1
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ No. $\nabla \cdot \vec{E}$ = $\frac{\rho}{\epsilon}$ is a well established maxwell's equation which was derived for electrostatic fields using the gauss's divergence theorem. You cannot just simply substitute volume charge density with line charge density $\lambda$ .......moreover, the dimensions will also not match up when you substitute $\lambda$ in place of $\rho$ $\endgroup$ Commented Oct 23, 2016 at 5:29
  • $\begingroup$ If you work in strictly two dimensions then such a substituition may be made, which is essentially a transformation from R^3 to R^2. However, here, you cannot take divergence w.r.t 3 variables and still use a line charge density $\endgroup$
    – Lelouch
    Commented Oct 23, 2016 at 9:30
  • $\begingroup$ Please use standard capitalization and punctuation, and mark up your math using mathjax. $\endgroup$
    – user4552
    Commented Oct 3, 2019 at 15:27

2 Answers 2

1
$\begingroup$

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!

$\endgroup$
-1
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ Then tell me what would be the genaral form of ∇.E for a line charge distribution and surface charge distribution?? $\endgroup$
    – user101134
    Commented Nov 2, 2016 at 9:32
  • $\begingroup$ "Divergence represents the volume density of the outward flux of a vector field from an infinitesimal volume around a given point." -- from wiki. $\endgroup$
    – Sofiane
    Commented Nov 6, 2016 at 16:56
  • $\begingroup$ and $\nabla\circ\vec{E}=\lim_{\Delta V\to0}\cfrac{1}{\Delta V}\oint_S{\vec{E}\circ d\vec{S}}$. So in geometry, using divergence corresponds to thinking over some volume $V$. $\endgroup$
    – Sofiane
    Commented Nov 6, 2016 at 17:14
  • $\begingroup$ I think that you cannot determine $\nabla\circ\vec{E}$ while just having line or surface charge distribution since you have to think over a volume. $\endgroup$
    – Sofiane
    Commented Nov 6, 2016 at 17:24
  • $\begingroup$ If you want to use the surface, linear or point quantities directly in the Maxwell equations, you have to use the equations in the sense of the distributions. It was fashionable in France in the 1970s. You can take a look at en.wikipedia.org/wiki/Distribution_(mathematics) $\endgroup$ Commented Jan 26, 2019 at 16:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.