Answer, Key – Homework 10 – David McIntyre
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The due time is Central time.
Chapter 26 problems.
001 (part 1 of 1) 0 points
A parallel-plate capacitor is charged by connecting it to a battery.
If the battery is disconnected and the separation between the plates is increased, what
will happen to the charge on the capacitor
and the electric potential a cross it?
1. The charge increases and the electric potential decreases.
2. The charge decreases and the electric potential increases.
Q
smaller. Thus the new potential V 0 = 0 is
C
larger.
002 (part 1 of 2) 5 points
Given: A capacitor network is shown below.
14 µF 20 µF
a
b
7 µF
116 V
Find the equivalent capacitance Cab between points a and b for the group of capacitors.
Correct answer: 15.2353 µF.
Explanation:
Given : C1
C2
C3
EB
3. The charge and the electric potential increase.
4. The charge decreases and the electric potential remains fixed.
5. The charge and the electric potential decrease.
1
C1
a
= 14 µF ,
= 20 µF ,
= 7 µF , and
= 116 V .
C2
C3
6. The charge remains fixed and the electric
potential increases. correct
7. The charge and the electric potential remain fixed.
8. The charge increases and the electric potential remains fixed.
9. The charge remains fixed and the electric
potential decreases.
Explanation:
Charge is conserved, so it must remain constant since it is stuck on the plates. With the
battery disconnected, Q is fixed.
C=²
A
d
A larger d makes the fraction smaller, so C is
b
EB
The capacitors C1 and C2 are in series, so
¶−1
µ
1
1
Cs =
+
C1 C2
C1 C2
=
.
C1 + C 2
Cs is parallel with C3 , so
Cab = Cs + C3
C1 C2
=
+ C3
C1 + C 2
(14 µF) (20 µF)
=
+ 7 µF
14 µF + 20 µF
= 8.23529 µF + 7 µF
= 15.2353 µF .
Answer, Key – Homework 10 – David McIntyre
003 (part 2 of 2) 5 points
What charge is stored on the 7 µF capacitor
on the lower portin of the parallel circuit?
Correct answer: 812 µC.
Explanation:
Since Cs and C3 are parallel, the same potential EB = V is across both, so
= 812 µC .
68 µF
35 µF
004 (part 1 of 2) 5 points
Four capacitors are connected as shown in the
figure.
c
µF
1
2
97 V
b
8
Find the capacitance between points a and
b.
Correct answer: 128.107 µF.
Explanation:
Given : C1
C2
C3
C4
E
= 21 µF ,
= 35 µF ,
= 68 µF ,
= 84 µF ,
= 97 V .
and
C2
a
b
C3
EB
d
C2
C4
b
C3
C4
C23 =
1
1
1
+
C2 C3
C2 C3
=
C2 + C 3
(35 µF) (68 µF)
=
35 µF + 68 µF
= 23.1068 µF .
Cab = C1 + C4 + C23
= 21 µF + 84 µF + 23.1068 µF
= 128.107 µF .
005 (part 2 of 2) 5 points
What is the charge on the 35 µF uppercentered capacitor?
Correct answer: 2241.36 µC.
Explanation:
The voltages across C2 and C3 , respectively,
(the voltage between a and b) are Vab = V23 =
97 V , and we have
c
C1
C1
a
The total capacitance Cab between a and b can
be obtained by calculating the capacitance in
the parallel combination of the capacitors C1 ,
C4 , and C23 ; i.e.,
F
4µ
d
A good rule of thumb is to eliminate junctions connected by zero capacitance.
Q
The definition of capacitance is C ≡ .
V
The series connection of C2 and C3 gives
the equivalent capacitance
q3 = C 3 V
= (7 µF) (116 V)
a
2
Q23 = Q3 = Q2
= Vab C23
= (97 V) (23.1068 µF)
= 2241.36 µC .
006 (part 1 of 2) 0 points
Answer, Key – Homework 10 – David McIntyre
Two capacitors of 21 µF and 4.1 µF are connected parallel and charged with a 131 V
power supply.
Calculate the total energy stored in the two
capacitors.
Correct answer: 0.215371 J.
Explanation:
CV2
2
= C1 + C2 .
