zapata (jz883) – HW #2 – Antoniewicz – (56495)
1
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23
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001
(part 1 of 4) 3.0 points
A charge
−
1
.
1
×
10
−
5
C is fixed on the
x
axis
at 9 m, and a charge 8
×
10
−
6
C is fixed on
the
y
axis at 5 m, as shown on the diagram.
9 m
−
1
.
1
×
10
−
5
C
5 m
8
×
10
−
6
C
Calculate the magnitude of the resultant
electric field
vector
E
at the origin.
Correct answer: 3124
.
29 N
/
C.
Explanation:
Let :
k
= 8
.
98755
×
10
9
N
·
m
2
/
C
2
,
Q
x
=
−
1
.
1
×
10
−
5
C
,
Q
y
= 8
×
10
−
6
C
,
Q
=
−
3
×
10
−
6
C
,
x
= 9 m
,
and
y
= 5 m
.
For charge
Q
x
,

E
x

=
k

Q
x

x
2
= (8
.
98755
×
10
9
N
·
m
2
/
C
2
)
×
vextendsingle
vextendsingle
−
1
.
1
×
10
−
5
C
vextendsingle
vextendsingle
(9 m)
2
= 1220
.
53 N
/
C
.
For charge
Q
y
,

E
y

=
k

Q
y

y
2
= (8
.
98755
×
10
9
N
·
m
2
/
C
2
)
×
8
×
10
−
6
C
(5 m)
2
= 2876
.
02 N
/
C
.
The resultant electric field at the origin is
shown on the diagram.
1220
.
53 N
/
C
2876
.
02 N
/
C
bardbl
vector
E
bardbl
Thus,
bardbl
vector
E
bardbl
=
radicalBig
E
2
x
+
E
2
y
=
radicalBig
(1220
.
53 N
/
C)
2
+ (2876
.
02 N
/
C)
2
=
3124
.
29 N
/
C
.
002
(part 2 of 4) 3.0 points
Calculate the electric potential at the origin.
Correct answer: 3395
.
3 V.
Explanation:
The electric potential of a point charge is
V
=
k
q
r
, so the electric potential at the origin
(due to charges
Q
x
and
Q
y
) is thus
V
=
k
parenleftbigg
Q
x
x
+
Q
y
y
parenrightbigg
= (8
.
98755
×
10
9
N
·
m
2
/
C
2
)
×
parenleftbigg
−
1
.
1
×
10
−
5
C
9 m
+
8
×
10
−
6
C
5 m
parenrightbigg
=
3395
.
3 V
.
003
(part 3 of 4) 2.0 points
A
−
3
×
10
−
6
C charge is brought from a very
distant point by an external force and placed
at the origin.
Calculate the magnitude of the electric
force on this charge.
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zapata (jz883) – HW #2 – Antoniewicz – (56495)
2
Correct answer: 0
.
00937286 N.
Explanation:
We know that
vector
F
=
q
vector
E
.
The electric force on
Q
is shown on the
diagram.
x
y
bardbl
vector
F
bardbl
bardbl
vector
F
bardbl
=

Q
 bardbl
vector
E
bardbl
=
vextendsingle
vextendsingle
−
3
×
10
−
6
C
vextendsingle
vextendsingle
(3124
.
29 N
/
C)
=
0
.
00937286 N
.
004
(part 4 of 4) 2.0 points
Calculate the work that had to be done by an
external force to bring
Q
to the origin from
the distant point.
Correct answer:
−
0
.
0101859 J.
Explanation:
The external force must do work equal in
magnitude to that done by the electric field;
i.e.
, equal to the change in electric potential
energy:
W
=
Q
Δ
V
= (
−
3
×
10
−
6
C) (3395
.
3 V
−
0)
=
−
0
.
0101859 J
.
005
10.0 points
Two charges,
−
2
Q
and +
Q
, are located on
the
x
axis, as shown below.
Point
P
, at a distance of 3
D
from the origin
O
, is one of two points on the positive
x
axis
at which the electric potential is zero.
x
0
−
2
Q
+
Q
P
D
D
D
How far
d
from the origin
O
is the other
point?
1.
d
=
9
5
D
2.
d
=
5
3
D
correct
3.
d
=
6
5
D
4.
d
=
7
4
D
5.
d
=
4
3
D
6.
d
=
3
2
D
7.
d
=
7
5
D
8.
None of these
9.
d
=
5
4
D
10.
d
=
8
5
D
Explanation:
Suppose the zeropotential point is at a
distance
d
from O.
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 Spring '10
 ANTONIEWICZ
 Physics, Charge, Energy, Potential Energy, Correct Answer, Electric charge, KE, Zapata

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