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)
