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Answer:
hope it help
Explanation:
E1=2Ecosθ=[x2+l2]3/22kq(l) alongAC;E2=2Ecosθ=[x2+l2]3/22kq(l) alongDB
So, E1⊥E2. Net field is
E0=E12+E22=[x2+l2]3/222kql=2πε0[x2+l2]3/2ql