Register to reply 
Doubt about differential Gauss's law 
Share this thread: 
#1
Aug1010, 09:41 PM

P: 109

We know that the Gauss's law expressed in the differential form is:
[tex]\mathbf{\nabla}\cdot\mathbf{E} = \frac{\rho}{\epsilon_0}[/tex], right? I read at wikipedia that [tex]\rho[/tex] 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
Aug1010, 10:11 PM

Sci Advisor
HW Helper
P: 1,276

[itex]\rho[/itex] isn't the total charge. It is the charge density.



#3
Aug1010, 10:18 PM

P: 4,663

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
Aug1010, 10:20 PM

Mentor
P: 11,615

Doubt about differential Gauss's law
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. 


#5
Aug1110, 12:35 PM

P: 109

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 explainme it more clearly? (I know it became a mathematics question but I think someone can help me here...) 


#6
Aug1110, 01:16 PM

P: 1,969

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. 


Register to reply 
Related Discussions  
Differential form of Gauss's Law  Advanced Physics Homework  2  
Differential Form of Gauss's Law  Classical Physics  5  
Gauss's Law in differential form  Advanced Physics Homework  1  
Gauss's law in differential forms  Introductory Physics Homework  19  
I have a doubt...  Introductory Physics Homework  1 