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Electric flux through cubical box
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[QUOTE="rude man, post: 4995556, member: 350494"] epsilon times the divergence of the E field equals the charge density (aka one of Maxwell's 4 equations). Solving that with your given E field would give you the charge distribution needed to effect the given E field. It was in answer to BvU who wondered how an E field like your given one could be produced. You could solve for ρ(x) if it amused you.epsilon = dielectric constant [/QUOTE]
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Electric flux through cubical box
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