Differential form of gauss's law.

In summary, the concept of charge density is defined as the charge in a small volume divided by the volume. The divergence of a vector-vector function is defined as the limit of the net flow of the vector field across the smooth boundary of a three-dimensional region divided by the volume as it shrinks to a point. This means that the divergence of an electric field integrated over a volume is equal to the surface integral of the normal component of the field enclosing the volume, which is equal to the charge of the volume divided by the permitivity constant. This explains the spreading-out of electric field lines from a point charge, and the Dirac delta "function" can be used to further understand this concept.
  • #1
bmrick
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I don't understand what charge density is meant in the equation: div E = constant times charge density. I have the derivation in front of me and the last step follows from accepting that the rate of change of the integral of the field divergence per change in volume is the same as the rate of change of the integral of charge density per that same change in volume, and as such the divergence of an e field is one over permitivity times the... charge density at that point? What does this mean? How does it account for divergence of an e field away from point charges?
 
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  • #2
Charge density is defined as the charge in a small volume divided by the volume. More exactly: the charge density at a point P is[tex]\rho=\lim_{\Delta V \rightarrow 0}\frac{\Delta Q}{\Delta V}[/tex]

On the other hand, the divergence of a vector-vector function is defined as
Definition of divergence

More rigorously, the divergence of a vector field F at a point p is defined as the limit of the net flow of F across the smooth boundary of a three-dimensional region V divided by the volume of V as V shrinks to p.
See http://en.wikipedia.org/wiki/Divergence
So the divergence of E integrated for a volume V surrounding the point P is equal to the surface integral of the normal component of E enclosing the volume, which is equal to ##Q/\epsilon _0##, Q being the charge of the volume, according to Gauss' Law. At the limit of ## \Delta V \rightarrow 0##,
## div E = \rho/\epsilon_0##

ehild
 
  • #3
bmrick said:
How does it account for divergence of an e field away from point charges?

Are you thinking about the spreading-out of electric field lines from a point charge? It might be instructive to calculate the divergence explicitly for the Coulomb's-law field of a point charge at the origin: [tex]\vec E = k \frac {\hat r}{|\vec r|^2} = k \frac {\vec r}{|\vec r|^3} = k \frac {x \hat i + y \hat j + z \hat k}{(x^2 + y^2 + z^2)^{3/2}}[/tex] Then consider what happens at ##\vec r = 0## and read up on the Dirac delta "function" if necessary (e.g. in Griffiths, section 1.5).
 
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What is the differential form of Gauss's Law?

The differential form of Gauss's Law is a mathematical equation that relates the electric field to the electric charge within a given space. It states that the divergence of the electric field is equal to the charge density divided by the permittivity of the medium.

How is the differential form of Gauss's Law different from the integral form?

The differential form of Gauss's Law is a local equation, meaning it applies to a specific point in space. On the other hand, the integral form is a global equation, meaning it applies to a closed surface enclosing a volume of space.

What are the advantages of using the differential form of Gauss's Law?

The differential form of Gauss's Law allows for a more precise analysis of the electric field at a specific point in space. It also makes it easier to solve complex problems involving non-uniform charge distributions.

How is the differential form of Gauss's Law used in electromagnetic theory?

The differential form of Gauss's Law is one of the four Maxwell's equations that form the basis of classical electromagnetic theory. It is used to describe the behavior of electric fields in the presence of electric charges.

Can the differential form of Gauss's Law be applied to any type of charge distribution?

Yes, the differential form of Gauss's Law can be applied to any type of charge distribution, as long as the medium is homogeneous and isotropic. However, for non-homogeneous or anisotropic media, modifications may be necessary.

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