mcconnell (kam2342) – homework10 – Turner – (60230)
1
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001
10.0 points
A dipole field pattern is shown in the figure.
Consider various relationships between the
electric potential at different points given in
the figure.
T
D
S
W
G
+

Notice:
Five potential relationships are
given below.
a)
V
D
=
V
W
> V
S
b)
V
D
=
V
W
=
V
S
c)
V
D
=
V
W
< V
S
d)
V
T
< V
S
< V
G
e)
V
T
> V
S
> V
G
Which relations shown above are correct?
1.
(
c
) only
2.
(
d
) only
3.
(
a
) and (
e
) only
4.
(
a
) only
5.
(
c
) and (
e
) only
6.
(
b
) and (
e
) only
7.
(
a
) and (
d
) only
8.
(
b
) and (
d
) only
correct
9.
(
e
) only
10.
(
c
) and (
d
) only
Explanation:
The electric potential due to one single
point charge at a distance
r
from the charge
is given by
V
=
k
q
r
.
For a dipole system, the total potential at
any place is the sum of potentials due to one
positive point charge and one negative point
charge (Superposition Principle).
From symmetry considerations, it is easy
to see that the electric field lines are perpen
dicular to a line which passes through the
midpoint
S
and points
D
and
W
.
No work needs to be done to move a positive
test charge along the midplane because the
force and the displacement are perpendicular
to each other.
V
D
=
V
S
=
V
W
, relation (
b
).
Furthermore, moving along the direction of
a electric field line (
i.e.
, moving in the direc
tion from positive charge to negative charge
along the electric field line) always lowers the
electric potential, because the electric field
will do positive work to a positive test charge
in order to lower its electric potential energy.
Therefore,
V
T
< V
D
by considering the line
going from
D
to
T
, and
V
W
< V
G
by consid
ering the line going from
G
to
W
.
V
T
< V
S
< V
G
, relation (
d
).
The correct choices are (
b
) and (
d
) only.
002
(part 1 of 4) 10.0 points
Two charges are located in the (
x, y
) plane
as shown in the figure below. The fields pro
duced by these charges are observed at the
origin,
p
= (0
,
0).
The
Coulomb
constant
is
8
.
98755
×
10
9
N
·
m
2
/
C
2
.
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mcconnell (kam2342) – homework10 – Turner – (60230)
2
x
y
+
Q
+
Q
p
b
a
a
Use Coulomb’s law to find the
x
component
of the electric field at
p
.
1.
E
x
=
2
k
e
Q a
(
a
2
+
b
2
)
3
/
2
2.
E
x
=
4
k
e
Q a
(
a
2
+
b
2
)
3
/
2
3.
E
x
=
2
k
e
Q
a
2
+
b
2
4.
E
x
=

k
e
Q a
(
a
2
+
b
2
)
3
/
2
5.
E
x
= 0
correct
6.
E
x
=
k
e
Q a
(
a
2
+
b
2
)
3
/
2
7.
E
x
=

2
k
e
Q
a
2
+
b
2
8.
E
x
=

2
k
e
Q a
(
a
2
+
b
2
)
3
/
2
9.
E
x
=

4
k
e
Q a
(
a
2
+
b
2
)
3
/
2
Explanation:
Let :
k
e
= 8
.
98755
×
10
9
N
·
m
2
/
C
2
.
+
Q
1
+
Q
2
p
b
a
a
r
1
=
radicalBig
x
2
1
+
y
2
1
=
radicalbig
a
2
+
b
2
.
r
2
=
radicalBig
x
2
2
+
y
2
2
=
radicalBig
(

a
)
2
+
b
2
=
radicalbig
a
2
+
b
2
,
so
r
2
=
r
1
=
r .
θ
θ
E
1
E
2
+
Q
1
+
Q
2
where

sin
θ

=
b
r
=
b
√
a
2
+
b
2

cos
θ

=
a
r
=
a
√
a
2
+
b
2
.
In the
x
direction, the contributions from
the two charges are
E
x
1
=

k
e
(+
Q
)
r
2
1

cos(
θ
)

(1)
=

k
e
(+
Q
)
(
a
2
+
b
2
)
a
√
a
2
+
b
2
=

k
e
Q a
(
a
2
+
b
2
)
3
/
2
E
x
2
=

k
e
(

Q
)
r
2
2

cos(
θ
)

(2)
=

k
e
(

Q
)
(
a
2
+
b
2
)
a
√
a
2
+
b
2
= +
k
e
Q a
(
a
2
+
b
2
)
3
/
2
E
x
=
E
x
1
+
E
x
2
= 0
.
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 Spring '10
 Turner
 Electrostatics, Electric charge, KE

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