homework 03 – KANUNGO, ANIL – Due: Sep 9 2007, 4:00 am
1
Question 1, chap 23, sect 1.
part 1 of 2
10 points
Four point charges are placed at the four
corners of a square,
where each side has
a length
a
.
The upper two charges have
identical positive 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.
1
√
2
(
i
−
j
)
2.
1
√
2
(
i
+
j
)
3. j
4.
−
1
√
2
(
i
−
j
)
5.
−
j correct
6. i
7.
−
1
√
2
(
i
+
j
)
8.
−
i
Explanation:
The direction is already clear:
all the
x

components cancel, and the lower charges at
tract and the top ones repel, so the direction
is
−
j
.
Question 2, chap 23, sect 1.
part 2 of 2
10 points
Determine the magnitude of the electric
field at point
P
.
1.
k
e
q
a
2
2.
√
2
k
e
q
a
2
3.
4
√
2
k
e
q
a
2
correct
4.
8
k
e
q
a
2
5.
2
√
2
k
e
q
a
2
6.
4
k
e
q
a
2
Explanation:
The electric field due to a point charge is
E
=
k
e
q
r
2
.
The field at
P
must be found through super
position of fields due to the four point parti
cles.
The direction of a field 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 fields
vector
E
1
and
vector
E
2
are directed as in the figure
P
vector
E
2
vector
E
1
The
x
components cancel.
The distance
from the center to one of the particles is
a
√
2
,
so the magnitudes are
E
1
y
=
E
2
y
=
1
√
2
k
e
q
parenleftbigg
a
√
2
parenrightbigg
2
=
√
2
k
e
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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homework 03 – KANUNGO, ANIL – Due: Sep 9 2007, 4:00 am
2
P
E
4
E
3
Similarly,
E
3
y
=
E
4
y
=
√
2
k
e
q
a
2
,
so the total magnitude at P (since the
x

components cancel) is
E
= 4
E
1
y
=
4
√
2
k
e
q
a
2
.
Question 3, chap 23, sect 1.
part 1 of 1
10 points
A charge of
−
3
.
39
μ
C is located at the
origin, and a charge of
−
3
.
44
μ
C is located
along the
y
axis at 1
.
22529 m.
The value of the Coulomb constant is 8
.
99
×
10
0
.
At what point along the
y
axis is the electric
field zero?
Correct answer: 0
.
610402
m (tolerance
±
1
%).
Explanation:
Let :
q
1
=
−
3
.
39
μ
C
,
q
2
=
−
3
.
44
μ
C
,
and
d
= 1
.
22529 m
.
Call the point where the fields cancel
y
. Since
the charges are of equal sign, the only place
y
can be is somewhere between them.
The field from the particle
q
1
at the origin is
E
1
=
k
e
q
1
y
2
pointing down (since
q
1
is negative).
The
field from the charge
q
2
at a point
d
along the
y
axis is
E
2
=
k
e
q
2
(
d
−
y
)
2
pointing up (since
q
2
is negative).
Thus we
have cancellation at
d
provided
E
1
equals
E
2
,
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 Fall '08
 Turner
 Charge, Work, Electric charge, KE, Anil, N/C

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