0
$\begingroup$

I'm trying to understand: $$ \nabla \cdot \vec{E} =\frac{\rho}{\epsilon_0}.$$

But taking the divergence of the vector field renders $0$ everywhere except where we have a point charge. At that region the E field is infinite. Why is it then equal to $\frac{\rho}{\epsilon_0}$ then?

I know it's possible to understand why, without re-writing the equation?

$\endgroup$
2
4
$\begingroup$

Gauss' law relates the divergence of the electric field to the charge density. If you are considering the exact point at which a charged particle is "located" you are considering a point of zero volume, within which the charge density is infinite. If you take a charged particle at a point $p$ you find that

$$\rho(p)=\lim_{V\rightarrow 0}\frac{Q}{V}=\infty.$$

Gauss' law is better illustrated when you take a volume integral of it and use the divergence theorem, giving

$$\iiint_{V}(\nabla\cdot \vec E)\,\mathrm dV=\iint_{\partial V}\vec E\cdot \mathrm d\vec n=\frac{Q_\text{total}}{\epsilon_0}.$$

This tells you that the flux out of a closed surface $\partial V$ is related to the amount of charge $Q_\text{total}$ inside of it.

$\endgroup$
3
$\begingroup$

You are right in saying that the electric field near a point charge is infinite, but that is not the case for continuous volume charge distributions. The best example is a charged solid sphere. Although the solid sphere has charge, still you can define electric field inside the sphere at any point and this electric field is finite, not infinite. And thus if you apply Gauss' Law to a surface which covers an infinitesimally small volume, you will arrive at the differential form of Gauss' Law.


Derivation

Let's take the surface of an infinitesimally small cuboid as our Gaussian surface. Let the dimensions of the cuboid be $\mathrm d x$, $\mathrm d y$ and $\mathrm d z$. Now let's find the flux through this cuboidal surface.

$$\phi= (E_{x+\mathrm d x}-E_x)\,\mathrm d y\,\mathrm d z + (E_{y+\mathrm d y}-E_y)\,\mathrm d z\,\mathrm d x +(E_{x+\mathrm d z}-E_z)\,\mathrm d x\,\mathrm d y$$

But by Gauss' Law,

$$\phi=\frac{q_{\text{enclosed}}}{\varepsilon_0}$$

Therefore,

$$(E_{x+\mathrm d x}-E_x)\,\mathrm d y\,\mathrm d z + (E_{y+\mathrm d y}-E_y) \,\mathrm d z\,\mathrm d x +(E_{x+\mathrm d z}-E_z)\,\mathrm d x\,\mathrm d y = \frac{\rho\,\mathrm d V}{\varepsilon_0}$$

But $\mathrm dV = \mathrm d x\,\mathrm d y\,\mathrm d z$ and $E_{q+\mathrm d q}-E_q=\mathrm dE_q \:\:\forall \: q = x,y,z$, so

$$\mathrm d x\,\mathrm d y\,\mathrm d z \left(\frac{\mathrm d E_x}{\mathrm dx}+\frac{\mathrm d E_y}{\mathrm dy}+\frac{\mathrm d E_z}{\mathrm dz}\right)= \mathrm d x\,\mathrm d y\,\mathrm d z \frac{\rho}{\varepsilon _0}$$

Also, here we will convert the differentials to partial differentials because $E_x$ might be a function of $y$ and $z$ as well and the same goes for other componenets. Therefore

$$\frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial y}+\frac{\partial E_z}{\partial z}=\frac{\rho}{\varepsilon_0}$$

But the left hand side of the above equation is nothing but the divergence of the electric field $\mathbf{E}$. Thus,

$$\nabla \cdot \mathbf{E}=\frac{\rho}{\varepsilon_0}$$

$\endgroup$

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