Past exam of the electromagnetism course of the University of Oxford / 2023 / 3 / c

Take the ideal electric dipole limit with a fixed smooth electric dipole moment $\mathbf{p}(t)$, so that ${\mathbf{r}}_0(t)=\mathbf{p}(t)/(2q)$; its required time derivatives are bounded on the time interval under consideration. Fix $r>0$ and put $\mathbf{n}=\mathbf{r}/r$, $t_0=t-r/c$. For $s=+1$ or $-1$, the source position is $s\mathbf{p}(t)/(2q)$. A Taylor expansion of the distance givesSubstitute this into the defining equation for retarded time:Boundedness first implies $t_r^s-t_0=O(q^{-1})$. The Taylor expansion $\mathbf{p}(t_r^s)=\mathbf{p}(t_0)+O(q^{-1})$ then changes the displayed correction only by $O(q^{-2})$. HenceThe nearer positive displacement reduces the propagation distance and consequently increases the retarded time. The Big O notation here is for $q\to \infty$ at a fixed observation event away from the origin. Keeping a fixed smooth $\mathbf{p}$ is essential: bounded amplitude alone, with arbitrarily increasing time frequency as $q$ varies, would not justify this Taylor expansion.

Use the sign $s=\pm 1$ to label a point charge $sq$ at $s\mathbf{p}/(2q)$, with velocity $s{\mathbf{p}}^{\prime }/(2q)$. All quantities on the right below are evaluated at $t_0=t-r/c$. To first order,The denominator in each Liénard–Wiechert potential therefore expands assoIn the sum of the two electric potentials, the $q/r$ terms cancel and the finite terms add. This gives the retarded potentials of a point electric dipole:For the magnetic vector potential, the signed electric charge and signed velocity multiply to give the same sign for both contributions:Since ${\mu }_0{ϵ}_0{c}^2=1$,The ${\mathbf{p}}^{\prime }$ term in the electric potential arises from the moving-source denominator and remains finite in the point-electric dipole limit; discarding it would fail to give the full retarded electric field.