oldhw10solution_pdf - klarin(sjk772 oldhomework 10...

klarin (sjk772) – oldhomework 10 – Turner – (56705) 1
This print-out should have 12 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 (part 1 of 4) 10.0 points
An air-filled capacitor consists of two parallel
plates, each with an area of 4 . 3 cm
2
, sepa-
rated by a distance 1 . 5 mm . A 15 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 =
V
d
. correct
2. E = V d .
3. E =
1
( V d )
2
.
4. E = ( V d )
2
.
5. E =
parenleftbigg
d
V
parenrightbigg
2
.
6. E =
1
V d
.
7. E =
d
V
.
8. E =
parenleftbigg
V
d
parenrightbigg
2
.
9. None of these
Explanation:
Since E is constant between the plates,
V =
integraldisplay
vector
E · d
vector
l = E d
E =
V
d
.
002 (part 2 of 4) 10.0 points
The magnitude of the surface charge density
on each plate is
1. σ = ǫ
0
V d
2. σ = ǫ
0
parenleftbigg
V
d
parenrightbigg
2
.
3. σ =
ǫ
0
V d
.
4. σ =
ǫ
0
( V d ) 2
.
5. σ = ǫ
0
( V d )
2
.
6. σ = ǫ
0
parenleftbigg
d
V
parenrightbigg
2
.
7. None of these
8. σ =
ǫ
0
V
d
. correct
9. σ =
ǫ
0
d
V
.
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 = ǫ
0
E =
ǫ
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) 10.0 points
Calculate the capacitance.

klarin (sjk772) – oldhomework 10 – Turner – (56705) 2
Correct answer: 2 . 5382 pF.
Explanation:
Let : A = 0 . 00043 m
2
,
d = 0 . 0015 m ,
V = 15 V , and
ǫ
0
= 8 . 85419 × 10
− 12
C
2
/ N · m
2
.
The capacitance is given by
C =
ǫ
0
A
d
= 8 . 85419 × 10
− 12
C
2
/ N · m
2
×
0 . 00043 m
2
0 . 0015 m
= 2 . 5382 × 10
− 12
F
= 2 . 5382 pF .
004 (part 4 of 4) 10.0 points
Calculate plate charge; i.e. , the magnitude of
the charge on each plate.
Correct answer: 38 . 073 pC.