GenericStudent – Homework 4 – savrasov – 39823 – Jan 02, 2008
This print-out should have 11 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering. The due time is Central
time.
001 (part 1 of 1) 1 points
A 18 pF capacitor is connected across a 68 V
source.
What charge is stored on it?
Correct answer: 1.224 × 10−9 C.
Explanation:
1
Explanation:
The definition of capacitance is
Q
V
Q=CV .
C=
Let V (∞) = 0. Then the potential of the
conducting sphere is
V =k
CV
Q
=k
,
r
r
so
Let :
r = kC.
C = 18 pF = 1.8 × 10−11 F and
V = 68 V .
The capacitance is
q
C=
V
q =CV
= (1.8 × 10−11 F) (68 V)
= 1.224 × 10−9 C
003 (part 2 of 2) 1 points
The Coulomb constant is 8.98755 ×
109 N · m2 /C2 .
If the potential on the surface of the
sphere is 5700 V and the capacitance is
7.56 × 10−11 F, what is the surface charge
density?
Correct answer: 74.278 nC/m2 .
Explanation:
keywords:
002 (part 1 of 2) 1 points
An isolated conducting sphere can be considered as one element of a capacitor (the other
element being a concentric sphere of infinite
radius).
1
If k =
, the capacitance of the system
4 π ²0
is C, and the charge on the sphere is Q, what
is the radius r of the sphere?
1. r =
C
k
2. r = k Q C
3. r = k C correct
4. r =
Q2 C
k
5. r = k Q
6. r =
kC
Q
Let : k = 8.98755 × 109 N · m2 /C2 ,
V = 5700 V , and
C = 7.56 × 10−11 F .
The charge is Q = C V .
Assuming a uniform surface charge density
we have
Q
4 π r2
CV
=
4 π (k C)2
1 V
=
4 π k2 C
σ=
=
1
4 π (8.98755 × 109 N · m2 /C2 )2
5700 V
×
7.56 × 10−11 F
= 74.278 nC/m2 .
GenericStudent – Homework 4 – savrasov – 39823 – Jan 02, 2008
keywords:
Cs is parallel with C3 , so
Cab = Cs + C3
C1 C2
=
+ C3
C1 + C 2
(20 µF) (24 µF)
+ 4 µF
=
20 µF + 24 µF
= 10.9091 µF + 4 µF
004 (part 1 of 2) 1 points
A capacitor network is shown below.
20 µF 24 µF
a
b
4 µF
= 14.9091 µF .
85 V
Find the equivalent capacitance Cab between points a and b for the group of capacitors.
Correct answer: 14.9091 µF.
Explanation:
Let :
C1
C2
C3
EB
C1
a
2
005 (part 2 of 2) 1 points
What charge is stored on the 4 µF capacitor
on the lower portion of the parallel circuit?
Correct answer: 340 µC.
Explanation:
Since Cs and C3 are parallel, the same potential EB = V is across both, so
= 20 µF ,
= 24 µF ,
= 4 µF , and
= 85 V .
q3 = C 3 V
= (4 µF) (85 V)
= 340 µC .
C2
C3
keywords:
b
EB
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.
The capacitors C1 and C2 are in series, so
¶−1
µ
C1 C2
1
1
=
Cs =
+
.
C1 C2
C1 + C 2
006 (part 1 of 4) 1 points
An infinite chain of capacitors is pictured below with C1 = 13.4 µF, C2 = 5.14 µF, and
C3 = 12.2 µF.
a
C1
C1
C2
C1
C2
C2
b
C3
C3
C3
What is Cab ?
Correct answer: 3.70922 µF.
Explanation:
Let : C1 = 13.4 µF ,
C2 = 5.14 µF ,
C3 = 12.2 µF .
and
GenericStudent – Homework 4 – savrasov – 39823 – Jan 02, 2008
The capacitance Ceq = Cab . Imagine points
a0 and b0 in the circuit just past the first
capacitor C2 . The equivalent circuit to the
right of points a0 and b0 is Ceq , as shown in the
figure below.
a
C1
A dielectric with constant 3.22 is only inserted
into the first capacitor labeled C2 and not into
the others.
What is Cab now?
Correct answer: 4.85549 µF.
