Husain, Zeena – Exam 1 – Due: Feb 17 2004, 11:00 pm – Inst: Sonia Paban
1
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printout
should
have
16
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before making your selection. The due time is
Central time.
001
(part 1 of 1) 10 points
Four point charges are placed at the four cor
ners of a square. Each side of the square has
a length
L
.
q
1
=

q
q
3
=
q
q
2
=
q
q
4
=
q
P
L
L
Find the magnitude of the electric force on
q
2
due to all three charges
q
1
,
q
3
and
q
4
. Given
L
= 1 m and
q
= 1
.
56
μ
C.
Correct answer: 0
.
0328082 N.
Explanation:
F
1
F
4y
F
4
F
3
F
4x
q
2
From the above figure, we see that
F
1
=

k q
2
L
2
ˆ
ı
=

(
8
.
98755
×
10
9
N
·
m
2
/
C
2
)
×
(1
.
56
×
10

6
C)
2
(1 m)
2
=

0
.
0218721 N
F
3
=

k q
2
L
2
ˆ
=

(
8
.
98755
×
10
9
N
·
m
2
/
C
2
)
×
(1
.
56
×
10

6
C)
2
(1 m)
2
=

0
.
0218721 N
F
4
x
=
k q
2
2
L
2
1
√
2
=
8
.
98755
×
10
9
N
·
m
2
/
C
2
2
·
(1 m)
2
×
(1
.
56
×
10

6
C)
2
√
2
= 0
.
00773296 N
F
4
y
=

k q
2
2
L
2
1
√
2
=

8
.
98755
×
10
9
N
·
m
2
/
C
2
2
·
(1 m)
2
×
(1
.
56
×
10

6
C)
2
√
2
=

0
.
00773296 N
k
~
F
k
2
= (
F
1
+
F
4
x
)
2
+ (
F
3
+
F
4
y
)
2
= (

0
.
0218721 N + 0
.
00773296 N)
2
+ (

0
.
0218721 N +

0
.
00773296 N)
2
so that
k
~
F
k
=
√
0
.
00107638 N
2
= 0
.
0328082 N
002
(part 1 of 1) 10 points
Consider two SOLID conducting spheres with
radii
r
1
and
r
2
, where
r
2
= 3
r
1
.
The two
spheres are separated by a large distance so
that the field and the potential at the surface
of sphere #1 only depend on the charge on
#1 and the corresponding quantities on #2
only depend on the charge on #2. There is a
wire connecting the two spheres.
r
1
r
2
#2
#1
Place a charge
q
on sphere #1. After equi
librium has been reached, what is the ratio
E
2
E
1
of the electric fields at the SURFACES of
the two solid spheres?
1.
∞
2.
27
3.
9
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Husain, Zeena – Exam 1 – Due: Feb 17 2004, 11:00 pm – Inst: Sonia Paban
2
4.
1
81
5.
81
6.
1
27
7.
1
3
correct
8.
3
9.
1
9
10.
1
Explanation:
When the spheres are connected by a wire,
charges will flow from one to the other until
the potential on both spheres is the same. In
this case,
k
q
1
r
1
=
k
q
2
r
2
, or
q
2
=
r
2
r
1
q
1
. We know
outside the sphere
E
(
r
) =
k
q
r
2
. Therefore,
E
2
E
1
=
q
2
q
1
·
(
r
1
r
2
)
2
=
r
2
r
1
·
(
r
1
r
2
)
2
=
3
1
·
(
1
3
)
2
=
1
3
003
(part 1 of 2) 10 points
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.
k
~
E
P
k
=
Q
²
0
A
2.
k
~
E
P
k
= 4
π ²
0
a Q
3.
k
~
E
P
k
=
²
0
Q A
4.
k
~
E
P
k
=
Q
4
π ²
0
a
2
5.
k
~
E
P
k
= 2
²
0
Q A
6.
k
~
E
P
k
= 4
π ²
0
a
2
Q
7.
k
~
E
P
k
=
Q
2
²
0
A
correct
8.
k
~
E
P
k
=
Q
4
π ²
0
a
9.
k
~
E
P
k
=
²
0
Q a
2
Explanation:
Basic Concepts
Gauss’ Law, electrostatic
properties of conductors.
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 Spring '08
 Turner
 Charge, Electrostatics, Electric charge, Husain, Sonia Paban

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