Gauss Divergence Theorem (or) Application of Gauss’s law to a differential volume element

Gauss Divergence theorem:-

A vector field changes from point to point throughout space, this change can be identified by two things Divergence and curl,the second is curl, which will be examined when magnetic fields are discussed, so Divergence , which is a scalar function and is somewhat similar to derivative of a function will be discussed in Gauss-Divergence theorem.

Generally, when the divergence ≠ 0 ⇒ that there exists sources or sinks in a particular region.

If divergence is positive ⇒ a source                                                                                    If divergence is negative ⇒ a sink

In static electric fields, there is a correspondence between positive divergence, sources and positive electric charge ‘Q’, since electric flux ψ by definition originates on a positive charge, thus a region which consists of positive charges contains the source of flux ψ.

∴ The divergence of electric flux density \overrightarrow{D} will be positive in that region.Divergence of a vector field \overrightarrow{A}/\overrightarrow{D} at a point ‘P’ is defined by 

div\overrightarrow{A}= \lim_{\Delta v->0}\frac{\oint \overrightarrow{A}.\overrightarrow{ds}}{\Delta v}            or      div\overrightarrow{D}= \lim_{\Delta v->0}\frac{\oint \overrightarrow{D}.\overrightarrow{ds}}{\Delta v}

here, the integration is over the surface of an infinitesimal volume \Delta v that shrinks at point P

Gauss-Divergence theorem:- (statement)

for a continuously differentiable vector field the net outward flux from a closed surface equals the volume integral of the divergence throughout the region bounded by that surface.

i.e, \oint_{s}\overrightarrow{D}.\overrightarrow{ds}=\int_{v}(\overrightarrow{\bigtriangledown}. \overrightarrow{D})dv   where \overrightarrow{D} be the vector field.

It relates volume integral with surface integral.

Proof of Divergence theorem:-

Now consider a slightly different type of situation to which Gauss’s law is to be applied, that is on the surface \overrightarrow{D} is not constant (or) zero and this surface doesn’t have any symmetry at all. For such type of surface the change in \overrightarrow{D} can be represented by Taylor series expansion by assuming \overrightarrow{D} as constant.

i.e, f(x+\frac{\Delta x}{2})=f(x)+\frac{\frac{\Delta x}{2}}{1!}{f}'(x)+\frac{\frac{\Delta x}{2}^{2}}{2!}{f}''(x)+..... over the surface.

This change in \overrightarrow{D} can be expressed by only first two terms 

i.e, f(x+\frac{\Delta x}{2})=f(x)+\frac{\frac{\Delta x}{2}}{1!}{f}'(x)   

for the differential volume element \large \Delta x =dx,\Delta y=dy and \Delta z=dz

Let us consider a point ‘P’ which is located at the center of the volume element, which is a differential volume with an assumption that \overrightarrow{D} is almost constant over this small volume dv and at P \overrightarrow{D} is given as 

\overrightarrow{D}=D_{xo}\overrightarrow{a}_{x} +D_{yo}\overrightarrow{a}_{y}+D_{zo}\overrightarrow{a}_{z}

Now, from Gauss’s law \oint_{s}\overrightarrow{D}.\overrightarrow{ds} =Q_{enclosed} , the surface integral is divided into ‘6’ integrals as \oint_{s}\overrightarrow{D}.\overrightarrow{ds} =\int_{front}+\int_{back}+\int_{left}+\int_{right}+\int_{top}+\int_{bottom}

\int_{front} = \overrightarrow{D}_{front}.\overrightarrow{d s}_{front}

\large \overrightarrow{D}_{front} = \overrightarrow{D}_{xfront}\overrightarrow{a}_{x} +\overrightarrow{D}_{yfront}\overrightarrow{a}_{y} +\overrightarrow{D}_{zfront}\overrightarrow{a}_{z}

\large \overrightarrow{d s}_{front} = \Delta y\Delta z \overrightarrow{a}_{x}

\large \overrightarrow{D}_{front}.\overrightarrow{ds}_{front}= (D_{xfront}\overrightarrow{a}_{x}+D_{yfront}\overrightarrow{a}_{y}+D_{zfront}\overrightarrow{a}_{z}).dydz\overrightarrow{a}_{x}

                                   \large = D_{xfront}dydz

now the component \large D_{xfront}=D_{xo}+\frac{dx}{2}\frac{\partial D_{x}}{\partial x}

\large \therefore \int_{front}=(D_{xo}+\frac{dx}{2}\frac{\partial D_{x}}{\partial x})dydz

simillarly,  \large \therefore \int_{back}=(D_{xo}-\frac{dx}{2}\frac{\partial D_{x}}{\partial x})(-dydz)

\large \int_{front}+\int_{back} = dxdydz \frac{\partial D _{x}}{\partial x}

then the remaining two integral values \large \int_{left}+\int_{right} = dxdydz \frac{\partial D _{y}}{\partial y}

\large \int_{top}+\int_{bottom} = dxdydz \frac{\partial D _{z}}{\partial z}

\large \oint_{s}\overrightarrow{D}.\overrightarrow{ds}=dxdydz \frac{\partial D _{x}}{\partial x}+dxdydz \frac{\partial D _{y}}{\partial y}+ dxdydz \frac{\partial D _{z}}{\partial z} —————>Equation(1)

According to the definition of div \large \overrightarrow{D}

\large \lim_{\Delta v->0}\frac{\oint \overrightarrow{D}.\overrightarrow{ds}}{\Delta v}=div\overrightarrow{D}=\overrightarrow{\bigtriangledown }.\overrightarrow{D}

from Equation(1) \large \oint_{s}\overrightarrow{D}.\overrightarrow{ds}=dv( \frac{\partial D _{x}}{\partial x}+ \frac{\partial D _{y}}{\partial y}+ \frac{\partial D _{z}}{\partial z})

by dividing the above equation with dv=Δv

\large \frac{\oint_{s}\overrightarrow{D}.\overrightarrow{ds}}{dv}=( \frac{\partial D _{x}}{\partial x}+ \frac{\partial D _{y}}{\partial y}+ \frac{\partial D _{z}}{\partial z})

apply the limit on both sides

\large \lim_{dv->0}\frac{\oint_{s}\overrightarrow{D}.\overrightarrow{ds}}{dv}=\lim_{dv->0}( \frac{\partial D _{x}}{\partial x}+ \frac{\partial D _{y}}{\partial y}+ \frac{\partial D _{z}}{\partial z})

\large \overrightarrow{\bigtriangledown }.\overrightarrow{D}= div \overrightarrow{D}=( \frac{\partial D _{x}}{\partial x}+ \frac{\partial D _{y}}{\partial y}+ \frac{\partial D _{z}}{\partial z})

hence proved.

The divergence of the  flux density \large \overrightarrow{D} (vector ) is the flow of flux from a small closed surface per unit volume as the volume shrinks to zero \large (\lim_{dv->0}).


1 Star2 Stars3 Stars4 Stars5 Stars (No Ratings Yet)



Author: Lakshmi Prasanna

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.