3
4.78 µF
a
5.6 µF
b
13.8 µF
17.1 V
U=
Cparallel
Each capacitor has voltage V, so
1
(C1 + C2 ) V 2
2
1
= (21 µF + 4.1 µF) (131 V)2
2
= 0.215371 J .
U=
007 (part 2 of 2) 0 points
What potential difference would be required
across the same two capacitors connected in
series in order for the combination to store
the same energy as in the first part?
Correct answer: 354.359 V.
Explanation:
When in series the equivalent capacitance
is
C1 C2
C1 + C 2
(21 µF) (4.1 µF)
=
21 µF + 4.1 µF
= 3.43028 µF .
Cseries =
Ceq V 02
, we have
2
s
2U
V0 =
Ceq
s
2 (0.215371 J) 106 µF
=
·
3.43028 µF
1F
= 354.359 V .
Since U =
008 (part 1 of 7) 2 points
Given: A capacitor network is shown in the
following figure.
What is effective capacitance Cab of the
entire capacitor network?
Correct answer: 16.3788 µF.
Explanation:
Given:
C1
C2
C3
EB
= 4.78 µF ,
= 5.6 µF ,
= 13.8 µF ,
= 17.1 V .
and
C1
a
C2
C3
b
EB
C1 and C2 are in series with each other, and
they are together are parallel with C3 . So
C1 C2
+ C3
C1 + C 2
(4.78 µF) (5.6 µF)
+ 13.8 µF
=
4.78 µF + 5.6 µF
Cab =
= 16.3788 µF .
009 (part 2 of 7) 2 points
What is the voltage across the 5.6 µF upper
right-hand capacitor?
Correct answer: 7.87457 V.
Explanation:
Since C1 and C2 are in series they carry the
same charge
C1 V1 = C 2 V2 ,
Answer, Key – Homework 10 – David McIntyre
and their voltages add up to V , voltage of the
battery
V1 + V 2 = V
C2 V2
+ V2 = V
C1
C2 V2 + C 1 V2 = V C 1
V C1
V2 =
C1 + C 2
(17.1 V)(4.78 µF)
=
4.78 µF + 5.6 µF
= 7.87457 V .
010 (part 3 of 7) 2 points
If a dielectric of constant 4.18 is inserted in
the 5.6 µF top right-hand capacitor (when
the battery is connected), what is the electric
potential across the 4.78 µF top left-hand
capacitor?
Correct answer: 14.2003 V.
Explanation:
Given : κ = 4.18 .
When the dielectric is inserted, the capacitance formerly C2 becomes
C20 = κ C2 ,
4
= (13.8 µF)(17.1 V)
= 235.98 µc
0
Q1 = Q02
0
= C12
V
= (3.96943 µF)(17.1 V)
= 67.8772 µc .
When we remove the dielectric, the sum of
the charges stays the same, and the voltages
on C3 and on C12 (where C12 is the equivalent
capacitance of C1 and C2 in series) are equal
to each other
Q001 + Q003 = Q01 + Q03
Q001
Q00
= 3.
C12
C3
Therefore
Q01 + Q03
C3
1+
C12
67.8772 µc + 235.98 µc
=
13.8 µF
1+
2.57881 µF
Q001 =
= 47.8416 µc .
and the new voltage across C1 is
V10
V C20
=
C1 + C20
κ V C2
=
C1 + κ C 2
(4.18)(17.1 V)(5.6 µF)
=
4.78 µF + (4.18)(5.6 µF)
= 14.2003 V .
011 (part 4 of 7) 1 points
If the battery is disconnected and then the
dielectric is removed, what is the charge on
4.78 µF top left-hand capacitor?
Correct answer: 47.8416 µc.
Explanation:
Immediately before the battery was disconnected the charges on the capacitors had been
Q03
= C3 V
012 (part 5 of 7) 1 points
What is now the voltage drop across the
5.6 µF top right-hand capacitor?
Correct answer: 8.54315 V.
Explanation:
Q002
C2
Q00
= 1
C2
47.8416 µc
=
5.6 µF
V200 =
= 8.54315 V .
013 (part 6 of 7) 1 points
What was the energy stored in the system
Answer, Key – Homework 10 – David McIntyre
before the dielectric was removed?
Correct answer: 0.00259798 J.
Explanation:
The total capacitance of the system before
the dielectric was removed had been
0
0
Cab
= C12
+ C3
= 3.96943 µF + 13.8 µF
= 17.7694 µF
so the energy stored in the system was
1 0 2
C V
2 ab
= 2597.98 µJ
U0 =
= 0.00259798 J .
