Since F = ma,
a =
Eq
F
=
m
m
PHYS 202 (DDE)
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(150 N/C)(1.6×10−19 C)
9.1×10−31 kg
= 2.6×1013 m/s2
=
(c) Drawing a free-body diagram, the upward electrical force, Eq is equal to the downward gravitational force, mg:
Eq = mg
Eq
m =
g
(150 N/C)(1.6×10−19 C)
=
9.8 m/s2
= 2.44×10−18 kg
This is a mass about equal to 1.5×109 protons, or a volume of water 1.3 µm on a side:
about as big as a bacterium.
8. Two identical 0.500 g insulating balls with equal charges hang on 20.0 cm long string
as shown in the figure. Find their charge.
Draw a free-body diagram:
T
10
T = tension
Fe = electric force
Fe
L = 20 cm
mg
We know
10
ΣFy = 0
so
10
T cos 10◦ = mg
and
q
q
q2
r
r2
Divide the bottom equation by the top one, and remember that sine/cosine = tangent:
ΣFx = 0
so
T sin 10◦ = Fe = k
PHYS 202 (DDE)
Solved problems in electricity
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T sin 10◦
kq 2 /r 2
=
T cos 10◦
mg
mgr 2 tan 10◦
q2 =
r k
mg
tan 10◦
q = r
k
But from the diagram (above right), r = 2(20 cm)sin 10◦ = 6.95 cm. Then
q = (0.0695
v
u
u (5.0×10−4
m)t
kg)(9.8 m/s2 )
tan 10◦
9.0×109 N m2 /C2
= 2.15×10−8 C
9. If the voltage between two points is zero, can a test charge be moved between them
with zero net work being done? Can this necessarily be done without exerting a force?
Explain.
Yes, a test charge can be moved between them with zero net work being done. However,
this cannot in general be done without exerting a force. For example, when you pick
a book up from a table and transfer it to another location on the table, the net work
done on the book is zero. But certainly a force was exerted; the potential energy of the
book went up when it was raised and back down when it was lowered onto the table.
In the same way, forces might have to be exerted to overcome electric fields present
between the two points and to increase the electrical potential energy of the charge
temporarily between the initial and final location.
10. What is the relationship between voltage and energy? More precisely, what is the
relationship between potential difference and electric potential energy?
The potential difference is the change in electrical potential energy per unit charge.
The unit for potential energy is J/C (= volt) while that for electric potential energy
is just J.
11. Car batteries are rated in ampere-hours (A · h). To what physical quantity do amperehours correspond, and what relationship do ampere-hours have to energy content?
Amp-hours correspond to charge, since A = C/s; when you multipy C/s by time you
get some number of coulombs. The energy content of a battery is its voltage multiplied
PHYS 202 (DDE)
Solved problems in electricity
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by its charge capacity. (To get joules, however, you have to use a charge capacity in
coulombs.)
12. Why do incandescent light bulbs grow dim late in their lives, particularly just before
their filaments break?
As a light bulb ages, the metal in the filament evaporates. That is why the glass of the
bulb ends up with a dark coating on it. As the filament evaporates, it gets thinner and
its resistance increases. This reduces the current through the bulb and it gets dimmer.
If the filament is not perfectly uniform, some part of the filament will be thinner than
the rest and will have a “locally” greater resistance. Since P = IR, this thin spot will
dissipate more power per unit length than the rest of the filament, and will therefore
be hotter. This “hot spot” will evaporate more quickly than the rest of the filament,
making it still thinner and hotter, until it gets so hot the filament melts and breaks in
two.
13. The 200 A current through a spark plug moves 0.300 mC of charge. How long does
the spark last?
We know that Q = It so
Q
0.30×10−3 C
t =
=
I
200 A
= 1.5×10−6 s or 1.5 microseconds.
14. A clock battery wears out after moving 10,000 C of charge through the clock at a rate
of 0.500 mA. (a) How long did the clock run? (b) How many electrons per second
flowed?
(a) One ampere is a coulomb per second. The charge is then found from Q = It. So
t =
=
Q
10,000 C
=
I
0.00050 A
2.0×107 s, or 231 days.
(b) There is 1.6×10−19 C per electron. Thus the flow in e− /s is
5.0×10−4 A
1.6×10−19 C/electron
= 3.13×1015 electrons/s
I =
PHYS 202 (DDE)
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15. The diameter of 0 gauge copper wire is 8.252 mm. Find the resistance of a 1.00 km
length of such wire used for power transmission.
We know
L
A
and we can find the value of ρ for copper in a table of resistivities. Then
R=ρ
R = (1.72×10−8 Ω-m)
1000 m
π(4.126×10−3 m)2
= 0.322 Ω
16. (a) Find the voltage drop in an extension cord having a 0.0600 Ω resistance and through
which 5.00 A of current is flowing. (b) A cheaper cord utilizes thinner wire and has a
resistance of 0.300 Ω. What is the voltage drop in it when 5.00 A flows? (The voltage
to whatever appliance is supplied by the cords is reduced by these amounts.)
