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A circular wire-loop of radius a carries a total charge Q distributed uniformly over its length. A small length dL of the wire is cut off. Find the electric field at the centre due to the remaining wire

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When we remove the small piece dL the net electric field at the centre will be that due to the small length only! Therefore, considering the charge per unit length is λ = q/2πa The charge (dq) on the fragment dL = λdL Hence, electric field intensity due to this fragment = (1/4πε)(dq/a²) or, (1/4πε)((qdL)/2πa³)

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