P22_012.PDF
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Pobierz
Chapter 22 - 22.12
12. (a) The distance between
q
1
and
q
2
is
r
12
=
(
x
2
−
x
1
)
2
+(
y
2
−
y
1
)
2
=
(
−
0
.
020
−
0
.
035)
2
+(0
.
015
−
0
.
005)
2
=0
.
0559 m
.
The magnitude of the force exerted by
q
1
on
q
2
is
F
21
=
k
|
q
1
q
2
|
r
12
=
8
.
99
×
10
9
3
.
0
×
10
−
6
4
.
0
×
10
−
6
=34
.
5N
.
0
.
0559
2
The vector
F
21
is directed towards
q
1
and makes an angle
θ
with the +
x
axis, where
θ
=tan
−
1
y
2
−
y
1
=tan
−
1
1
.
5
−
0
.
5
=
−
10
.
3
◦
.
x
2
−
x
1
−
2
.
0
−
3
.
5
(b) Let the third charge be located at (
x
3
,y
3
), a distance
r
from
q
2
.Wenotethat
q
1
,
q
2
and
q
3
must
be colinear; otherwise, an equilibrium position for any one of them would be impossible to find.
Furthermore, we cannot place
q
3
on the same side of
q
2
wherewealsofind
q
1
, since in that region
both forces (exerted on
q
2
by
q
3
and
q
1
) would be in the same direction (since
q
2
is attracted
to both of them). Thus, in terms of the angle found in part (a), we have
x
3
=
x
2
−
r
cos
θ
and
r
sin
θ
(which means
y
3
>y
2
since
θ
is negative). The magnitude of force exerted on
q
2
by
q
3
is
F
23
=
k
|
q
2
q
3
|
/r
2
, which must equal that of the force exerted on it by
q
1
(found in part (a)).
Therefore,
r
=
r
12
q
3
k
|
q
2
q
3
|
r
2
=
k
|
q
1
q
2
|
r
12
=
⇒
q
1
=0
.
0645 cm
.
Consequently,
x
3
=
x
2
−
r
cos
θ
=
−
2
.
0cm
−
(6
.
45 cm) cos(
−
10
.
3
◦
)=
−
8
.
4cmand
y
3
=
y
2
−
r
sin
θ
=
1
.
5cm
−
(6
.
45 cm) sin(
−
10
.
3
◦
)=2
.
7cm
.
y
3
=
y
2
−
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