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Chapter 23: Gauss' Law
Problem 1
The square surface shown in Fig. 2326
measures 3.2 mm on each side. It is
immersed in a uniform electric field with
E = 1800 N/C. The field lines make an
angle of 35
°
with a normal to the surface,
as shown. Take the normal to be outward
as if the surface were one face of a box.
Calculate the electric flux through the
surface.
Normal
35
o
Figure 2326
Answer
 1.51
×
10
2
Nm
2
/C.
Solution
Flux is defined to be
Φ
=
E
⋅
d
A
∫
=
E cos
(
θ
)
d
A
∫
.
In this case E and
θ
are constant over the area so that
Φ
=
E cos(
θ
)
d
A
∫
= EA cos(
θ
), where A is the area of the square.
Thus
Φ
= (1800 N/C)*(3.2 mm)
2
cos(145
°
), (Note that the angle is 145
°
)
Φ
=  1.51
×
10
2
Nm
2
/C.
Problem 3
The cube in Fig. 2327 has edge lengths of
1.40 m and is oriented as shown in a region of
uniform electric field. Find the electric flux
through the right face if the electric field, in
newtons per coulomb, is given by (a) 6.00
i
, (b)
–2.00
j
, and (c) –3.00
i
+ 4.00
k
. (d) What is
the total flux through the cube for each of
these fields?
Answer
(a) zero. (b) 3.92 Nm
2
/C. (c) zero. (d) zero
for each field.
Figure 2327
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Chapter 23: Gauss’ Law
Solution
The
vector area
of the right face is
A
= (1.4 m)
2
j
= 1.96 m
2
j
. Since E is a constant in each
of the cases the flux is just
Φ
=
E
⋅
A
.
(a)
Here
E
= 6 N/C
i
. Therefore
Φ
=
E
⋅
A
Φ
= (6 N/C
i
)
⋅
(1.96 m
2
j
) = 0.
(b)
Here
E
= 2 N/C
j
. Therefore
Φ
=
E
⋅
A
Φ
= (2 N/C
j
)  (1.96 m
2
j
) = 3.92 Nm
2
/C.
(c)
Here
E
= 3 N/C
i
+ 4 N/C
k
. Therefore
Φ
=
E
⋅
A
Φ
= (3 N/C
i
+ 4 N/C
k
)
⋅
(1.96 m
2
j
) = 0.
(d)
The answer is zero in each case. The total flux is just the sum of the flux for each of
the six faces
Φ
tot
=
Φ
top
+
Φ
bottom
+
Φ
left
+
Φ
right
+
Φ
front
+
Φ
back
Φ
tot
=
E
⋅
A
+
E
⋅
A
bottom
+
E
⋅
A
left
+
E
⋅
A
right
+
E
⋅
A
front
+
E
⋅
A
Φ
tot
=
E
⋅
(
A
+
A
bottom
+
A
left
+
A
right
+
A
front
+
A
)
This last step is justified since
E
is the same on every face of the cube. The sum of the
vector areas is zero (all have same size and they occur in pairs of antiparallel vectors).
Problem 5
A point charge of 1.8
μ
C is at the center of a cubical Gaussian surface 55 cm on edge.
What is
Φ
through the surface?
Answer
2.03
×
10
5
Nm
2
/C.
From Gauss' law the flux through a closed surface is
Φ
=
E
⋅
d
A
∫
= Q/
ε
o
,
where Q is the net charge inside the Gaussian surface. If you know the charge inside the
surface it is not necessary to know about the geometry of the surface in order to calculate
the flux  it is just Q/
ε
o
. Therefore the flux through the cube is just
Chapter 23: Gauss’ Law
233
Φ
= Q/
ε
o
Φ
= (1.8
μ
C)/
ε
o
= 2.03
×
10
5
Nm
2
/C.
Notice that we are not assuming that the electric field is constant on the surface or anything
else about the field. This solution only tells us about the flux.
Problem 9
It is found experimentally that the electric field in a certain region of the Earth's atmosphere
is directed vertically down. At an altitude of 300 m the field has magnitude 60.0 N/C; at an
altitude of 200 m, the magnitude is 100 N/C. Find the net amount of charge contained in a
cube 100 m on edge, with horizontal faces at altitudes of 200 and 300 m.
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This note was uploaded on 08/03/2009 for the course PHYSICS 2049 taught by Professor C during the Spring '09 term at Vanderbilt.
 Spring '09
 C
 Physics, Gauss' Law

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