Version One – Homework 4 – Savrasov – 39821 – Apr 16, 2007
This print-out should have 8 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering. The due time is Central
time.
Air Filled Capacitor S
26:02, trigonometry, numeric, > 1 min, fixed.
001 (part 1 of 1) 1 points
An air-filled capacitor consists of two parallel
plates, each with an area A, separated by a
distance d. A potential difference V is applied
to the two plates. The magnitude of the
surface charge density on the inner surface of
each plate is
1. σ = ²0 (V d)2
2. σ = ²0 V d
1
Alternatively, we could just recall this result
for an infinite conducting plate (meaning we
neglect edge effects) and apply it.
keywords:
Capacitance of a Sphere
26:02, trigonometry, numeric, > 1 min, normal.
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 = k Q C
²0 d
3. σ =
V
4. σ = ²0
5. σ = ²0
2. r =
µ
V
d
µ
d
V
¶2
3. r = k C correct
kC
Q
Q2 C
5. r =
k
4. r =
¶2
6. r = k Q
²0 V
correct
d
²0
7. σ =
Vd
6. σ =
8. σ =
Explanation:
The definition of capacitance is
Q
V
Q=CV .
C=
²0
(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 = ²0 E =
C
k
²0 V
d
Let V (∞) = 0. Then the potential of the
conducting sphere is
V =k
CV
Q
=k
,
r
r
so
r = kC.
003 (part 2 of 2) 1 points
Version One – Homework 4 – Savrasov – 39821 – Apr 16, 2007
The Coulomb constant is 8.98755 ×
109 N · m2 /C2 .
If the potential on the surface of the sphere
is 1000 V and the capacitance is 8.5 × 10−11 F,
what is the surface charge density?
Correct answer: 11.5901 nC/m2 .
Explanation:
Let : k = 8.98755 × 109 N · m2 /C2 ,
V = 1000 V , and
C = 8.5 × 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
1000 V
×
8.5 × 10−11 F
= 11.5901 nC/m2 .
keywords:
Plate Separation
26:02, trigonometry, numeric, > 1 min, normal.
004 (part 1 of 1) 2 points
A parallel-plate capacitor has a plate area of
12 cm2 and a capacitance of 7 pF.
The permittivity of a vacuum is 8.85419 ×
−12 2
10
C /N · m2 .
What is the plate separation?
Correct answer: 0.00151786 m.
Explanation:
Let : A = 12 cm2 = 0.0012 m2 ,
C = 7 pF = 7 × 10−12 F , and
²0 = 8.85419 × 10−12 C2 /N · m2 .
2
²0 A
d
²0 A
d=
¢
¡C
8.85419 × 10−12 C2 /N · m2
=
7 × 10−12 F
¡
¢
× 0.0012 m2
C=
= 0.00151786 m .
keywords:
Infinite Capacitor Chain 02
26:03, trigonometry, numeric, > 1 min, normal.
005 (part 1 of 1) 2 points
An infinite chain of capacitors is pictured
below with C1 = 10 µF, C2 = 5 µF, and
C3 = 15 µF.
a
C1
C1
C2
C1
C2
C2
b
C3
C3
C3
What is Cab ?
Correct answer: 3.5208 µF.
Explanation:
Let : C1 = 10 µF ,
C2 = 5 µF , and
C3 = 15 µF .
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'
C2
Ceq
b'
b
C3
Version One – Homework 4 – Savrasov – 39821 – Apr 16, 2007
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
C2
C1 + C 3
=
Ceq (C2 + Ceq )
C1 C3
1. E0 =
C1 C2 C3
2
= Ceq
+ C2 Ceq
C1 + C 3
−C2 +
Ceq =
r
2. E0 =
C1 C2 C3
=0
C1 + C 3
C22 +
4 C 1 C2 C3
C1 + C 3
2
3. E0 =
4. E0 =
.
5. E0 =
Under the radical,
4 C 1 C2 C3
C1 + C 3
4 (10 µF) (5 µF) (15 µF)
= (5 µF)2 +
(10 µF) + (15 µF)
2
= 145 (µF) ,
C22 +
so
Cab = Ceq
1
1
= − (5 µF) +
2
2
= 3.5208 µF .
q
`
A = `2
d
Without the dielectrics, what is the field E0
between the plates?
1
1
1
1
−
=
+
Ceq C2 + Ceq
C1 C3
2
Ceq
+ C2 Ceq −
dielectric κ2
constant
left
dielectric
constant κ1
right
1
1
1
1
=
+
+
Ceq
C1 C2 + Ceq C3
3
145 (µF)2
6. E0 =
7. E0 =
Explanation:
Solution:
Q
From Gauss’ Law, we know that Φ = encl .
²0
For a flat surface, like a plate, the flux is the
normal component of the electric field times
the area. Also, for a parallel plate capacitor,
the field is only in the gap (neglecting edge effects), so
keywords:
Two Dielectric Slabs
26:05, trigonometry, numeric, > 1 min, normal.
006 (part 1 of 3) 1 points
A parallel plate capacitor has capacitance C0 ,
plate separation d and plate area A. Two dielectric slabs of dielectric constants κ1 and κ2 ,
d
each of thickness , are inserted between the
2
plates. Charges Q and −Q are put on the
upper and lower plates.
Q
correct
²0 A
Q
2²0 A
Q
4 ²0 A
2Q
²0 A
Q
d
2 ²0 A
Q
d
4 ²0 A
2Q
d
²0 A
Φ = E0 A
E0 =
Φ
Q
=
.
A
²0 A
007 (part 2 of 3) 1 points
With the dielectrics inserted, the fields in κ1
and κ2 are respectively
1. E1 =
E0 κ1 κ2
E0 κ1 κ2
and E2 =
κ1 + κ2
κ1 + κ2
2. E1 = E0 κ1 and E2 = E0 κ2
Version One – Homework 4 – Savrasov – 39821 – Apr 16, 2007
·
¸
(2) (3)
=2
E0 κ2
E0 κ1
2+3
and E2 =
3. E1 =
¡
¢
κ1 + κ2
κ1 + κ2
8.854 × 10−12 F/m (0.5 m2 )
E0
E0
×
4. E1 =
and E2 =
correct
0.00025 m
κ1
κ2
= 4.24992 × 10−8 F .
E0
E0
5. E1 =
and E2 =
κ2
κ1
6. E1 = E0 κ2 and E2 = E0 κ1
E0 κ1
E0 κ2
and E2 =
κ1 + κ2
κ1 + κ2
Explanation:
In any region with dielectric constant κ, the
permittivity ²0 becomes κ²0 , so the fields in
κ1 and κ2 are
7. E1 =
Q
E0
=
κ1 ²0 A
κ1
Q
E0
E2 =
=
.
κ2 ²0 A
κ2
E1 =
008 (part 3 of 3) 1 points
Let κ1 = 2, κ2 = 3, A = 0.5 m2 , and d =
0.25 mm.
What is the resultant capacitance?
Correct answer: 4.24992 × 10−8 F.
Explanation:
The potential difference between the two
plates is the sum of the potential differences
across the two dielectrics. The potential difference across a region with a constant electric
field is the product of the field and the distance, so
V = V 1 + V2
E0 d E0 d
=
+
κ1 2
κ 2
¶
µ2
Q d
1
1
.
=
+
²0 A 2 κ1 κ2
The capacitance is given by
Q
V
Ã
1
2 ²0 A
=
1
d
κ1 +
C=
=2
1
κ2
κ1 κ2 ²0 A
·
κ1 + κ2
d
!
keywords:
4