(d) For
x
= 26.0 mm = 2.60
×
10
−
2
m, we take a Gaussian surface of the same shape and
orientation, but with
x
>
d
/2, so the left and right faces are outside the slab. The total flux
through the surface is again
2
2
Ea
Φ=
but the charge enclosed is now
q = a
2
d
ρ
. Gauss’
law yields 2
ε
0
Ea
2
=
a
2
d
, so
15
3
3
6
12
2
2
0
(5.80 10
C/m )(9.40 10 m)
3.08 10 N/C.
22
(
8
.
8
5
1
0
C
/
N
m
)
d
E
ε
−−
−
−
××
==
=×
×⋅
43. We use a Gaussian surface in the form of a box with rectangular sides. The cross
section is shown with dashed lines in the diagram below. It is centered at the central plane
of the slab, so the left and right faces are each a distance
x
from the central plane. We
take the thickness of the rectangular solid to be
a
, the same as its length, so the left and
right faces are squares.
The electric field is normal to the left and right faces and is uniform
over them. Since
= 5.80 fC/m
3
is positive, it points outward at
both faces: toward the left at the left face and toward the right at the
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

Click to edit the document details