terry (ect328) – oldhomework 03 – Turner – (59130)
1
This
printout
should
have
10
questions.
Multiplechoice questions may continue on
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before answering.
001
10.0 points
Four point charges are placed at the four cor
ners of a square, where each side has a length
a
. The upper two charges have identical pos
itive charge, and the two lower charges have
charges of the same magnitude as the first two
but opposite sign. That is,
q
1
=
q
2
=
q
and
q
3
=
q
4
=
−
q
where
q >
0.
q
4
q
2
q
3
q
1
P
j
i
Determine the direction of the electric field
at the center, point P.
1.
−
i
2.
−
1
√
2
(
i
−
j
)
3. i
4. j
5.
−
1
√
2
(
i
+
j
)
6.
1
√
2
(
i
−
j
)
7.
−
j correct
8.
0
9.
1
√
2
(
i
+
j
)
Explanation:
The direction is already clear:
all the
x

components cancel, and the lower charges at
tract and the top ones repel, so the answer is
−
j
.
002
10.0 points
A uniformly charged insulating rod of length
13
.
5 cm is bent into the shape of a semicircle
as in the figure.
The
value
of
the
Coulomb
constant
is
8
.
98755
×
10
9
N m
2
/
C
2
.
1
3
.
5
c
m
−
5
.
86
μ
C
O
If the rod has a total charge of
−
5
.
86
μ
C,
find the horizontal component of the electric
field at
O
, the center of the semicircle. Define
right as positive.
Correct answer:
−
1
.
81573
×
10
7
N
/
C.
Explanation:
Let :
L
= 13
.
5 cm = 0
.
135 m
and
q
=
−
5
.
86
μ
C =
−
5
.
86
×
10
−
6
C
.
Call the length of the rod
L
and its charge
q
. Due to symmetry
E
y
=
integraldisplay
dE
y
= 0
and
E
x
=
integraldisplay
dE
sin
θ
=
k
e
integraldisplay
dq
sin
θ
r
2
,
where
dq
=
λ dx
=
λ r dθ
, so that
y
x
θ
E
x
=
−
k
e
λ
r
integraldisplay
3
π/
2
π/
2
cos
θ dθ
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terry (ect328) – oldhomework 03 – Turner – (59130)
2
=
−
k
e
λ
r
(sin
θ
)
vextendsingle
vextendsingle
vextendsingle
vextendsingle
3
π/
2
π/
2
= 2
k
e
λ
r
,
where
λ
=
q
L
and
r
=
L
π
.
Therefore,
E
x
=
2
k
e
q π
L
2
=
2 (8
.
98755
×
10
9
N m
2
/
C
2
)
(0
.
135 m)
2
×
(
−
5
.
86
×
10
−
6
C)
π
=
−
1
.
81573
×
10
7
N
/
C
.
Since the rod has negative charge, the field is
pointing to the left (towards the charge dis
tribution). A positive test charge at O would
feel an attractive force from the semicircle,
pointing to the left.
003
(part 1 of 2) 10.0 points
Consider a square with side
a
. Four charges
+
q
, +
q
,
−
q
, and +
q
are placed at the corners
A
,
B
,
C
, and
D
, respectively.
+
+
−
+
D
C
A
B
a
The magnitude of the electric field at
D
due
to the charges at
A
,
B
, and
C
is given by
1.
bardbl
vector
E
bardbl
=
5
4
k q
a
2
2.
bardbl
vector
E
bardbl
=
√
2
k q
a
2
3.
bardbl
vector
E
bardbl
=
7
2
k q
a
2
4.
bardbl
vector
E
bardbl
=
3
4
k q
a
2
5.
bardbl
vector
E
bardbl
=
k q
a
2
6.
bardbl
vector
E
bardbl
= 2
k q
a
2
7.
bardbl
vector
E
bardbl
=
5
2
k q
a
2
8.
bardbl
vector
E
bardbl
=
3
2
k q
a
2
correct
9.
bardbl
vector
E
bardbl
= 3
k q
a
2
10.
bardbl
vector
E
bardbl
=
9
4
k q
a
2
Explanation:
The magnitudes of the electric fields at
D
due to
A
and
C
are
E
A
=
E
C
=
k q
a
2
since they are at a distance
a
from
d
, whereas
E
B
=
k q
(
a
√
2)
2
=
k q
2
a
2
since
B
is at a distance
√
2
a
from
d
.
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 Fall '09
 Turner
 Electrostatics, Electric charge, Terry, KE

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