ragsdale (zdr82) – HW4 – ditmire – (58335)
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before answering.
001 (part 1 of 4) 10.0 points
An air-filled capacitor consists of two parallel
plates, each with an area of 1.3 cm2 , separated by a distance 1.8 mm. A 25 V potential
difference is applied to these plates.
The permittivity of a vacuum is 8.85419 ×
10−12 C2 /N · m2 . 1 pF is equal to 10−12 F.
The magnitude of the electric field between
the plates is
2. E =
3. E =
V
d
2
002 (part 2 of 4) 10.0 points
The magnitude of the surface charge density
on each plate is
1. None of these
2. σ =
ǫ0 V
. correct
d
3. σ = ǫ0 V d
4. σ =
1. None of these
ǫ0
.
Vd
5. σ = ǫ0 (V d)2 .
.
6. σ = ǫ0
V
. correct
d
7. σ =
2
4. E = (V d) .
8. σ = ǫ0
d
6. E = .
V
9. σ =
1
.
Vd
8. E =
9. E =
d
V
2
1
(V d)2
V
d
2
.
2
.
ǫ0 d
.
V
5. E = V d .
7. E =
1
d
V
ǫ0
.
(V d)2
Explanation:
Use Gauss’s Law. We find that a pillbox of
cross section S which sticks through the surface on one of the plates encloses charge σ S.
The flux through the pillbox is only through
the top, so the total flux is E S. Gauss’ Law
gives
.
.
σ = ǫ0 E =
Explanation:
Since E is constant between the plates,
Z
~ · d~l = E d
V = E
E=
V
.
d
ǫ0 V
d
Alternatively, we could just recall this result
for an infinite conducting plate (meaning we
neglect edge effects) and apply it.
003 (part 3 of 4) 10.0 points
ragsdale (zdr82) – HW4 – ditmire – (58335)
Calculate the capacitance.
Correct answer: 0.639469 pF.
Explanation:
Let : A = 0.00013 m2 ,
d = 0.0018 m ,
V = 25 V , and
ǫ0 = 8.85419 × 10−12 C2 /N · m2 .
The capacitance is given by
ǫ0 A
C=
d
= 8.85419 × 10−12 C2 /N · m2
0.00013 m2
×
0.0018 m
= 6.39469 × 10−13 F
= 0.639469 pF .
004 (part 4 of 4) 10.0 points
Calculate plate charge; i.e., the magnitude of
the charge on each plate.
Correct answer: 15.9867 pC.
Explanation:
The charge Q on one of the plates is simply
Q=CV
= (6.39469 × 10−13 F) (25 V)
= 1.59867 × 10−11 C
= 15.9867 pC .
005 10.0 points
By what factor does the capacitance of a
metal sphere increase if its volume is tripled?
1. 1.41 times
2. 1.5 times
3. 1.73 times
4. 1.33 times
5. 2 times
2
6. 2.5 times
7. 27 times
8. 3 times
9. 1.44 times correct
10. 9 times
Explanation:
Let : V ′ = 3 V .
4
C = 4 π ǫ0 R , and V = π R3 , so
3
C ∝ V 1/3 .
√
3
Thus √
C ′ ∝ (3 V )1/3 = 3 V 1/3 is larger than
3
C by 3 = 1.44 times.
006 (part 1 of 2) 10.0 points
A coaxial cable has a charged inner conductor
(with charge +8.6 µC and radius 2.219 mm)
and a surrounding oppositely charged conductor (with charge −8.6 µC and radius
7.255 mm).
The Coulomb constant is 8.98755 ×
109 N · m2 /C2 . Assume the region between
the conductors is air, and neglect end effects.
61 m
2.219 mm
+8.6 µC
7.255 mm
−8.6 µC
b
What is the magnitude of the electric field
halfway between the two cylindrical conductors?
Correct answer: 5.34979 × 105 V/m.
Explanation:
Let : ke = 8.98755 × 109 N · m2 /C2 ,
ragsdale (zdr82) – HW4 – ditmire – (58335)
a = 2.219 mm = 0.002219 m ,
b = 7.255 mm = 0.007255 m ,
Q = 8.6 µC = 8.6 × 10−6 C , and
ℓ = 61 m .
The charge per unit length is λ ≡
r=b
−Q
b
Z
Q
. Since
ℓ
b
~ · d~s
E
Z b
dr
= −2 ke λ
a r Q
b
= −2 ke ln
,
ℓ
a
V =−
ℓ
r=a
+Q
3
a
the capacitance of a cylindrical capacitor is
given by
Q
V
ℓ
=
2 ke
C≡
The linear charge density is
Q
ℓ
8.6 × 10−6 C
=
61 m
= 1.40984 × 10−7 C/m .
