Physics 107 – Homework #2 Solutions
19.52
E
=
σ
ε
0
=
0
E
=
8.85
×
10
−
12
C
2
N
⋅
m
2
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
7.0
×
10
5
N
C
⎛
⎝
⎜
⎞
⎠
⎟ =
6.2
×
10
−
6
C/m
2
19.55
The Gaussian surface is chosen to be a cylinder with radius
r
and length
L
, and its axis is along the
wire.
Φ=
EA
=
E
(2
π
rL
)
is assumed to be parallel to the end caps, so only the area of the curved surface is considered. The
total charge enclosed by the Gaussian surface is
q
r
E
=
λ
L
. Apply Gauss’s law.
q
0
=
L
0
=
E
rL
) .
Solve for
E
.
E
=
2
πε
0
r
20.13
13. (a)
W
=
q
Δ
V
=
e
Δ
V
=
(1.60
×
10
−
19
C)(0.070 V)
=
1.1
×
10
−
20
J
(b)
The work required to “pump” a Na
ion out of the cell is independent of the thickness of the cell wall, so the
answer to part (a)
+
20.15
(a)
Δ
K
=−Δ
U
=−
(
U
C
−
U
A
)
=
K
A
U
A
−
U
C
=
K
A
U
C
=
U
A
−
K
A
Δ
K
(
U
C
−
U
B
)
=
2
K
A
U
B
−
U
C
=
2
K
A
U
C
=
U
B
−
2
K
A
So,
U
B
−
2
K
A
=
U
A
−
K
A
.
Find
K
A
.
U
B
−
2
K
A
=
U
A
−
K
A
U
B
−
U
A
=
K
A
Find
V
.
C
U
C
=
U
A
−
K
A
=
U
A
−
(
U
B
−
U
A
)
=
2
U
A
−
U
B
V
C
=
2
V
A
−
V
B
=
2(352 V)
−
129 V
=
575 V
Also accept V
C
= 2(332V) – 149 V =
515 V
(b)
K
A
=
U
B
−
U
A
eV
B
−
(
−
eV
A
)
=
e
(
V
A
−
V
B
)
=
(1.60
×
10
−
19
C)(352 V
−
129 V)
=
3.57
×
10
−
17
J
Also accept K
A
= (1.6 x 10 –19 C ) (332 V – 149 V) =
2.93x 10
–17
J
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View Full Document20.17
1
2
mv
2
=
e
Δ
V
v
=
2
e
Δ
V
m
=
2(1.60
×
10
−
19
C)(25,000 V)
9.11
×
10
−
31
kg
=
9.4
×
10
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 Spring '07
 CHEN
 Electrostatics, Work, Magnetic Field, Electric charge

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