SOLUTIONS, ASSIGNMENT 6
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
(
29
(
29
2
6
5
2
1.25 10 N/C 4
0.150 m
3.53 10 N m /C
E
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
=
P
E
XECUTE
:
Solving for
q
and substituting the numbers gives
(
29
(
29
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

=
=
×
=
×
×
⋅
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
r
. The area of each face is
2
L
, where
0.300 m
L
=
.
E
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 Fall '10
 dfs
 Flux, Qencl

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