λ=
The halfway point is
a+b
2
0.002219 m + 0.007255 m
=
2
= 0.004737 m .
r=
The electric field of a cylindrical capacitor is
given by
2 ke λ
E=
r
2 8.98755 × 109 N · m2 /C2
=
0.004737 m
× (1.40984 × 10−7 C/m)
= 5.34979 × 105 V/m .
007 (part 2 of 2) 10.0 points
What is the capacitance of this cable?
Correct answer: 2.86467 nF.
Explanation:
1
b
ln
a
61 m
=
2 (8.98755 × 109 N · m2 /C2 )
1
1 × 109 nF
× 7.255 mm
1F
ln
2.219 mm
= 2.86467 nF .
008 10.0 points
Consider the capacitor circuit
a
9 µF
8 µF
2 µF
c 4 µF
b
EB
Calculate the equivalent capacitance Cab
between points a and b.
Correct answer: 5.73913 µF.
Explanation:
Let : C1
C2
C3
C4
= 9 µF ,
= 8 µF ,
= 2 µF ,
= 4 µF .
and
ragsdale (zdr82) – HW4 – ditmire – (58335)
b
EB
19 µF
C4
19 µF
19 µF
c
C3
19 µF
y
14 µF
a
C2
137 V
C1
4
z
For capacitors in series,
1
Cseries
Vseries
X 1
C
X i
=
Vi ,
=
and the individual charges are the same.
For parallel capacitors,
X
Cparallel =
Ci
X
Qparallel =
Qi ,
19 µF
19 µF
What is the equivalent capacitance between
points y and z of the entire capacitor network?
Correct answer: 20.9091 µF.
Explanation:
Let :
and the individual voltages are the same.
Let c be the point between the two parallel
sections.
Let c be the point between the two parallel
sections. The capacitance between a an c is
Ca
Cb
Cc
Cd
Ce
Cf
=C
=C
=C
=C
=C
=C
= 19 µF ,
= 19 µF ,
= 19 µF ,
= 19 µF ,
= 19 µF ,
= 19 µF ,
Cx = 14 µF = 1.4 × 10−5 F
EB = V = 137 V .
Cl = C1 + C3 = 9 µF + 2 µF
= 11 µF .
Ca
y
and
Cb
Thus the capacitance between a and b is
1
Cr + Cl
1
1
=
=
+
Cab
Cl Cr
Cl Cr
z
Cf
For capacitors in series,
1
Cseries
Cl Cr
Cl + Cr
(11 µF) (12 µF)
=
11 µF + 12 µF
Cab =
= 5.73913 µF .
009 (part 1 of 2) 10.0 points
A capacitor network is shown below.
Cc
Ce
Cx
Cr = C2 + C4 = 8 µF + 4 µF
= 12 µF .
ER
The capacitance between b and c is
Vseries
Cd
X 1
C
X i
=
Vi ,
=
and the individual charges are the same.
For parallel capacitors,
X
Cparallel =
Ci
X
Qparallel =
Qi ,
and the individual voltages are the same.
ragsdale (zdr82) – HW4 – ditmire – (58335)
5
The capacitors Cb , Cc , and Cd are in series,
so
010 (part 2 of 2) 10.0 points
What is the charge on the 14 µF capacitor
centered on the left directly between points y
and z?
Correct answer: 0.001918 C.
Explanation:
Cbcd
Ce
ER
Cx
1
1
1
3
1
= + + =
Cbcd
C C C
C
1
Cbcd = C .
3
This reduces the circuit to
Ca
y
q
V
q = Cx V
= (1.4 × 10−5 F) (137 V)
C≡
z
ER
Cx
Cbcde
Cf
The capacitors Ce and Cbcd are parallel, so
1
4
Cbcde = C + Cbcd = C + C = C .
3
3
This reduces the circuit to
Ca
y
z
Cabcdef
Cx
ER
Cf
The capacitors Ca , Cbcde and Cf are in series,
so
1
3
1
11
1
+ =
= +
Cabcdef
C 4C C
4C
4
Cabcdef =
C.
11
This reduces the circuit to
y
= 0.001918 C .
011 (part 1 of 4) 10.0 points
The capacitors in the figure are initially uncharged and are connected as in the diagram.
Then switch S1 closed and switch S2 is left
open.
c
92
F
µ
µF
23
a
92
µF
137 V
S2
b
µF
23
d
S1
After a long wait, what is the magnitude of
the potential difference Vcd ?
