HPHY 253 Workout 24
Gauss°s Law for Electricity
Dr. A. E. Bak
Name
1. Serway & Jewett Problem
24
:
4
.
Consider a closed triangular box resting within a horizontal electric
°eld of magnitude
E
= 7
:
80
°
10
4
N=
C
as shown in Figure
P
24
:
4
.
We take all area vectors here to
point exterior to the box.
(a)
Calculate the electric ±ux through the vertical rectangular surface.
This box side is °at, having
area
A
= (0
:
30
m
) (0
:
10
m
) = 0
:
030
m
2
:
Because the electric ±eld is constant on this °at surface, we obtain
Z
E
±
da
=
E
±
A
=
EA
cos 180
°
=
°
7
:
80
°
10
4
±
(0
:
030) (
²
1) =
²
2340
N
±
m
2
=
C
as the electric °ux through the vertical rectangular surface.
(b)
Calculate the electric ±ux through the slanted surface.
This box side is °at, having area
A
= (0
:
30
m
)
²
0
:
10
m
cos 60
°
³
= 0
:
060
m
2
:
Because the electric ±eld is constant on this °at surface, we obtain
Z
E
±
da
=
E
±
A
=
EA
cos 60
:
0
°
=
°
7
:
80
°
10
4
±
(0
:
060)
²
1
2
³
= +2340
N
±
m
2
=
C
as the electric °ux through the slanted rectangular surface.
(c)
Calculate the electric ±ux through the entire surface of the box.
The box²s remaining sides are all
°at. Because the electric ±eld is constant on each of these sides, we obtain
Z
E
±
da
=
E
±
A
=
EA
cos 90
°
= 0
for each remaining side
:
Therefore, we obtain
I
box
E
±
da
=
²
2340 + 2340 + [
terms of zero
] = 0
as the electric °ux through the entire box surface.
FYI:
It follows from Gauss²s law for electricity
that the net charge within the box²s volume is zero.
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 Spring '09
 bak
 Electric charge, charge density, electric …eld

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