SOLUTIONS, ASSIGNMENT 7
22.13.
(a) I
DENTIFY
and
S
ET
U
P
:
It is rather difficult to calculate the flux directly from
d
Φ =
⋅
E
A
r
r
ú
since the
magnitude of
E
r
and its angle with
d
A
r
varies over the surface of the cube. A much easier approach is
to use Gauss's law to calculate the total flux through the cube. Let the cube be the Gaussian surface.
The charge enclosed is the point charge.
E
XECUTE
:
6
6
2
encl
0
12
2
2
9.60
10
C
/
1.084
10
N m /C.
8.854
10
C / N m
E
Q


×
Φ
=
=
=
×
⋅
×
⋅
P
By symmetry the flux is the same through each of the six faces, so the flux through one face is
(
29
6
2
5
2
1
6
1.084
10
N
m
/C
1.81
10
N m /C.
×
⋅
=
×
⋅
(b) E
VALUATE
:
In part (a) the size of the cube did not enter into the calculations. The flux through one
face depends only on the amount of charge at the center of the cube. So the answer to (a) would not
change if the size of the cube were changed.
22.16.
I
DENTIFY
:
Apply the results of Example 22.5.
S
ET
U
P
:
At a point 0.100 m outside the surface,
0.550 m
r
=
.
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 Fall '10
 dfs
 Charge, Electric charge, point charge, gaussian surface

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