# Doubt about differential Gauss's law

1. Aug 10, 2010

### Taturana

We know that the Gauss's law expressed in the differential form is:

$$\mathbf{\nabla}\cdot\mathbf{E} = \frac{\rho}{\epsilon_0}$$,

right?

I read at wikipedia that $$\rho$$ is: the total charge density including dipole charges bound in a material.

I don't understand...

The left side of equation is the divergence of the field vector E (electric field), right?

The divergence is the measure of the flux density at a given point in space (so it's a function of x,y,z considering 3D), right?

So the flux density at any point in the electric field will be different (unless we have uniform field), because in some regions the field lines are more (convergent? next, near, you got it) and in other regions the field lines are more separate, right?

The the right side of the equation is a constant. It is the total charge density divided by the permittivity... So this is telling me that the flux density is the same for ALL points in the space, isn't it?

Or is the density on the right side the density of the point I'm calculating he divergence?

Where am I wrong?

I appreciate the help,
Thank you

2. Aug 10, 2010

### nicksauce

$\rho$ isn't the total charge. It is the charge density.

3. Aug 10, 2010

### Bob S

If you integrate both sides over a volume, and apply Stokes theorem, the total charge enclosed in the volume is related to the net flux through the surface of the volume. See

http://en.wikipedia.org/wiki/Divergence_theorem

Bob S

4. Aug 10, 2010

### Staff: Mentor

No, it's a measure of the flux that is "created" or "destroyed" at a point, rather than simply passing through it. You can have a very high flux density (field strength) at a point, with zero divergence.

Consider the electric field of a solid sphere with a uniform charge distribution. The electric field outside the sphere is just like the field of an ideal point charge located at the center of the sphere. The magnitude of the field decreases as 1/r^2 where r is the distance from the center of the sphere. But the divergence of the field is zero at all points outside the sphere, and so is the charge density.

Inside the sphere the magnitude of the field increases linearly with r, reaching a maximum at the surface of the sphere. But the divergence of the field has the same value at all points inside the sphere, just like the charge density.

Last edited: Aug 10, 2010
5. Aug 11, 2010

### Taturana

quote from wikipedia: More technically, the divergence represents the volume density of the outward flux of a vector field from an infinitesimal volume around a given point.

It's the same to say that divergence represents the flux density around a given point, isnt it?

I don't know if I get it but then the divergence of a vector field at a point represents how is this contributing with the field "generation"? Could you explain-me it more clearly? (I know it became a mathematics question but I think someone can help me here...)

Yes, sorry, but the question stills the same...

6. Aug 11, 2010

### nasu

The charge density is a function of position, ro(x,y,z).
The word "total" here means that you add contributions from free charges, bound charges, etc. All these contributions are functions of position in general.
It does not mean total charge in a finite volume.
So the divergence of E at a given point depends on the charge density at that point.