## Friday, April 24, 2009

### 2.2 Gauss's Law of electrostatic

Previous sections:
1. Introduction
2. Electrostatic
2.1 Coulomb's Law

2.2 Gauss's Law of electrostatic

In the previous section, we have came to conclusion that E field is q/(4πεr^2). But that was due to a point charge and assuming that the field spreads out equally in all direction, i.e. in sphere-like manner. What if we want to know the field generated by an arbitrary collection (or shape) of charges and the field distribution other than a sphere?

This is where calculus comes in handy. We could always start with a small surface, dA and small charge, dQ and then sum it (integration) to obtain a more general equation for the electric field.

Recall that we obtain 4πr^2 from the surface of a sphere. Instead of using 4πr^2, let's use dA to denote an arbitrary small surface and ρdV to denote an arbitrary small volume of charge (where ρ is the charge density and dV is the small volume occupied by this charge). Then the E field is

E=(ρ * dV )/(dA * ε) ;

(quiz question: why don't we multiply a density function for dA like we did for dV?).

Rearranging and integrating,
ε∫E.dA = ∫ ρ dV

Here, it is important to introduce a very useful and important mathematical tool called Gauss' theorem. We need not concern ourselves how to derive this theorem. All we need to know is that this theorem converts a surface integral of a vector field into a volume integral, and vice versa.

∫F.dA = ∫ ∇. F dV

where F is any vector field and the sign ∇, called nabla or del, is the vector (d/dx, d/dy, d/dz). The term ∇.F is also called the divergence of the vector field F.

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At this stage, it may be important to ask what does divergence and the symbol nabla physically mean. The origins of nabla, divergence, and the curl of a field (not yet touched upon here) are used so often in field theory that I would dedicate a section to explain them. Please refer to this section for more information.
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Hence, applying Gauss' theorem on the LHS,
ε∫∇.E dV = ∫ ρ dV

which implies
ε∇.E=ρ or ∇.E=ρ/ε

Here, I will introduce yet another quantity called the displacement field, D. I will explain in more detail what it means in the next section. But for now, it's suffice just to remember that D=εE.

Hence, the above equation becomes
∇.D=ρ
which is the first equation of electrostatic.

Gauss theorem, is not a law of nature as some students may misunderstood. It is really just a fancy mathematical tool to convert a surface integral to a volume integral. Another way to see it, is that it converts the equation from a 2D one, to a 3D one and vice versa. For those who are interested you can look up for Green's theorem, which is a more general theorem to Gauss' and Stoke's.

It's always useful to take a step back and think how we've come to this equation to understand the physical significance of this equation. What ∇.E=ρ/ε really means is that the flux of electric field through a closed surface is equal to the total charges contained within the closed surface, multiplied by a constant. In a simpler form, EA=kQ, where E is the electric field, A is the area crossed by the electric field, Q is the amount of charge contained within A, and k is a constant. The product of E and A is also known as the electric flux.

The original meaning of flux is flow - as in flow of water. In science it usually means the rate of change of a particular 'thing' over an area. E.g. the flow of water in our pipes; flow of temperature; or in this case, the 'flow' of electric field. So, one can think of electric charges as sources where electric field will 'flow' from. And that from the principle of conservation of field (or matter), the source must equal to the resultant flow, i.e. total flux=source=kQ, which is the same as above (the constant k is just a scaling factor and can be easily set equal to 1 if appropriate units for Q and flux are used. Refer to previous section on how it was decided to use 1/ε as the constant). Just like our water supply, the amount of water that has flowed out from the pipe must equals to how much water is lost at the supply tank. All the fancy equations about fields come down this simple analogy of water flowing from a tap! Anonymous said...

shinliang said...

hi. 我听过你的名字，也看过你的留言。 Anonymous said...

The World 记忆里的空间Of Memory said...