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Griffiths states that given the multipole expansion: $$V(\vec{r})=\frac{1}{4\pi\epsilon_o}\sum_{n=0}^\infty\frac{1}{r^{(n+1)}}\int(\vec{r}')^nP_n(cos(\theta')\rho(\vec{r}')d\tau'$$ for large $r$ the multipole expanison is dominated by the monopole potential: $$V_{m}(\vec{r})=\frac{1}{4\pi\epsilon_o}\frac{Q}{r} \ni Q=\int \rho d\tau$$

This is all well and good, where I am getting confused is the text statement.

'if the total charge is zero then the dominant term is the dipole potential (unless this potential also vanishes)'. So: $$V_{D}(\vec{r})=\frac{1}{4\pi\epsilon_o}\frac{1}{r^2}\int r'cos(\theta')\rho(\vec{r}')d\tau$$ How is this term not also zero if the total charge is zero?

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If $\int f(x) dx = 0$, that does not imply that $\int g(x) f(x) dx = 0$ for some other function $g$.

Similarly, "the total charge is zero" means that $\int \rho(\vec r') d^3r' = 0$. This certainly does not imply that $\int r' \cos(\theta') \rho(\vec r') d^3r' = 0$.

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  • $\begingroup$ Oh duh, i knew it was something obvious lol. Thanks! $\endgroup$ – Alex Sampson Dec 2 '18 at 3:33

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