Explanation:
After the dielectric is added denote symbols
0
with primes, C2 → C20 , Cab → Cab
, etc.
a'
Ceq
C2
C20 = κ C2 = (3.22)(5.14 µF) = 16.5508 µF .
b'
b
C3
That is, we are replacing the circuit to the
right of points a0 and b0 with the circuit we
are trying to resolve.
Solving for Ceq
= 4.85549 µF .
1
1
1
1
−
=
+
Ceq C2 + Ceq
C1 C3
008 (part 3 of 4) 1 points
A 3.81 V battery is placed between the terminals ab.
What is the charge on the capacitor containing the dielectric?
Correct answer: 15.1125 µC.
Explanation:
Let
C1 + C 3
C2
=
Ceq (C2 + Ceq )
C1 C3
C1 C2 C3
2
= Ceq
+ C2 Ceq
C1 + C 3
−C2 +
Ceq =
r
C1 C2 C3
=0
C1 + C 3
C22 +
4 C 1 C2 C3
C1 + C 3
2
.
Under the radical,
4 C 1 C2 C3
C1 + C 3
4 (13.4 µF) (5.14 µF) (12.2 µF)
= (5.14 µF)2 +
(13.4 µF) + (12.2 µF)
2
= 157.714 (µF) ,
C22 +
so
Cab = Ceq
1
1
= − (5.14 µF) +
2
2
= 3.70922 µF .
q
157.714 (µF)2
007 (part 2 of 4) 1 points
1
1
1
1
C1 + C20 + Ceq + C3
C1 (C20 + Ceq )C3
=
C1 + C20 + Ceq + C3
(13.4 µF)(16.5508 µF + 3.70922 µF)(12.2 µF)
=
13.4 µF + 16.5508 µF + 3.70922 µF + 12.2 µF
0
Cab
=
1
1
1
1
=
+
+
Ceq
C1 C2 + Ceq C3
2
Ceq
+ C2 Ceq −
3
0
V
Q0ab = Cab
= (4.85549 µF) (3.81 V)
= 18.4994 µC
be the charge on the equivalent capacitance
of the network at ab.
Let
Ca0 0 b0 = C20 + Ceq
= 16.5508 µF + 3.70922 µF
= 20.26 µF
be the equivalent capacitance of capacitors C20
and Ceq , which are parallel.
Capacitors C1 , C3 , and Ca0 0 b0 are in series
and the charge on all capacitors in series are
equal, so
Q1 = Q3 = Q0a0 b0 = Q0ab = 18.4994 µC .
GenericStudent – Homework 4 – savrasov – 39823 – Jan 02, 2008
a
Q1
4
a'
Q'2
Q'eq
b'
b
Q3
Therefore
Q02 + Qeq
Ca0 0 b0
Q0 0 0
= a0 b
C a0 b 0
18.4994 µC
=
20.26 µF
= 0.9131 V .
V a0 b 0 =
and
Q02 = C20 Va0 b0
= κ C 2 V a0 b 0
= (3.22)(5.14 µF)(0.9131 V)
= 15.1125 µC .
009 (part 4 of 4) 1 points
The battery is removed.
How much work will then be required to
pull the dielectric out of the capacitor?
Correct answer: 10.8907 µJ.
Explanation:
1 Q2
. Once the
The stored energy is U =
2 C
battery is disconnected we must consider the
two capacitors parallel before and after the
dielectric is removed.
Before removal,
Ca0 0 b0 = C20 + Ceq
= 16.5508 µF + 3.70922 µF
= 20.26 µF ,
and after,
Ca0 b0 = C2 + Ceq
= 5.14 µF + 3.70922 µF
= 8.84922 µF .
(Q0ab )2
∆U =
2
µ
1
Ca0 0 b0
−
1
C a0 b 0
¶
µC)2
(18.4994
¶
µ 2
1
1
−
×
20.26 µF 8.84922 µF
= −10.8907 µJ
=
W = −∆U = 10.8907 µJ .
keywords:
010 (part 1 of 2) 1 points
Consider a air-filled parallel-plate capacitor
with plates of length 9.1 cm, width 6.81 cm,
spaced a distance 1.38 cm apart.
