Energy of a dipole

I have a dipole such as:



[itex]\rho(\vec{r}) = q \delta(\vec{r} - \vec{a}) - q \delta(\vec{r} + \vec{a})[/itex]



with [itex]\vec{a} = a \vec{e}_x[/itex].



I have to show that the energy in a constant external field [itex]\vec{E}[/itex] is:



[itex]V = - 2 q \vec{a} \vec{E}[/itex]



My calculations so far:



With the formula: [itex]\phi(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \int \! \rho(\vec{r}') \frac{1}{|\vec{r} - \vec{r}'|} \, d^3r' [/itex]



I have calculated for the potential:



[itex]\phi(\vec{r}) = \frac{q}{4 \pi \epsilon_0} (\frac{1}{|\vec{r} - \vec{a}|} - \frac{1}{|\vec{r} + \vec{a}|})[/itex]



The energy [itex]V(\vec{r}) = \phi(\vec{r}) q[/itex]



Thus I get for the energy:



[itex]V(\vec{r}) = \frac{q^2}{4 \pi \epsilon_0} (\frac{1}{|\vec{r} - \vec{a}|} - \frac{1}{|\vec{r} + \vec{a}|})[/itex]



However I don't understand why that is equal to [itex]-2q \vec{a} \vec{E}[/itex].



Any ideas?