P22_005.PDF
(
58 KB
)
Pobierz
Chapter 22 - 22.5
5. We put the origin of a coordinate system at the lower left corner of the square and take +
x
rightward
and +
y
upward. The force exerted by the charge +
q
on the charge +2
q
is
F
1
=
k
q
(2
q
)
a
2
(
−
j)
.
The force exerted by the charge
−
q
on the +2
q
charge is directed along the diagonal of the square and
has magnitude
F
2
=
k
q
(2
q
)
(
a
√
2)
2
which becomes, upon finding its components (and using the fact that cos 45
◦
=1
/
√
2),
F
2
=
k
2
√
2
a
2
i+
k
2
√
2
a
2
j
.
q
(
2
q
)
Finally, the force exerted by the charge
−
2
q
on +2
q
is
F
3
=
k
(2
q
)(2
q
)
a
2
i
.
(a)Therefore, the horizontal component of the resultant force on +2
q
is
F
x
=
F
1
x
+
F
2
x
+
F
3
x
=
k
q
2
a
2
1
√
2
+4
10
9
1
.
0
10
−
7
2
0
.
050
2
1
√
2
+4
=0
.
17 N
.
=
8
.
99
×
×
(b)The vertical component of the net force is
F
y
=
F
1
y
+
F
2
y
+
F
3
y
=
k
q
2
a
2
−
2+
1
√
2
=
−
0
.
046 N
.
q
(
2
q
)
Plik z chomika:
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Inne pliki z tego folderu:
P22_036.PDF
(54 KB)
P22_034.PDF
(52 KB)
P22_001.PDF
(53 KB)
P22_003.PDF
(51 KB)
P22_005.PDF
(58 KB)
Inne foldery tego chomika:
chap01
chap02
chap03
chap04
chap05
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