A wire of length l is bent to form a semicircle. If it has charge q, then electric field intensity at the centre of semicircle is
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0
Answer:
lollul
Explanation:
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Answered by
0
Answer
length of the wire l=πr
∴ Radius of wire R=
π
l
In a semicircle the horizontal components of E get cancelled
∴ E due to a small element dl. at angle dθ is
∫
0
E
dE=∫
0
π
lR
2
KQ⋅Rdθ
sinθ
E=
lR
KQ
[−cosθ]
0
π
E=
4πε
0
1
lR
Q
2(−
y
^
)
E=
2ε
0
l
2
Q
(−
y
^
)
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