Problem on gauss law I have read about the electric field in electrostatic and electric field has a property that $\nabla \times \mathbf E = 0$, and $\nabla \cdot \mathbf E = \frac{\rho}{ε}$.
But the second formula we had derive from gauss divergence theorem and where we deal with volume.
From gauss law, $$\oint \mathbf E\cdot \mathrm d\mathbf S= \frac{Q}{ε}$$
And if $$Q = \rho\iiint \mathrm dV$$
$\rho$ = volume charge density
And so from gauss divergence theorem, $$\oint\mathbf E\cdot\mathrm d\mathbf S = \iiint (\nabla\cdot \mathbf E )\mathrm dV = \iiint \frac{\rho}{ε} \mathrm dV$$
So $\nabla \cdot \mathbf E = \frac{\rho}{ε}$
Hence we get the relation.
But here we deal with volume charge density ($\rho$).
But if there is a line charge ,then we should have deal with
Line charge density ($\lambda$).
My question is: Will that relation be true from line charge also?
Because I can't get that relation from the electric field on any line charge, say charged circular loop.
For charge circular loop electric field at a height $z$ from its center is $$E = \frac{\lambda}{2ε} \frac{zR}{(z^2+R^2)^{3/2}}$$
{$R=$ radius}
If I calculate $\nabla \cdot \mathbf E$, I can't get $\frac{\lambda}{ε}$
So is the relation valid for line charge?
 A: I don't completely agree with the earlier answer: there is a simple way to represent volume charge densities in terms of surface and line charge densities so that we can use them in Maxwell's Equations. How useful it might be, however, is debatable. For practical reasons, it's usually easier to use the integral form.
Consider an infinite plane of charge, with some surface charge density $\sigma$. What would the corresponding volume charge density $\rho$ be? Well, all the charge is constrained to be in (say) the $xy-$plane, and we can use a Dirac Delta function to enforce this! Therefore, for this problem, (if the plate in at $z=0$) $$\rho = \sigma \delta(z).$$
Notice that because the Dirac Delta has dimensions 1/Length, this equation is also dimensionally consistent. Gauss's Law in differential form is then just $$\vec{\nabla}\cdot\vec{E}= \frac{\sigma \delta(z)}{\epsilon_0}$$
It's a nice exercise to use this definition and Poisson's Equation to show that the Electric Field must have a discontinuity of $\sigma/\epsilon_0$ on either side of the plate.
Similarly, we can define a volume charge density for a wire (aligned along the $z$ axis) as being $$\rho = \lambda \delta(x)\delta(y) = \lambda \frac{\delta(r)}{2\pi r}, \quad\quad \text{where  } r^2 = x^2 + y^2.$$
A logical conclusion of this is to realise that the "volume charge density" of a point charge is simply $$\rho = q \delta^3(r).$$
Plugging this into the divergence equation we get the important result that
$$\vec{\nabla}\cdot \left( \frac{\hat{r}}{r^2}\right) = 4\pi \delta^3(r).$$
So while it may not be "useful" to solve simple problems, you can certainly extract a lot of information about surface, line, and point charge distributions by writing their volume densities appropriately!
A: The answer is no. The first Maxwell equation in its local (or differential) form is valid only for volume charge density. The integral form describes the flux of an electric field through a closed surface (which gives a volume $v$). This flux is expressed in terms of the total charge $Q$ in the volume $v$.
If there is a problem in which you have a line or surface charge, you have absolutely to use the integral form of the first Maxwell equation, never the local one.
