le (jml2943) – Homework 02 – de lozanne – (56785)
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001(part1of2)10.0points
A uniformly charged conducting plate with
area
A
has a total charge
Q
which is positive.
The figure below shows a crosssectional view
of the plane and the electric field lines due to
the charge on the plane.
The figure is not
drawn to scale.
E
E
+
Q
P
Find the magnitude of the field at point P,
which is a distance
a
from the plate. Assume
that
a
is very small when compared to the
dimensions of the plate, such that edge effects
can be ignored.
1.
bardbl
vector
E
P
bardbl
=
ǫ
0
Q a
2
2.
bardbl
vector
E
P
bardbl
=
Q
4
π ǫ
0
a
3.
bardbl
vector
E
P
bardbl
= 4
π ǫ
0
a Q
4.
bardbl
vector
E
P
bardbl
=
Q
2
ǫ
0
A
5.
bardbl
vector
E
P
bardbl
=
ǫ
0
Q A
6.
bardbl
vector
E
P
bardbl
=
Q
4
π ǫ
0
a
2
7.
bardbl
vector
E
P
bardbl
= 2
ǫ
0
Q A
8.
bardbl
vector
E
P
bardbl
=
Q
ǫ
0
A
9.
bardbl
vector
E
P
bardbl
= 4
π ǫ
0
a
2
Q
002(part2of2)10.0points
Two uniformly charged conducting plates are
parallel to each other.
They each have area
A
.
Plate #1 has a positive charge
Q
while
plate #2 has a charge
−
3
Q
.
+
Q
#1
−
3
Q
#2
P
x
y
Using the superposition principle find the
magnitude of the electric field at a point
P
in
the gap.
1.
bardbl
vector
E
P
bardbl
=
3
Q
2
ǫ
0
A
2.
bardbl
vector
E
P
bardbl
=
Q
3
ǫ
0
A
3.
bardbl
vector
E
P
bardbl
=
2
Q
ǫ
0
A
4.
bardbl
vector
E
P
bardbl
=
4
Q
ǫ
0
A
5.
bardbl
vector
E
P
bardbl
=
Q
ǫ
0
A
6.
bardbl
vector
E
P
bardbl
=
Q
ǫ
0
7.
bardbl
vector
E
P
bardbl
= 0
8.
bardbl
vector
E
P
bardbl
=
3
Q
ǫ
0
A
9.
bardbl
vector
E
P
bardbl
=
Q
2
ǫ
0
A
10.
bardbl
vector
E
P
bardbl
=
5
Q
ǫ
0
A
003
10.0points
A large flat sheet of charge has a charge per
unit area of 7
.
7
μ
C
/
m
2
.
Find the electric field intensity just above
the surface of the sheet, measured from its
midpoint.
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 Spring '10
 TSOI
 Physics, Charge, Work, Electric charge, 10 m, 2.8%, 31 kg, 8 k, 1.6 cm

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