A circular wire loop of radius r carries a total charge Q distributed uniformly over its length. A small length dl of the wire is cut off. The electric field at the centre due to the remaining wire is:-
A
Qdl8π2ϵ0r3
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B
Qdl2π2ϵ0r3
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C
Qdl8πϵ0r3
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D
Qdl4π2ϵ0r3
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Solution
The correct option is AQdl8π2ϵ0r3 Linear charge density (λ) will be
λ=Q2πr
For a complete loop with uniform charge Q, electric field at the centre will be zero.
→Ec=0
Now dl is being cut from circumference so that the system will have some electric field at centre C.
Let, →Edl= electric field due to dl
→Erw= electric field due to remaining wire
∴→Edl+→Erw=0
⇒→Erw=−→Edl
Considering only magnitude, we get
⇒|→Erw|=|→Edl|...........(1)
Let's say dq charge is removed
So, dq=λdl
From Coulomb's law, considering dq is a point charge