221
G
AUSS
’
S
L
AW
22.1.
(a) IDENTIFY
and
SET UP:
cos
,
E
Ed
A
φ
Φ=
∫
where
is the angle between the normal to the sheet ˆ
n
and the
electric field
.
E
G
EXECUTE:
In this problem
E
and cos
are constant over the surface so
()
22
cos
cos
14 N/C cos60
0.250 m
1.8 N m /C.
E
A
E
A
φφ
=
=
°
=
⋅
∫
(b) EVALUATE:
E
Φ
is independent of the shape of the sheet as long as
and
E
are constant at all points on the sheet.
(c) EXECUTE:
(i)
cos
.
EE
EA
Φ
is largest for
0,
so
cos
1
and
.
E
EA
=°
=
(ii)
E
Φ
is smallest for
90 , so cos
0 and
0.
E
=
EVALUATE:
E
Φ
is 0 when the surface is parallel to the field so no electric field lines pass through the surface.
22.2.
IDENTIFY:
The field is uniform and the surface is flat, so use
cos
E
EA
.
SET UP:
is the angle between the normal to the surface and the direction of
E
G
, so
70
=
°
.
EXECUTE:
2
(75.0 N/C)(0.400 m)(0.600 m)cos70
6.16 N m /C
E
=
⋅
°
EVALUATE:
If the field were perpendicular to the surface the flux would be
2
18.0 N m /C.
E
EA
=
⋅
The flux in
this problem is much less than this because only the component of
E
G
perpendicular to the surface contributes to the
flux.
22.3.
IDENTIFY:
The electric flux through an area is defined as the product of the component of the electric field
perpendicular to the area times the area.
(a)
SET UP:
In this case, the electric field is perpendicular to the surface of the sphere, so
2
(4
)
E
EA
E
r
π
=
.
EXECUTE:
Substituting in the numbers gives
2
65
2
1.25 10 N/C 4
0.150 m
3.53 10 N m /C
E
×
=
×
⋅
(b)
IDENTIFY:
We use the electric field due to a point charge.
SET UP:
2
0
1
4
q
E
r
=
P
EXECUTE:
Solving for
q
and substituting the numbers gives
2
26
6
0
92
2
1
4
0.150 m
1.25 10 N/C
3.13 10 C
9.00 10 N m /C
qr
E
−
==
× =
×
×⋅
P
EVALUATE:
The flux would be the same no matter how large the circle, since the area is proportional to
r
2
while
the electric field is proportional to 1/
r
2
.
22.4.
IDENTIFY:
Use Eq.(22.3) to calculate the flux for each surface. Use Eq.(22.8) to calculate the total enclosed charge.
SET UP:
ˆˆ
( 5.00 N/C m)
(3.00 N/C m)
xz
−⋅
⋅
E
=i
+
k
G
. The area of each face is
2
L
, where
0.300 m
L
=
.
EXECUTE:
11
1
ˆ
0
sS
A
−
⇒
⋅
=
n= j
En
G
.
2
2
ˆ
(3.00 N C m)(0.300 m)
(0.27 (N C) m)
SS
Az
z
+
⇒
Φ= ⋅
=
⋅
=
⋅
n= k
G
.
2
2
(0.27 (N/C)m)(0.300 m)
0.081 (N/C) m
=
⋅
.
33
3
ˆ
0
A
+
⇒
=
G
.
44
4
ˆ
(0.27 (N/C) m)
0 (since
0).
z
−
⇒
=−
⋅
=
=
n=k
G
55
2
5
ˆ
( 5.00 N/C m)(0.300 m)
(0.45 (N/C) m) .
Ax
x
+
⇒
⋅
⋅
n=i
G
2
5
(0.45 (N/C) m)(0.300 m)
(0.135 (N/C) m ).
Φ=−
⋅
⋅
66
6
ˆ
(0.45 (N/C) m)
0 (since
0).
x
−
⇒
=+
⋅
=
=
G
22
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View Full Document222
Chapter 22
(b)
Total flux:
22
25
(0.081 0.135)(N/C) m
0.054 N m /C.
Φ=Φ +Φ =
−
⋅
=−
⋅
Therefore,
13
0
4.78 10
C.
q
−
=Φ
=
−
×
P
EVALUATE:
Flux is positive when
E
G
is directed out of the volume and negative when it is directed into the
volume.
22.5.
IDENTIFY:
The flux through the curved upper half of the hemisphere is the same as the flux through the flat circle
defined by the bottom of the hemisphere because every electric field line that passes through the flat circle also must
pass through the curved surface of the hemisphere.
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 Spring '11
 Shaefer
 Charge, Electrostatics, Electric charge

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