Now imagine that a dielectric slab with dielectric constant 3.8 is inserted a length 3.6 cm
into the capacitor. The slab has the same
width as the plates. The capacitor is completely filled with dielectric material down to
a length of 3.6 cm.
First a battery is connected to the plates so
that they are at a potential of 0.57 V , then the
battery is disconnected while the dielectric is
inserted.
dielectric
constant 3.8
top
3.6 cm
9.1 cm
dielectric
none - air 1.0
bottom
1.38 cm
What is the ratio of the new potential energy to the potential energy before the insertion of the dielectric? Edge effects can be
neglected.
Correct answer: 2.10769 .
Explanation:
Let :
w = 6.81 cm ,
d = 1.38 cm ,
L = 9.1 cm ,
` = 3.6 cm ,
GenericStudent – Homework 4 – savrasov – 39823 – Jan 02, 2008
5
Therefore, we have
κ = 3.8 , and
V = 0.57 V .
The given capacitor can be considered as
two capacitors in parallel, as shown below.
C1
²0 w
L
U0
C
d o
= 0 =n
²0 w
U
C
L + ` [κ − 1]
d
L
h
i
=
L+` κ−1
=
(5)
(9.1 cm)
h
i
(9.1 cm) + (3.6 cm) (3.8) − 1
= 0.474453 .
C2
When the dielectric is fully inserted ` = L ,
The potential energy in a capacitor is
L
1
U0
h
i= .
=
U
κ
L+L κ−1
1 Q2
U=
,
2 C
(6)
011 (part 2 of 2) 1 points
From this, we can deduce that if a person
inserts the slab part way into the conductor,
and then releases it, the capacitor will exert a
force on the dielectric in what direction?
so the ratio is
Q0 2 C
U0
= 0
.
U
C Q2
Since the charge on the capacitor is constant,
1. pushing it out of the capacitor
U0
C
= 0.
U
C
(1)
The capacitance of a parallel plate capacitor is
C=
²0 w L
²0 · area
=
.
separation
d
(2)
After the insertion of the dielectric, the capacitor can be considered in two parts: the
part filled with dielectric, and the part with
no dielectric.
Since the dielectric is inserted a length ` ,
C 0 = C10 + C2 , where
²0 w κ `
C10 =
d
²0 w (L − `)
C2 =
.
d
(3)
= (L − ` + κ `)
²0 w
.
d
3. toward the negatively charged plate
4. toward the positively charged plate
5. there will be no force exerted on the slab
Explanation:
Force is the negative gradient of potential;
i.e., since potential energy U is decreased as `
increases, the dielectric will pull itself into the
capacitor.
We found earlier that the potential energy
stored in the capacitor is given by
1 0 2
C V
2
·
¸
1 2 ²0 w
L
= V
.
2
d
L + ` (κ − 1)
U0 =
The total capacitance of the new system is
C 0 = C10 + C2
2. pulling it further into the capacitor correct
(4)
Note: κ is always greater than one in a dielectric, so U 0 has to increase with increasing
`.
GenericStudent – Homework 4 – savrasov – 39823 – Jan 02, 2008
The capacitor exerts a force in the direction
of decreasing potential energy, so it will push
the slab out of the capacitor.
IF BATTERY STAYS CONNECTED
If the battery remains connected while the dielectric is inserted, the force on the dielectric
will be in the opposite direction; i.e., “pushing
it out of the capacitor”.
The potential energy in a capacitor is
U=
1
CV2,
2
so the ratio is
C 0 V 02
U0
=
.
U
CV2
Since the electric potential difference (voltage) across the capacitor is constant,
U0
C0
=
,
U
C
(7)
which is the inverse of Eq. 1.
Therefore (using Eq. 2 and 7), we have
n
o² w
0
L
+
`
[κ
−
1]
0
0
U
C
d
=
=
²0 w
U
C
L
d
h
i
`
=1+
κ−1
(8)
L
i
(3.6 cm) h
=1+
(3.8) − 1
(9.1 cm)
1
(see part 1) .
= 2.10769 =
0.474453
When the dielectric is fully inserted ` = L,
so
i
U0
Lh
=1+
κ − 1 = κ,
U
L
which is also the inverse of Eq. 6.
keywords:
(9)
6