Proving Maxwell's equation, divergence of $E$

Question

For a sourceless medium perturbed as shown below, how can I prove that divergence of $vec E$ and $vec H$ equal to zero?
$$vec D = varepsilon vec E -j xi vec B$$
$$vec H = dfrac{1}{mu} vec B -j xi vec E$$
My initial intuition was finding this by applying a vector identity.
$$nablatimesnablatimes vec E = nabla (nabla bullet vec E)-nabla^2 vec E$$
We know Maxwell's equations for a sourceless medium as
$$nabla times vec E = -jomega vec B$$
and
$$nabla times vec H = jomega vec D$$
For deriving for the $ vec E$ field, we know the wave equation as $nablatimesnablatimes vec E-2muomegaxi nablatimes vec E-omega^2 mu varepsilon vec E = 0$
So, $nabla (nabla bullet vec E)-nabla^2 vec E-2muomegaxi nablatimes vec E-omega^2 mu varepsilon vec E = 0$
As we found the divergence term, we should be able to prove $nablabullet vec E =0$.
How can I derive the rest of the problem?

Shubham Kumar · Accepted Answer

Electric field at any arbitrary point $mathcal{r}$ due to a charge density $rho$ present in a given volume $mathcal{V}$ is given by Coulomb's law as -
$$vec{E} = frac{1}{4piepsilon_{0}}int_{mathcal{V}}frac{rho hat{r} }{r^2} dtau$$
We will evaluate the surface integral of the $vec{E}$ over a sphere containing the source charges.
$$oint vec{E} cdot dvec{a} = frac{1}{4piepsilon_{0}}intfrac{rho hat{r} }{r^2} cdot r^2sin (theta) dphi dtheta hat{r} dtau$$
or, $$oint vec{E} cdot dvec{a} = intfrac{rho}{epsilon_{0}}dtau$$
Invoking Fundamental theoram of gradient -
$$int nabla cdot vec{E} dtau = oint vec{E}cdot dvec{a}$$
So,$$int nabla cdot vec{E} dtau = intfrac{rho}{epsilon_{0}}dtau$$
Since this is true for any given volume , thus we get -
$$nabla cdot vec{E} = frac{rho}{epsilon_0}$$
Now in a charge free region $rho = 0 $ so you get $nabla cdot vec{E} = 0$.
A similar exercise with Biot-Savart law will give you the desired result for the magnetic field.

Proving Maxwell's equation, divergence of $E$

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