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Question

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 A Qdl8π2ϵ0r3
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

Edl=14πϵ0dqr2=14πϵ0λdlr2
Edl=14πϵ0Qdl2πr1r2=Qdl8π2ϵ0r3

putting the value in eq.(1), we get

|Erw|=Qdl8π2ϵ0r3

Ans (a)

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