Platt, David – Homework 4 – Due: Sep 14 2005, 4:00 am – Inst: Ken Shih
1
This
printout
should
have
12
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
Again! For 005, you need to answer in unit
of uC (miu Coulomb,*E6C).
001
(part 1 of 3) 10 points
An electric field of magnitude 3450 N
/
C is
applied along the
x
axis.
Calculate the electric flux through a rect
angular plane 0
.
214 m wide and 0
.
41 m long
if the plane is parallel to the
yz
plane.
Correct answer: 302
.
703 N
·
m
2
/
C.
Explanation:
Let :
E
= 3450 N
/
C
,
θ
= 0
◦
,
L
= 0
.
41 m
,
and
W
= 0
.
214 m
.
The area of the rectangle is
A
=
L W
= (0
.
41 m) (0
.
214 m)
= 0
.
08774 m
2
.
Since the field is uniform,the flux is
φ
=
E
·
A
=
E A
cos
θ
= (3450 N
/
C) (0
.
08774 m
2
) (cos 0
◦
)
=
302
.
703 N
·
m
2
/
C
.
002
(part 2 of 3) 10 points
Calculate the electric flux through the same
rectangle, if it is parallel to the
xy
plane.
Correct answer: 0 N
·
m
2
/
C.
Explanation:
The angle is
θ
= 90
◦
,cos
θ
= 0
,
and
φ
=
E
·
A
=
E A
cos
θ
= (3450 N
/
C) (0
.
08774 m
2
) (0)
=
0 N
·
m
2
/
C
.
003
(part 3 of 3) 10 points
Calculate the electric flux through the same
rectangle, but now the rectangle contains the
y
axis and its normal makes an angle of 26
◦
with the
x
axis.
Correct answer: 272
.
068 N
·
m
2
/
C.
Explanation:
Let
θ
= 26
◦
φ
=
E
·
A
=
E A
cos
θ
= (3450 N
/
C) (0
.
08774 m
2
) (cos 26
◦
)
=
272
.
068 N
·
m
2
/
C
.
004
(part 1 of 1) 10 points
A (9
.
31 m by 9
.
31 m) square base pyramid
with height of 3
.
32 m is placed in a vertical
electric field of 59
.
9 N
/
C.
9
.
31 m
3
.
32 m
59
.
9 N
/
C
Calculate the total electric flux which goes
out through the pyramid’s four slanted sur
faces.
Correct answer: 5191
.
9 N m
2
/
C.
Explanation:
Let :
s
= 9
.
31 m
,
h
= 3
.
32 m
,
and
E
= 59
.
9 N
/
C
.
By Gauss’ law,
Φ =
~
E
·
~
A
Since there is no charge contained in the pyra
mid, the net flux through the pyramid must
be 0 N/C. Since the field is vertical, the flux
through the base of the pyramid is equal and
opposite to the flux through the four sides.
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 Fall '08
 Turner
 Physics, Electrostatics, Work, Electric charge, Gaussian curvature

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