maini (nm7637) – hw3 – Shneidman – (12108)
1
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001
(part 1 oF 2) 10.0 points
±our 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 frst two
but opposite sign. That is,
q
1
=
q
2
=
q
and
q
3
=
q
4
=
−
q
where
q >
0.
b
b
b
b
b
q
4
q
2
q
3
q
1
P
j
i
Determine the direction oF the electric feld
at the center (point
P
).
1. i
2.
−
j correct
3. j
4.
1
√
2
(
i
−
j
)
5.
1
√
2
(
i
+
j
)
6.
−
i
7.
−
1
√
2
(
i
+
j
)
8.
−
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
(part 2 oF 2) 10.0 points
Determine the magnitude oF the electric feld
at point
P
.
1.
8
k
e
q
a
2
2.
4
k
e
q
a
2
3.
k
e
q
a
2
4.
√
2
k
e
q
a
2
5.
4
√
2
k
e
q
a
2
correct
6.
2
√
2
k
e
q
a
2
Explanation:
The electric feld due to a point charge is
E
=
k q
r
2
.
The feld at
P
must be Found through super
position oF felds due to the Four point par
ticles. Recall that the direction oF a feld at
a point
P
is the direction in which a
positive
test charge at that point would Feel a Force.
The top two charges (
q
1
and
q
2
) give rise to
repulsive Forces on a positive test charge at
P
.
The felds
E
1
and
E
2
are directed as in the
fgure
P
E
2
E
1
We see that the
x
components cancel. The
distance From the center to one oF the particles
is
a
√
2
. ThereFore, the magnitudes are
E
1
y
=
E
2
y
=
1
√
2
k q
p
a
√
2
P
2
=
√
2
k q
a
2
The bottom two charges are negative, so the
Forces on a positive test charge at the center
are attractive:
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2
P
E
4
E
3
Similarly,
E
3
y
=
E
4
y
=
√
2
k q
a
2
so the total magnitude at P (since the
x

components cancel) is
E
= 4
E
1
y
=
4
√
2
k q
a
2
.
003
(part 1 of 3) 10.0 points
Coulomb constant is 8
.
98755
×
10
9
N m
2
/
C
2
.
The 1
.
68
μ
C charge is at the origin and a
−
8
.
54
μ
C charge is 10 cm to the right, as
shown in the Fgure.
x
O
I
II
III
y
1
.
68
μ
C
−
8
.
54
μ
C
10 cm
Identify the direction of
v
E
in the region II
(0
< x <
10 cm, along the
x
axis).
1.
all possibilities: right, left, or zero
2.
down
3.
None of these
4.
left
5.
right
correct
6.
up
Explanation:
Let :
q
1
= 1
.
68
μ
C
,
q
2
=
−
8
.
54
μ
C
,
and
a
= 10 cm
.
The direction of the electric Feld at a point
P
is the direction that a positive charge would
move if placed at
P
. A positive charge placed
in region II would be attracted to
q
2
and
repelled by
q
1
. Thus the direction is to the
right.
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 Spring '08
 moro
 Electron, Charge, Electric charge, Fundamental physics concepts

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