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An electric dipole is a system of two opposite point charges when their separation goes to zero and their charge goes to infinity in a way that the product of the charge and the separation remains finite.

  1. How can we have a continuous volume charge distribution from such a collection of point charges?

  2. In the article 'Electric dipole' at Knowino it is said that

    The charge distribution is written in terms of Dirac delta functions: $$\rho (\mathbf{r})=q_1 \delta (\mathbf{r}-\mathbf{r}_1)+q_2 \delta (\mathbf{r}-\mathbf{r}_2)$$

Here $\mathbf{r}_1$ and $\mathbf{r}_2$ are the position vectors of $q_1$ and $q_2$ and $\mathbf{r}=\mathbf{r}_1-\mathbf{r}_2$. Please explain why do we need Dirac delta in describing dipolar charge distribution?

Emilio Pisanty
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Alfred
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1 Answers1

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You don't need Dirac delta functions to describe a dipolar charge distribution. There's a very wide range of charge distributions whose electric fields are exactly dipolar (this Q&A describes one such example) and an even wider class whose electric fields are dominated by a dipolar asymptotic when you're away from the support of the charge distribution (basically: every neutral charge distribution with a nonzero dipole moment).

The expression you've asked about is simply the correct encapsulation into a single analytical formula for the charge density $\rho(\mathbf r)$ that corresponds to two point charges $q_1$ and $q_2$ at positions $\mathbf r_1$ and $\mathbf r_2$. This needs to be fairly singular (in technical language, it's a distribution, not a function) because point charges are not strictly describable as (continuous) volumetric charge densities.

Emilio Pisanty
  • 137,480