Correct answer: 82.2 V.
Explanation:
z
These capacitors are parallel, so
Cyz = Cx + Cabcdef
4
= Cx +
C
11
4
(19 µF)
= 14 µF +
11
= 20.9091 µF .
Let : C1
C2
C3
C4
Eab
= 23 µF ,
= 92 µF ,
= 92 µF ,
= 23 µF ,
= 137 V .
and
ragsdale (zdr82) – HW4 – ditmire – (58335)
c
C1
a
C3
S2
EB
C2
C4
Redrawing the figure, we have
C1
C2
c
a
EB
C3
b
C4
d
Q12 = C12 Vab =
Q34 = C34 Vab
= (18.4 µF) (137 V)
= 2520.8 µC .
Since C1 and C2 are in series, Q1 = Q2 =
Q12 = 2520.8 µC . C3 and C4 are in series, so
Q3 = Q4 = Q34 = 2520.8 µC . Therefore
Vac
= −82.2 V .
b
EB
The two equivalent capacitors C12 and C34
are parallel, so
Cab = C12 + C34
= 18.4 µF + 18.4 µF
= 36.8 µF .
Q3
C3
2520.8 µC
=
92 µF
= 27.4 V , and
Q1
=
C1
2520.8 µC
=
23 µF
= 109.6 V , and
= Vad − Vac
Vad =
Vcd
C12
C34
EB
Because C34 = C12 , the charge is equally
divided between the two parallel capacitors
To get the potential V across each capacitor
C, we need each charge Q (Q = C V ).
To get Qab , we need Cab (see figure below).
Switch S2 is open. Capacitors C1 and C2 are
in series, and C3 and C4 are in series.
Since C1 = C4 and C2 = C3
1
1
1
1
1
1
=
=
+
=
+
,
C12
C34
C1 C2
C3 C4
so,
C2 C1
=
C12 =
C1 + C2
C3 C4
C34 =
C3 + C4
(92 µF) (23 µF)
=
92 µF + 23 µF
= 18.4 µF .
a
b
Qab = Cab Vab
= (36.8 µF) (137 V)
= 5041.6 µC .
S1
d
Cab
a
b
6
012 (part 2 of 4) 10.0 points
The potential difference Vcd is
1. negative. correct
2. positive.
Explanation:
Vcd = −82.2 V , which is negative.
ragsdale (zdr82) – HW4 – ditmire – (58335)
a
013 (part 3 of 4) 10.0 points
The switch S2 is closed.
c
92
F
µ
µF
23
a
92
S2
µF
137 V
S1
d
Explanation:
A good rule of thumb is to eliminate junctions connected by zero capacitance.
Switch S2 is closed.
C1
C2
C3
c
d
b
C4
Capacitors C1 and C3 are parallel and C2
and C4 are parallel, so
C13 = C1 + C3 =
C24 = C2 + C4
= 23 µF + 92 µF
= 115 µF .
a
Qab
C13
7877.5 µC
=
115 µF
= 68.5 V , and
Vad = V1 = V3 =
Qab
C24
7877.5 µC
=
115 µF
Vdb = V2 = V4 =
014 (part 4 of 4) 10.0 points
What magnitude of charge flowed through the
switch S2 when it was closed?
Correct answer: 4726.5 µC.
Explanation:
When switch S2 is open (before the switch
S2 is closed).
Q1 = Q2 = Q3 = Q4 = 2520.8 µC .
C24
b
EB
Capacitors C13 and C24 are in series. Since
1
1
1
+
, then
=
Cab
C13 C24
C13 C24
C13 + C24
(115 µF) (115 µF)
=
115 µF + 115 µF
= 57.5 µF .
Cab =
Qab = Q13 = Q24 = Cab Vab
= (57.5 µF) (137 V)
= 7877.5 µC .
= 68.5 V .
EB
C13
b
Therefore,
After closing the switch S2 , what is the
potential difference Vad ?
Correct answer: 68.5 V.
a
Cab
EB
b
µF
23
7
After switch S2 is closed
Q1 = C1 Vad =
Q4 = C4 Vdb
= (23 µF) (68.5 V)
= 1575.5 µC , and
Q2 = C2 Vdb =
Q3 = C3 Vad
= (92 µF) (68.5 V)
= 6302 µC .
ragsdale (zdr82) – HW4 – ditmire – (58335)
Therefore
∆Q = |1575.5 µC − 6302 µC|
= 4726.5 µC
of charge flowed from point d to point c when
switch S2 was closed.
