Self Energy of two small Dipoles aligned in a straight line

Self Energy of two small dipoles

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Potential due to a dipole V is used to find the potential energy of the two charges in the other dipole, \[V(r)=k\frac{p_1}{r^2} \] Here recall that in the potential derivation dipole length $a\ll r$ so $a^2$ term ignored. So $a_1$ does not appear in this expression. \[U_-=k \frac{p_1}{(r-a_2)^2} \, \left(-q_2\right) \] \[U_+=k \frac{p_1}{(r+a_2)^2} \, \left(+q_2\right)\] \[U=k \, p_1q_2\left[\frac{1}{(r+a_2)^2}-\frac{1}{(r-a_2)^2}\right]\] \[U= - k \, p_1q_2\frac{2\times 2a_2r}{(r^2-a_2^2)^2}\]

\[\boxed{U= - k \frac{2 \, p_1 p_2 \, r}{(r^2-a_2^2)^2}}\] Special case: For small dipoles, $a_2 \ll r$ \[\boxed{U=-k\frac{2p_1p_2}{r^3}=-\vec p_1 . \vec E_{axial 2} =-\vec p_2 . \vec E_{axial 1} }\]

This is the interaction energy of two small collinear electric dipoles aligned parallel to each other.

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\[\boxed{ }\]