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Assignment 2 Solutions
Chpt 23
2.
We use
A
d
E
r
r
⋅
=
Φ
∫
and note that the side length of the cube is (3.0 m–1.0 m) = 2.0 m.
(a) On the top face of the cube
y
= 2.0 m and
( )
ˆ
j
dA
dA
=
r
. Therefore, we have
()
2
ˆˆ
4i
3 2.0
2 j
4i 18j
E
=−
+ =−
r
ˆ
ˆ
. Thus the flux is
() ( )
( )
2
22
top
top
top
ˆ
4i 18j
j
18
18
2.0
N m C
72 N m C.
EdA
dA
Φ=
⋅
=
−
⋅
= −
⋅
⋅
∫∫
∫
r
r
(b) On the bottom face of the cube
y
= 0 and
)
ˆ
)(
(
j
dA
A
d
−
=
r
. Therefore, we have
j
i
j
i
E
ˆ
6
ˆ
4
ˆ
)
2
0
(
3
ˆ
4
2
−
=
+
−
=
r
. Thus, the flux is
( )
2
bottom
bottom
bottom
ˆ
4
i 6
j
j 6
62
.
0 Nm C
2
4
Nm C
.
⋅
=
−
⋅
− =
=
⋅
=+
⋅
∫
r
r
(c) On the left face of the cube
( )
ˆ
i
dA
dA
=
−
r
. So
( )
( )
2
left
left
bottom
ˆ
ˆ
4i
j
i
4
4 2.0
N m C
16 N m C.
y
E dA
E
dA
dA
⋅
=
+
⋅
− =−
⋅
⋅
∫
r
(d) On the back face of the cube
( )
ˆ
k
dA
dA
=
−
r
. But since
E
r
has no
z
component
. Thus,
Φ
= 0.
0
⋅=
r
r
(e) We now have to add the flux through all six faces. One can easily verify that the flux
through the front face is zero, while that through the right face is the opposite of that
through the left one, or +16 N·m
2
/C. Thus the net flux through the cube is
Φ
= (–72 + 24 – 16 + 0 + 0 + 16) N·m
2
/C = – 48 N·m
2
/C.
20.
Using Eq. 2311, the surface charge density is
( )
51
2
2
2
0
2.3 10 N C 8.85 10
C / N m
2.0 10 C/m .
E
σε
−−
==×
×
×
6
2
27.
We assume the charge density of both the conducting cylinder and the shell are
uniform, and we neglect fringing effect. Symmetry can be used to show that the electric
1
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View Full Documentfield is radial, both between the cylinder and the shell and outside the shell. It is zero, of
course, inside the cylinder and inside the shell.
(a) We take the Gaussian surface to be a cylinder of length
L
, coaxial with the given
cylinders and of larger radius
r
than either of them. The flux through this surface is
where
E
is the magnitude of the field at the Gaussian surface. We may
ignore any flux through the ends. Now, the charge enclosed by the Gaussian surface is
q
enc
=
Q
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 Spring '08
 Parsons

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