In vacuum we have $$\nabla \cdot \mathbf{E} = \frac {\rho}{\varepsilon_0}.$$
Can we still use this formula when there's dielectric material in space? Where $\rho$ is total charge density.
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asked Mar 21, 2016 at 21:33
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1 Answer
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4
There is a modified form of Gauss's law for dielectrics.
Inside a dielectric there is an induced charge that screens the electric field. The bound charge density $\rho_b = - \vec\nabla . \vec P$ where $\vec P$ is the polarisation vector.
This bound charge acts as a source of electric field, so Gauss' law reads
$\vec \nabla . \vec E = \frac{1}{\epsilon_0}(\rho_f + \rho_b)$
We replace the electric field with a so called displacement vector $\vec D = \epsilon_0 \vec E + \vec P$ and thus we can write Gauss' law as
$\vec \nabla . \vec D = \rho_f$
This is the form of Gauss' law to be used inside a dielectric.
See chapter 4 in Griffith's Introduction to Electrodynamics for more information.
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$\begingroup$ Thanks, so that means the original equation still holds right? Since the new one in dielectric is derived from the old one in vacuum. $\endgroup$ Commented Mar 21, 2016 at 21:55
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$\begingroup$ The original equation is not true inside a dielectric. The presence of the dielectric has changed the electric field. $\endgroup$– jm22bCommented Mar 21, 2016 at 22:08
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$\begingroup$ Hmm, so what's the basis to derive the new equation since the original one doesn't work anymore in dielectric. $\endgroup$ Commented Mar 21, 2016 at 23:38
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1$\begingroup$ Sorry, I think I misunderstood your question. Gauss's law does apply, but you need to take into account both the free and bound charge densities of the dielectric. $\endgroup$– jm22bCommented Mar 21, 2016 at 23:45