# Dipole moment of a displaced charge distribution [duplicate]

This one is part b) of the problem I asked a day ago. Special thanks to MarkMitchison and sbp on the first part. I finally got it. (https://physics.stackexchange.com/questions/217229/how-can-i-prove-that-the-dipole-moment-is-zero).

Let $\rho(r)$ be an arbitary charge distribution now. Under what circumstances does the dipole moment of the displaced charge distribution $\rho '(\vec{r})=\rho(\vec{r}-\vec{b})$ differ from the one not displaced at all.

Now this one is even nastier (excuse my language) than the one before. In hindsight, the one before was easy because I could express the integral in spherical coordinates.

But how do I approach this one? Am I supposed to solve $p=\int r'\cos\theta' \rho d^3r'$ in carthesian coordinates? And what would the boundaries be like?

## marked as duplicate by Qmechanic♦Nov 8 '15 at 23:19

The dipole moment of a continuous charge distribution is given by $$\mathbf{p} = \int\mathrm{d}^3\mathbf{r} \; \mathbf{r} \rho(\mathbf{r}),$$ (the moment is taken with respect to the point $\mathbf{r} = 0$). For a displaced charge distribution $\rho'(\mathbf{r}) = \rho(\mathbf{r} - \mathbf{b})$, you can use a change of integration variables to show that $$\mathbf{p}' = \int\mathrm{d}^3\mathbf{r}\; \mathbf{r} \rho(\mathbf{r} - \mathbf{b}) = \int\mathrm{d}^3\mathbf{r}\; (\mathbf{r} + \mathbf{b}) \rho(\mathbf{r}) = \mathbf{p} + Q\mathbf{b},$$ where $Q$ is the total charge $$Q = \int\mathrm{d}^3\mathbf{r}\; \rho(\mathbf{r}).$$ So displacing a globally neutral system of charges ($Q=0$) does not change the total dipole moment. In other words, calculating the dipole moment of a neutral system with respect to any point always gives the same answer.