This preview shows pages 1–2. Sign up to view the full content.
PHY108 FALL 08
Solutions for HW3(Due on 9/24)
3. We use
E A
, where
A
A
.
j
m
j
2
140
b
g
.
(a)
2
ˆˆ
6.00 N C i 1.40 m
j
0.
(b)
2
2
2.00 N C j 1.40 m
j
3.92 N m C.
(c)
2
ˆ
ˆ
ˆ
3.00 N C i
400N C k
1.40 m
j
0
.
(d) The total flux of a uniform field through a closed surface is always zero.
7. To exploit the symmetry of the situation, we imagine a closed Gaussian surface in the
shape of a cube, of edge length
d
, with a proton of charge
19
1.6 10
C
q
situated at
the inside center of the cube. The cube has six faces, and we expect an equal amount of
flux through each face. The total amount of flux is
net
=
q
/
0
, and we conclude that the
flux through the square is onesixth of that. Thus,
=
q
/6
0
= 3.01
10
–9
N
m
2
/C.
9. Let
A
be the area of one face of the cube,
E
u
be the magnitude of the electric field at
the upper face, and
E
l
be the magnitude of the field at the lower face. Since the field is
downward, the flux through the upper face is negative and the flux through the lower face
is positive. The flux through the other faces is zero, so the total flux through the cube
surface is
(
).
u
A E
E
The net charge inside the cube is given by Gauss’ law:
12
2
2
2
00
6
(
)
(8.85 10
C / N m )(100 m) (100 N/C 60.0 N/C)
3.54 10 C
3.54
C.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 03/26/2010 for the course PHY 108 taught by Professor Iashvili during the Spring '08 term at SUNY Buffalo.
 Spring '08
 IASHVILI
 Physics

Click to edit the document details