Before the switch S2 was closed the net
charge on the conductor on the upper side
(at point c) is zero and the net charge on the
conductor on the lower side (at point d) is
zero. However, after the switch S2 is closed,
the net charge on the upper side (at point c)
is −4726.5 µC and the net charge on the lower
side (at point d) is +4726.5 µC . Consequently,
4726.5 µC of charge must flow through the
switch S2 .
Note: By the upper side (at point c) we
mean the right plate of C1 plus the left plate
of C2 . By the lower side (at point d) we mean
the right plate of C3 plus the left plate of C4 .
015 (part 1 of 3) 10.0 points
Given a spherical capacitor with radius of the
inner conducting sphere a and the outer shell
b. The outer shell is grounded. The charges
are +Q and −Q. A point C is located at
R
r = , where R = a + b.
2
+Q
−Q
a
A
C
B
b
~ at C is
The magnitude of electric field, E,
given by
ke Q
. correct
R2
ke Q 2
.
2. E = 4
R
ke Q 2
3. E = 4
.
R2
1. E = 4
ke Q
.
R
ke Q 2
5. E =
.
R2
ke Q
6. E =
.
R
ke Q 2
7. E = 2
.
R2
ke Q
.
8. E = 2
R
Explanation:
The magnitude of E at C is given by
4. E = 4
ke Q
4 ke Q
E = 2 =
.
R2
R
2
016 (part 2 of 3) 10.0 points
The potential V at C is given by
Q
.
a+b
Q
2. V = −ke .
b
ke Q 1 1
.
−
3. V =
2
a b
2
1
4. V = ke Q
. correct
−
a+b b
ke Q 1 1
.
+
5. V =
2
a b
2
.
6. V = ke Q
a+b
1 1
.
+
7. V = ke Q
a b
1 1
.
−
8. V = ke Q
a b
Explanation:
The potential V at C is given by
Z C
VC − VB = −
E dr
B
Z r
ke Q
=−
dr
r2
b
r= a+b
ke Q 2
=
r b
1. V = ke
8
∆V = Va − Vb
1 1
−0
−
= ke Q
a b
since Vb is grounded. The charge on the
inside of the shell doesn’t affect the grounded
potential.
The capacitance of this spherical capacitor
is
C=
Q
∆V
Q
1 1
−
ke Q
a b
1
.
= 1 1
−
ke
a b
96 V
What is the effective capacitance of the
circuit?
Correct answer: 29.5 µF.
Explanation:
Let : C1
C2
C3
C4
EB
= 34 µF ,
= 25 µF ,
= 26 µF ,
= 33 µF and
= 96 V .
For capacitors in series,
X 1
1
=
Cseries
C
X i
Vseries =
Vi ,
and the individual charges are the same.
For parallel capacitors,
X
Cparallel =
Ci
X
Qparallel =
Qi ,
and the individual voltages are the same.
C1
C2
C3
C4
=
keywords:
9
33 µF
26 µF
1
.
ke (a − b)
ke
.
2. C =
b
a+b
3. C =
.
ke
1
. correct
4. C = 1 1
−
ke
a b
ke
5. C =
.
a
1
6. C =
.
ke (a + b)
b
.
7. C =
ke
a
8. C =
.
ke
Explanation:
1. C =
25 µF
017 (part 3 of 3) 10.0 points
The capacitance of this spherical capacitor is
34 µF
ragsdale (zdr82) – HW4 – ditmire – (58335)
2
1
= ke Q
.
−
018 (part 1 of 2) 10.0 points
a+b b
A capacitor network is shown below.
EB
Parallel connections:
Cparallel = C1 + C2 + C3 + · · ·
Series connections:
1
1
1
1
=
+
+
+ ···
Cseries
C1 C2 C3
ragsdale (zdr82) – HW4 – ditmire – (58335)
C1 and C2 are in parallel, so
10
1
= C4
2
C12 = C1 + C2
= 34 µF + 25 µF
= 59 µF
EB
2
1
= (33 µF)
2
2
96 V
2
2
= 38.016 mJ .
C3 and C4 are in parallel, so
C34 = C3 + C4
= 26 µF + 33 µF
= 59 µF .
C12
EB
C34
020 (part 1 of 4) 10.0 points
Determine the total energy stored in a conducting sphere with charge Q.
Hint: Use the capacitance formula for
a spherical capacitor which consists of two
spherical shells, and take the outer shell to
have an infinite radius.
