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.


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.


  • $\begingroup$ Thanks, so that means the original equation still holds right? Since the new one in dielectric is derived from the old one in vacuum. $\endgroup$ – Salmonella mayonnaise Mar 21 '16 at 21:55
  • $\begingroup$ The original equation is not true inside a dielectric. The presence of the dielectric has changed the electric field. $\endgroup$ – jm22b Mar 21 '16 at 22:08
  • $\begingroup$ Hmm, so what's the basis to derive the new equation since the original one doesn't work anymore in dielectric. $\endgroup$ – Salmonella mayonnaise Mar 21 '16 at 23:38
  • 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. $\endgroup$ – jm22b Mar 21 '16 at 23:45

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