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Physics
Classical Physics
Gauss' theorem and inverse square law
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[QUOTE="vanhees71, post: 5550051, member: 260864"] Gauss's Law reads (in Heaviside-Lorentz units) $$\vec{\nabla} \cdot \vec{E}=\rho.$$ For electrostatics you also have $$\vec{\nabla} \times \vec{E}=0$$ and this implies that there exists (in any simply connected region of space) an electric potential such that $$\vec{E}=-\vec{\nabla} \phi.$$ This implies $$\Delta \phi=-\rho,$$ and the Treen's function for the (negative) Laplace operator in 3D Euclidean space is $$G(\vec{x},\vec{x}')=\frac{1}{4 \pi |\vec{x}-\vec{x}'|}.$$ From a long distance you can characterize any charge distribution just by its total charge, and the corresponding monopole contribution to the potential reads $$\phi(\vec{x})=\frac{Q}{4 \pi |\vec{x}|} \; \Rightarrow \; \vec{E}=-\vec{\nabla} \phi=\frac{Q}{4 \pi |\vec{x}|^2} \frac{\vec{x}}{|\vec{x}|}.$$ So the leading order of the multipole expansion indeed always decays with ##1/r^2## with distance (in three dimensions). Of course, this does not hold for the full Maxwell equations, i.e., for time-dependnet fields. There the leading order of the em. waves goes like ##1/r##. [/QUOTE]
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Gauss' theorem and inverse square law
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