22-1
G
AUSS
’
S
L
AW
22.1.
(a) I
DENTIFY
and
S
ET
U
P
:
cos
,
E
E
dA
φ
Φ
=
∫
where
φ
is the angle between the normal to the sheet
ˆ
n
and the
electric field
.
E
G
E
XECUTE
:
In this problem
E
and
cos
φ
are constant over the surface so
(
)(
)
(
)
2
2
cos
cos
14 N/C
cos60
0.250 m
1.8 N
m /C.
E
E
dA
E
A
φ
φ
Φ
=
=
=
°
=
⋅
∫
(b) E
VALUATE
:
E
Φ
is independent of the shape of the sheet as long as
φ
and
E
are constant at all points on the sheet.
(c) E
XECUTE
:
(i)
cos
.
E
E
E
A
φ
Φ
=
Φ
is largest for
0 , so cos
1 and
.
E
EA
φ
φ
=
°
=
Φ
=
(ii)
E
Φ
is smallest for
90 , so cos
0 and
0.
E
φ
φ
=
°
=
Φ
=
E
VALUATE
:
E
Φ
is 0 when the surface is parallel to the field so no electric field lines pass through the surface.
22.2.
I
DENTIFY
:
The field is uniform and the surface is flat, so use
cos
E
EA
φ
Φ
=
.
S
ET
U
P
:
φ
is the angle between the normal to the surface and the direction of
E
G
, so
70
φ
=
°
.
E
XECUTE
:
2
(75.0 N/C)(0.400 m)(0.600 m)cos70
6.16 N
m /C
E
Φ
=
=
⋅
°
E
VALUATE
:
If the field were perpendicular to the surface the flux would be
2
18.0 N
m /C.
E
EA
Φ
=
=
⋅
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.
I
DENTIFY
:
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)
S
ET
U
P
:
In this case, the electric field is perpendicular to the surface of the sphere, so
2
(4
)
E
EA
E
r
π
Φ
=
=
.
E
XECUTE
:
Substituting in the numbers gives
(
)
(
)
2
6
5
2
1.25
10
N/C
4
0.150 m
3.53
10
N m /C
E
π
Φ
=
×
=
×
⋅
(b)
I
DENTIFY
:
We use the electric field due to a point charge.
S
ET
U
P
:
2
0
1
4
q
E
r
π
=
P
E
XECUTE
:
Solving for
q
and substituting the numbers gives
(
)
(
)
2
2
6
6
0
9
2
2
1
4
0.150 m
1.25
10
N/C
3.13
10
C
9.00
10
N m /C
q
r E
π
−
=
=
×
=
×
×
⋅
P
E
VALUATE
:
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.
I
DENTIFY
:
Use Eq.(22.3) to calculate the flux for each surface. Use Eq.(22.8) to calculate the total enclosed charge.
S
ET
U
P
:
ˆ
ˆ
(
5.00 N/C
m)
(3.00 N/C
m)
x
z
−
⋅
⋅
E =
i +
k
G
. The area of each face is
2
L
, where
0.300 m
L
=
.
E
XECUTE
:
1
1
1
ˆ
ˆ
ˆ
0
s
S
A
−
⇒ Φ =
⋅
=
n
=
j
E n
G
.
2
2
2
2
ˆ
ˆ
ˆ
(3.00 N C m)(0.300 m)
(0.27 (N C) m)
S
S
A
z
z
+
⇒ Φ
=
⋅
=
⋅
=
⋅
n
=
k
E n
G
.
2
2
(0.27 (N/C)m)(0.300 m)
0.081 (N/C)
m
Φ =
=
⋅
.
3
3
3
ˆ
ˆ
ˆ
0
S
S
A
+
⇒ Φ =
⋅
=
n
=
j
E n
G
.
4
4
4
ˆ
ˆ
ˆ
(0.27 (N/C)
m)
0 (since
0).
S
S
A
z
z
−
⇒ Φ
=
⋅
= −
⋅
=
=
n
=
k
E n
G
5
5
2
5
ˆ
ˆ
ˆ
(
5.00 N/C
m)(0.300 m)
(0.45 (N/C)
m) .
S
S
A
x
x
+
⇒ Φ =
⋅
= −
⋅
= −
⋅
n
=
i
E n
G
2
5
(0.45 (N/C)
m)(0.300 m)
(0.135 (N/C)
m ).
Φ = −
⋅
= −
⋅
6
6
6
ˆ
ˆ
ˆ
(0.45 (N/C)
m)
0 (since
0).
S
S
A
x
x
− ⇒ Φ
=
⋅
= +
⋅
=
=
n
=
i
E n
G
22

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