Felder, Jacob – Homework 6 – Due: Oct 17 2006, 9:00 pm – Inst: Vitaly
1
This
printout
should
have
23
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
001
(part 1 of 4) 10 points
An airfilled capacitor consists of two parallel
plates, each with an area of 6
.
6 cm
2
,
sepa
rated by a distance 3 mm
.
A 23 V potential
difference is applied to these plates.
The permittivity of a vacuum is 8
.
85419
×
10

12
C
2
/
N
·
m
2
. 1 pF is equal to 10

12
F
.
The magnitude of the electric field between
the plates is
1.
E
=
1
V d
.
2.
E
=
d
V
¶
2
.
3.
E
=
1
(
V d
)
2
.
4.
E
=
V
d
.
correct
5.
E
=
V d .
6.
E
=
d
V
.
7.
E
=
V
d
¶
2
.
8.
E
= (
V d
)
2
.
9.
None of these
Explanation:
Since
E
is constant between the plates,
V
=
Z
~
E
·
d
~
l
=
E d
E
=
V
d
.
002
(part 2 of 4) 10 points
The magnitude of the surface charge density
on each plate is
1.
σ
=
²
0
d
V
¶
2
.
2.
σ
=
²
0
V d
3.
σ
=
²
0
V
d
.
correct
4.
None of these
5.
σ
=
²
0
(
V d
)
2
.
6.
σ
=
²
0
(
V d
)
2
.
7.
σ
=
²
0
d
V
.
8.
σ
=
²
0
V d
.
9.
σ
=
²
0
V
d
¶
2
.
Explanation:
Use Gauss’s Law. We find that a pillbox of
cross section
S
which sticks through the sur
face 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
=
²
0
E
=
²
0
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 points
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Felder, Jacob – Homework 6 – Due: Oct 17 2006, 9:00 pm – Inst: Vitaly
2
Calculate the capacitance.
Correct answer: 1
.
94792 pF.
Explanation:
Let :
A
= 0
.
00066 m
2
,
d
= 0
.
003 m
,
V
= 23 V
,
and
²
0
= 8
.
85419
×
10

12
C
2
/
N
·
m
2
.
The capacitance is given by
C
=
²
0
A
d
= 8
.
85419
×
10

12
C
2
/
N
·
m
2
×
0
.
00066 m
2
0
.
003 m
= 1
.
94792
×
10

12
F
=
1
.
94792 pF
.
004
(part 4 of 4) 10 points
Calculate plate charge;
i.e.
, the magnitude of
the charge on each plate.
Correct answer: 44
.
8022 pC.
Explanation:
The charge
Q
on one of the plates is simply
Q
=
C V
= (1
.
94792
×
10

12
F) (23 V)
= 4
.
48022
×
10

11
C
=
44
.
8022 pC
.
keywords:
005
(part 1 of 2) 10 points
An isolated conducting sphere can be consid
ered as one element of a capacitor (the other
element being a concentric sphere of infinite
radius).
If
k
=
1
4
π ²
0
, the capacitance of the system
is
C
, and the charge on the sphere is
Q
, what
is the radius of the sphere?
1.
r
=
C
k
2.
r
=
k C
Q
3.
r
=
Q
2
C
k
4.
r
=
k Q C
5.
r
=
k Q
6.
r
=
k C
correct
Explanation:
If the radius of the sphere is
r
, then the
potential difference for the charged sphere is
V
=
k Q
r
and for a capacitor,
C
=
Q
V
=
r
k
.
Thus
r
=
k C .
006
(part 2 of 2) 10 points
The
Coulomb
constant
is
8
.
98755
×
10
9
N
·
m
2
/
C
2
.
If the potential on the surface of the sphere
is 2603 V and the capacitance is 6
.
2
×
10

11
F,
what is the surface charge density?
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 Fall '08
 Opyrchal
 Charge, Work, Correct Answer, Electric charge, Felder, Dielectric

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