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Ahmed1029

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Ahmed1029

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$$\Delta \Phi(\vec{x})=-\frac{1}{\epsilon_0} \rho(\vec{x})$$

is given by "summing up Coulomb potentials", i.e.,

$$\Phi(\vec{x})=\int_{B_R} \mathrm{d}^3 x' \frac{\rho(\vec{x}')}{4 \pi \epsilon_0 |\vec{x}-\vec{x}'|},$$

where ##B_R## is the solid sphere ("ball") of radius ##R## around the origin, which by definition contains all the charge there is.

Helmholtz's theorem in this form is for sure also valid under the conditions you quote, because if ##\rho(\vec{x}) = \mathcal{O}(r^{-\alpha})## for ##|\vec{x}|\rightarrow \infty## and ##\alpha>2##, then

$$\mathrm{d}^3 x' \frac{\rho(\vec{x}')}{|\vec{x}-\vec{x}'|} = \mathrm{d} r' \mathrm{d\vartheta'} \mathrm{d} \varphi' \sin \vartheta' \mathcal{O}(r^{\prime -\alpha+1}),$$

i.e., the integrand falls faster with ##r'## than ##1/r##, and thus the integral over ##r'## converges for ##r' \rightarrow \infty##. The integrals over the angles is over finite regions (##\vartheta \in (0,\pi)##, ##\varphi \in (0,2 \pi)##) and thus don't diverge.

The theorem even works if ##\rho=\mathcal{O}(1/r^\beta)## with ##\beta>1##, because the potential is defined only up to an additive constant anyway. In this case you just subtract a clever constant, defining

$$\Phi(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\rho(\vec{x}')}{4 \pi \epsilon_0} \left (\frac{1}{|\vec{x}-\vec{x}|}-\frac{1}{|\vec{x}_0-\vec{x}'|} \right).$$

The expression in the parentheses goes like ##\mathcal{O}(r^{\prime 2})## for ##r' \rightarrow \infty##, and thus the integral converges even under the more general condition on ##\rho##.

- #3

Ahmed1029

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got it thanks

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