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For a region devoid of charge, maxwell's equation yields $\nabla \cdot \mathbf{E} = 0$ which still allows a constant field. So why is in electrostatics for the vacuum always $\mathbf{E} = 0$ assumed?

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up vote 2 down vote accepted

Maxwell's equation $\nabla \cdot \mathbf{E} = 0$ only states that the electic field does not change in a region devoid of charge. The assumption $\mathbf E = 0$ states that the electric field actually vanishes.

For example between the plates of a condensator the region is devoid of charge and the electric field is constant but non-zero. Such a behaviour is of course unacceptable for an unlimited region, such as empty space in a general electrostatics problem.

So electodynamics is defined by Maxwell's equations with the additional requirement that the solutions be finite.

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Why is it unacceptable? – artistoex May 15 '13 at 10:54
It would require an infinite ammount of energy. – Neuneck May 15 '13 at 10:57
I don't follow you. – artistoex May 15 '13 at 10:58
A test charge in an infinitely streched vacuum with a nonzero electric field will recieve an infinitly long constant acceleration. This would drive its kinetic energy to infinity. Maybe it will help you to look into Maxwell's equations for potentials rather than fields in order to understand the issue. There you have $\Delta \phi = 0$. This would allow for a linear potential, but then such a potential would not be limited and thereby unphysical. So it must be zero. – Neuneck May 15 '13 at 11:00
Electrostatics alone does settle the equations if you require that your results are physical, i.e. $\phi(x) \to 0$ for $x \to \infty$. Such a relation should be mentioned in any good book on ES. – Neuneck May 15 '13 at 11:05

The unique solution to $\nabla^2 \phi = 0$ with the boundary condition $\phi(\infty) = 0 $ is $\phi(\mathbf r) = 0$ for all $\mathbf r$. The solution to a PDE is in general unique only given some boundary conditions. The equation $\nabla^2 = 0$ has infinitely many solutions, but a standard result in the theory of PDE guarantees that only one of them vanishes at infinity.

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void is devoid of charge carrier because there is no dust particle in void region and only electrons and ions are there which neutralize themselves because of which no electric potential exist there...

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