⇒(r−a)21​−(r+a)21​=(a2+r2)3/22a​(r2−a2)24ar​=(a2+r2)3/22a​(r2−a2)2=2r(a2+r2)3/2(1−r2a2​)2=2(1+r2a2​)3/2(1−x2)2=2(1+x2)3/2(x=ia​)(1+x2)3/2(1−x2)2​=2Now for x=3We get 1010​64​≈2⇒ra​≈3 [But for a>r point charge will between the dipole where Eâ†’î€ =0 ]