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If I have a dipole of two opposite charges separated by a distance $d$, then by taking the leading non-zero term in multipole expansion centered about their midpoint, I find the potential to be:

$$ V= \frac{ k \vec{p} \cdot \hat{r} }{r^2}$$

Where $\hat{r}$ is the radial vector from the set of circles centered at the midpoint of the dipole.

Clearly this expression is undefined at $r=0$ i.e: at the midpoint itself but if I calculate the potential by considering the potential due to each point charge, I find the midpoint has potential zero. Why am I getting contradicting answers in each method?

Does this have any relation to the distinction between Laurent vs Taylor series? In the Laurent expansion of a function, I have heard that it is not necessary that the expansion agree with function about the point of expansion itself.

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2 Answers 2

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The dipole potential $V=\frac{\vec{p}.\hat{r}}{r^2}$ is a consequence when we assume charge separation $d\ll r$, so we are considering dipole as a point source. The actual potential is given by: $$ V_{\text{actual}}=k\left(\frac{q}{r_{+}}-\frac{q}{r_{-}}\right)$$ where $$(r_{\pm})^2=r^2+(d/2)^2\mp rd\cos\theta$$

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The formula used is an approximation, valid if $r$ is large compared to the distance between the charges, $d$.

The potential a distance $r$ along the line of the dipole is

$$V= \frac{1}{4\pi \epsilon_0} \frac{Q}{r} - \frac{1}{4\pi \epsilon_0} \frac{Q}{{(r+d)}}\tag1$$

If $d$ is small compared to $r$

$$V= \frac{1}{4\pi \epsilon_0} \frac{Q[{(r+d)}-r]}{r^2}\tag2$$

$$V= \frac{1}{4\pi \epsilon_0} \frac{2Qd}{r}\tag3$$

Equation 3) is the equivalent of your formula, but if $r$ is small equation 1) should be used

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