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AMU Engineering 2018 solved paper Solved Paper
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© examsnet.com
Question : 26
Total: 150
Four charges each equal to
(
−
Q
)
are placed at the four corners of a square and a charge
q
is placed at its centre. If the system of charges is in equilibrium, the value of
q
is
Q
4
(
2
√
2
−
1
)
Q
4
(
2
√
2
+
1
)
Q
2
(
2
√
2
−
1
)
Q
2
(
2
√
2
+
1
)
Validate
Solution:
Four charges having magnitude
(
−
Q
)
are placed on the four corner of square having side '
a
′
each.
q
be the charge placed on centre '
O
′
Electrostatic force between charges at
A
and
B
F
A
B
=
1
4
π
ε
0
.
Q
2
a
2
Similarly,
F
B
C
=
1
4
π
ε
0
.
Q
2
a
2
and
F
D
B
=
1
4
π
ε
0
.
Q
2
(
√
2
a
)
2
[
∵
D
B
=
√
a
2
+
a
2
=
√
2
a
]
F
D
B
=
1
4
π
ε
0
.
Q
2
2
a
2
Total force on the charge
(
−
Q
)
at
B
due to charge at
A
,
C
and
D
.
F
2
=
F
1
+
F
D
B
∴
(where,
F
1
is the resultant force of
F
A
B
and
F
B
C
)
=
√
F
A
B
2
+
F
B
C
2
+
1
4
π
ε
0
.
Q
2
2
a
2
F
2
=
√
(
1
4
π
ε
0
.
Q
2
a
2
)
2
+
(
1
4
π
ε
0
.
Q
2
a
2
)
2
+
1
4
π
ε
0
.
Q
2
2
a
2
F
2
=
1
4
π
ε
0
.
√
2
Q
2
a
2
+
1
4
π
ε
0
.
Q
2
2
a
2
=
1
4
π
ε
0
.
Q
2
a
2
[
√
2
+
1
2
]
=
1
4
π
ε
0
.
Q
2
a
2
[
2
√
2
+
1
2
]
.........(i)
Electrostatic force between charge
q
at
O
and
(
−
Q
)
at
B
F
O
B
=
1
4
π
ε
0
.
Q
q
(
√
2
a
2
)
2
=
1
4
π
ε
0
.
Q
q
a
2
2
F
O
B
=
1
4
π
ε
0
.
2
Q
q
a
2
.........(ii)
Since, system of charge are in equilibrium, hence
F
OB
=
F
2
1
4
π
ε
0
.
2
Q
q
a
2
=
1
4
π
ε
0
.
Q
2
a
2
(
2
√
2
+
1
2
)
2
q
=
Q
(
2
√
2
+
1
2
)
q
=
Q
4
(
2
√
2
+
1
)
© examsnet.com
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