We note that U¢ (q) differs from U(q) by a quantity which is just a constant for a given dipole. Since a constant is insignificant for potential energy, we can drop the second term in Eq. (2.34) and it then reduces to Eq. (2.32). We can now understand why we took q0=π/2. In this case, the work done against the external field E in bringing +q and – q are equal and opposite and cancel out, i.e., q [V (r1) – V (r2)]=0. _____________________________________________________________________ Lol, that's from some Physics E-book I was reading..