(a)
V
= IR = (5.0 A)(0.060 Ω)
= 0.30 V
V
= IR = (5.0 A)(0.30 Ω)
= 1.5 V
(b)
17. Find the power dissipated in each of the extension cords in the previous problem
(a)
P = I 2 R = (5.0 A)2 (0.060 Ω)
= 1.5 W
(b)
P = I 2 R = (5.0 A)2 (0.30 Ω)
= 7.5 W
PHYS 202 (DDE)
Solved problems in electricity
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18. If two household light bulbs rated 60 W and 100 W are connected in series to household
power, which will be brighter?
The 60 W bulb will be brighter. The power ratings are valid only for bulbs connected
to the usual household voltage of 115 V. Since P = V 2 /R, the 60 W bulb has a higher
resistance. When the bulbs are hooked in series, they both pass the same current, and
since P = I 2 R, the higher resistance bulb (the 60 W one) will dissipate the most power
and be brightest.
19. (a) What is the resistance of a 100 Ω, a 2.50 k Ω, and a 4.00 k Ω resistor connected in
series? (b) In parallel?
(a)
Req = 100Ω + 2500Ω + 4000Ω
= 6600 Ω or 6.6 k Ω
(b)
1
1
1
1
=
+
+
Req
100Ω 2500Ω 4000Ω
= 0.01065Ω−1
Req = 94 Ω
20. What resistance do you need to connect to a 137 Ω resistor to get a total of 53.0 Ω?
We must use a parallel connection to get a lower resistance.
1
Req
1
53.0Ω
1
R2
R2
1
1
+
R1 R2
1
1
=
+
137Ω R2
1
1
=
−
= 0.01157Ω−1
53.0Ω 137Ω
= 86.4 Ω
=
PHYS 202 (DDE)
Solved problems in electricity
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21. Standard automobile batteries have six lead-acid cells in series, creating a total emf of
12.0 V. What is the emf of an individual lead-acid cell?
Each cell supplies 2.0 V. When in series, the voltages add.
22. (a) Given a 48.0 V battery, and 24.0 Ω and 96.0 Ω resistors, find the current and power
for each when connected in series. (b) Repeat when the resistances are in parallel.
(a)
V
48 V
48 V
=
=
R
24 Ω + 96 Ω
120 Ω
= 0.40 A for both resistors.
I =
Now, P = I 2 R, so the power dissipated is
P24 = (0.4 A)2 (24 Ω) = 3.84 W for the 24 Ω resistor.
P96 = (0.4 A)2 (96 Ω) = 15.4 W for the 96 Ω resistor.
P = IV = (0.4 A)(48 V) = 19.2 W supplied by the battery.
(b)
I24 =
I96 =
Ibatt =
P24 =
P96 =
P =
48 V
= 2.0 A for the 24 Ω resistor.
24 Ω
48 V
= 0.50 A for the 96 Ω resistor.
96 Ω
2.0 A + 0.5 A = 2.5 A supplied by the battery.
V2
(48 V)2
=
= 96 W for the 24 Ω resistor.
R
24 Ω
(48 V)2
V2
=
= 24 W for the 96 Ω resistor.
R
96 Ω
IV = (2.5 A)(48 V) = 120 W supplied by the battery.
+
23. What is the internal resistance of an automobile battery that has an emf of 12.0 V and
a terminal voltage of 15.0 V while a current of 8.00 A is charging it?
12.0 V
terminal
ri
The voltage drop across the internal resistance is 15.0 V - 12.0 V = 3.0 V. Therefore
terminal
V
3.0 V
R=
=
= 0.38 Ω
I
8.0 A
+
15.0 V
PHYS 202 (DDE)
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24. A 1.58 V alkaline cell with a 0.200 Ω internal resistance is supplying 8.50 A to a load.
(a) What is its terminal voltage? (b) What is the value of the load resistance? (c)
What is unreasonable about these results? (d) Which assumptions are unreasonable
or inconsistent?
(a) The voltage drop across ri is
V = Iri = (8.5 A)(0.20 Ω);
V = 1.70 V across ri .
This would make the terminal voltage a negative 0.12
volts!
(b) The load resistance would have to be negative also,
which does not make sense. You would have
+
1.58 V
−
RL
ri
V
-0.12 V
=
= -0.014 Ω
I
8.50 A
(c) The terminal voltage cannot be negative if the load is just a resistor.
R=
(d) The current is far too large and/or the internal resistance is unreasonable high.
PHYS 202 (DDE)
Solved problems in electricity
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25. In the diagram below, find the current through the 20-ohm resistor.
6Ω
4Ω
20 Ω
−
+
8Ω
12.0 V
Solution:
4Ω
6Ω
+
12.0 V
−
8Ω
−
+
8Ω
6Ω
6Ω
24 Ω
+
20 Ω
−
6Ω
12.0 V
12.0 V
The parallel resistance is found from
1
1
1
4
=
+
=
Rp
24 Ω 8 Ω
24 Ω
so Rp = 6 Ω
The total current is I = V /R = (12.0 V)/(12.0 Ω) = 1.0 A.
Therefore, the voltage drop across the parallel combination is V = IR = (1.0 A)(6 Ω)
= 6.0 V. The current through the 20 Ω plus 4 Ω series combination is
I=
V
6.0 V
=
= 0.25 A
R
24 Ω