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JEE Main Physics Class 12 Electric Charges and Fields Questions
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© examsnet.com
Question : 31
Total: 108
Find the electric field at point
P
(as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length
L
carrying a charge
Q
. The distance of the point
P
from the centre of the rod is
a
=
√
3
2
L
[26 Feb 2021 Shift 1]
√
3
Q
4
π
ε
0
L
2
Q
3
π
ε
0
L
2
Q
2
√
3
π
ε
0
L
2
Q
4
π
ε
0
L
2
Validate
Solution:
Given, length of conductor
=
L
Charge on conductor
=
Q
According to figure,
O
P
=
a
=
√
3
2
L
,
O
Q
=
L
2
Let
P
Q
=
r
=
√
O
P
2
+
O
Q
2
⇒
P
Q
=
√
(
√
3
2
L
)
2
+
(
L
2
)
2
=
√
3
4
L
2
+
L
4
=
L
and
E
be the electric field at point
P
.
Since,
E
(due to finite wire)
=
k
λ
a
(
sin
φ
1
+
sin
φ
2
)
. . . (i)
where,
k
=
Coulomb's constant
=
1
4
π
ε
0
λ
=
linear charge density
=
Q
L
and
sin
φ
=
sin
φ
2
=
L
∕
2
L
=
1
2
Substituting the above value in Eq. (i), we get
E
=
k
λ
a
=
1
4
π
ε
0
Q
L
⋅
√
3
L
2
=
1
2
√
3
π
ε
0
L
2
© examsnet.com
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