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Nightmare

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  1. 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.

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    Lol, that's from some Physics E-book I was reading..

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