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For a pure dipole situated at the origin, I don't understand why the dipole moment vector is $$ \vec{p} = p\cos \theta \hat{r} - p\sin \theta \hat{\phi} $$ Since the vector does not have any $r$ in it, the value of the vector is same any distance away from the origin if the angle is constant, which doesn't make sense to me. It would be very helpful if someone could derive it because Griffiths uses this value without explanation.

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up vote 2 down vote accepted

The position dependence comes in via the $R$ in the following expression¹:

$$ \vec E = \frac{3 \vec p \cdot \vec R}{4 \pi \varepsilon_0 R^3} \vec R - \frac{\vec p}{4 \pi \varepsilon_0 R^3} $$

This makes sense, as the charge $q$ also does not depend on distance - it’s effect on the electric field $\vec E$, however, does depend on said distance and the expression for $E$, not $q$, hence includes $R$.

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