014 (part 7 of 7) 1 points
What is the energy stored in the system after
the dielectric was removed?
Correct answer: 0.00281856 J.
Explanation:
Since the battery is disconnected, the total
charge is conserved:
Q00 = Q0
0
= Cab
V
= (17.7694 µF)(17.1 V)
= 303.857 µc .
Therefore, the energy stored is
Q002
U =
2 Cab
= 2818.56 µJ
5
3. The capacitance decreases.
4. The charge on the capacitor plates decreases.
5. The electric field between the capacitor
plates increases.
Explanation:
Since the capacitor is isolated, the charge
on the capacitor plates remains the same. On
the other hand, the capacitance is increased
by introducing a dielectric. Therefore, from
Q
V = , the potential difference across the
C
capacitor is decreased.
016 (part 1 of 1) 0 points
A parallel plate capacitor is attached to a
battery which maintains a constant potential
difference of V between the plates. While
the battery is still connected, a glass slab is
inserted so as to just fill the space between
the capacitor plates.
V
V
The stored energy will
1. remain the same.
2. increase. correct
00
= 0.00281856 J .
015 (part 1 of 1) 0 points
A sheet of mica is inserted between the plates
of an isolated charged parallel-plate capacitor.
Which of the following statements is true?
3. decrease.
Explanation:
The energy stored in the capacitor is given
by
Q2
κC 2
U=
=
V
2 κC
2
Without the glass, the stored energy is
U0 =
1
CV2
2
1. The potential difference across the capacitor decreases. correct
After inserting the glass, it becomes
2. The energy of the capacitor does not
change.
U=
1
κC V 2
2
Answer, Key – Homework 10 – David McIntyre
Since κ > 1, the stored energy will increase.
017 (part 1 of 6) 2 points
A parallel-plate capacitor has a plate area of
105 cm2 and a plate separation of 2.25 mm. A
potential difference of 5.87 V is applied across
the plates with only air between the plates.
The battery is then disconnected, and a piece
of glass (κ = 8.53) is inserted to completely
fill the space between the plates.
What is the capacitance before the dielectric is inserted?
Correct answer: 4.13195 × 10−11 F.
Explanation:
Given :
²0 = 8.85419 × 10−12 C2 /N m2 ,
A = 105 cm2 , and
d = 2.25 mm .
The capacitance before the dielectric is inserted is
²0 A
¡d
¢
8.85419 × 10−12 C2 /N m2 (105 cm2 )
=
(2.25 mm)
C1 =
= 4.13195 × 10−11 F .
018 (part 2 of 6) 2 points
What is the capacitance after the dielectric is
inserted?
Correct answer: 3.52456 × 10−10 F.
Explanation:
Given : κ = 8.53 .
The capacitance after the dielectric is inserted
is
κ ²0 A
d
(8.53) (²0 ) (105 cm2 )
=
(2.25 mm)
C2 =
= 3.52456 × 10−10 F .
019 (part 3 of 6) 2 points
6
What is the charge on the plates before the
dielectric is inserted?
Correct answer: 2.42546 × 10−10 C.
Explanation:
Given : V = 5.87 V .
The charge on the plates before the dielectric
is inserted is given by
Q1 = C 1 V
= (4.13195 × 10−11 F) (5.87 V)
= 2.42546 × 10−10 C .
020 (part 4 of 6) 2 points
What is the charge on the plates after the
dielectric is inserted?
Correct answer: 2.42546 × 10−10 C.
Explanation:
The charge on the plates doesn’t change after
the dielectric is inserted, so it is given by
Q2 = Q1 = 2.42546 × 10−10 C .
021 (part 5 of 6) 1 points
What is the potential difference across the
plates before the dielectric is inserted?
Correct answer: 5.87 V.
Explanation:
The potential difference across the plates
before the dielectric is inserted is given by
V1 = V = 5.87 V .
022 (part 6 of 6) 1 points
What is the potential difference across the
plates after the dielectric is inserted?
Correct answer: 0.688159 V.
Explanation:
The potential difference across the plates
after the dielectric is inserted is given by
V2 =
V
5.87 V
=
= 0.688159 V .
κ
8.53