Q2
16 π ǫ0 a
Q2 π
2. U =
8 ǫ0 a
Q2
3. U =
a
Q2
4. U =
8 π ǫ0 a2
Q2
5. U =
correct
8 π ǫ0 a
Q
6. U =
8 π ǫ0 a
Q2 a
7. U =
4 π ǫ0
Q2
8. U =
4 π ǫ0 a
Explanation:
The capacitance formula for a spherical capacitor of inner radius a and outer radius b
is
ab
C=
.
ke (b − a)
1. U =
C12 and C34 are in series with the battery,
so
C1234
−1 −1
1
C12 + C34
1
+
=
=
C12 C34
C12 C34
C12 C34
=
C12 + C34
(59 µF) (59 µF)
=
59 µF + 59 µF
= 29.5 µF .
019 (part 2 of 2) 10.0 points
For the circuit above what is the total energy
stored by the 33 µF capacitor on the righthand side of the circuit?
Correct answer: 38.016 mJ.
Explanation:
In the simplified circuit, C12 and C34 are in
series with the battery. Since C34 = C12 we
have
EB
V34 = V12 =
.
2
Note: V34 is the potential across both C3 and
C4 since they are parallel.
Finally,
1
U = C4 (V4 )2
2
If we let b → ∞, we find we can neglect a in
the denominator compared to b, so
C→
a
= 4 π ǫ0 a .
ke
The total energy stored is
U=
Q2
Q2
=
.
2C
8 π ǫ0 a
ragsdale (zdr82) – HW4 – ditmire – (58335)
021 (part 2 of 4) 10.0 points
Find the energy stored in a capacitor of charge
Q filled with dielectric, use Cκ = κ C.
Q2
κC
Q
2. U =
4κC
Q
3. U =
2κC
Q2
4. U =
3κC
Q2
5. U =
3 (κ − 1) C
Q2
6. U =
correct
2κC
Q2
7. U =
3C
Q2
8. U =
2C
Q2
9. U =
2 (κ − 1) C
Explanation:
The energy stored in a capacitor of capacitance Cκ is
1. U =
U=
Q2
Q2
=
.
2 Cκ
2κC
022 (part 3 of 4) 10.0 points
Work = Uf − Ui , where “i” is the initial state
where there is a slab in the gap and “f” is the
final state where there is no slab in the gap.
Find the work done in pulling a dielectric
slab of dielectric constant κ from the gap of a
parallel plate capacitor of plate charge Q and
capacitance C.
Q2
1. Wif =
2C
Q2 1
2. Wif =
−1
2C κ
Q2
(1 − κ)
3. Wif =
2κC
1
Q2
κ−
4. Wif =
2C
κ
11
Q2
2κC
Q2
1
6. Wif =
κ+
2C
κ
Q2
1
7. Wif =
1−
correct
2C
κ
Q2
8. Wif =
5κC
Q2
9. Wif =
(1 − κ)
2C
Explanation:
The energy stored in the capacitor with the
dielectric is
Q2
Ui =
.
2κC
5. Wif =
After pulling the slab out, the charge on the
capacitor remains the same (charge conservation). The energy stored is
Uf =
Q2
,
2C
so the work that must be done to pull the slab
out is
W = Uf − Ui =
or
Q2
W =
2C
Q2
Q2
−
,
2C 2κC
1
1−
.
κ
Since κ > 1, this work is positive, meaning we
must supply energy to actually pull the slab
out. This is expected, since edge effects on
the capacitor tend to pull the slab into the
capacitor.
023 (part 4 of 4) 10.0 points
Consider a capacitor which is connected to a
battery with an emf V . Denote the energy
stored in the capacitor in the absence of the
dielectric to be U and in the presence of the
dielectric to be U ′ . ′
U
as the potential across
Find the ratio
U
the plates is held at a constant value by the
battery.
ragsdale (zdr82) – HW4 – ditmire – (58335)
U′
=1
U
U′ √
= 2
2.
U
1
U′
=
3.
U
3
′
U
1
4.
=
U
2κ
′
U
1
5.
=
U
2
′
U
= κ correct
6.
U
1
U′
=
7.
U
κ
′
U
= 2κ
8.
U
Explanation:
Here charge can be supplied by the battery,
so Q is not constant. On the other hand, V
is constant now, so we use the expression for
the energy in terms of C and V instead of Q
and C:
CV2
U=
.
2
With the dielectric inserted, C → κ C, so
1.
U′ =
κC V 2
,
2
and the ratio is
U′
κ C V 2 /2
=
= κ.
U
C V 2 /2
It can be seen from part 3 that the corre1
sponding ratio for keeping Q fixed is .
κ
12