How two charges become a dipole in the limit of vanishing separation.The electric field of a dipole can be seen as the result of two charges approaching each other. Learn in this problem how to use the Taylor expansion of \(1/\left|\mathbf{r}-\mathbf{r}^{\prime}\right|\) to calculate this field. Find out how this series expansion yields multipole moments, a very powerful description of the electric field.

The field of two point charges becomes that of a dipole in the limit of vanishing separation.

Problem Statement

Find the electric field of an electric dipole using the following steps:

  • Find the electrostatic potential of two point charges \(q_{1/2}=\pm q\) at \(\mathbf{r}_{1/2}=\pm\frac{1}{2}\mathbf{d}\).
  • Take the limit \(\left|\mathbf{d}\right|\rightarrow0\) of the potential holding \(\mathbf{p}:=q\mathbf{d}\) constant.
  • Calculate the electric field from the electrostatic potential.
  • additional: Verify that the charge distribution \(\rho\left(\mathbf{r}\right)=-\mathbf{p}\cdot\nabla\delta\left(\mathbf{r}\right)\) yields the correct dipole potential.

Background: The Dipole as Limiting Process

As we have found in The Electric Field of two Point Charges, the electric field of two opposite charges is fundamentally different from that of a single charge decreasing with \(1/r^{3}\) at large distances, not only with \(1/r^{2}\). This is not just a coincidence but stems from the so-called multipole expansion as we have outlined earlier.

For now, let us try to find the dipole term by the limiting process of the given problem!

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