Gausss - law - in - Integral - and - Differential - form
1 Answer

The Gauss's law states that, the total outward electric displacement through any closed surface surrounding charges is equal to the total charge enclosed.

Consider a charge Q at some location in space (Fig.1).The electric displacement density D at some point P at a distance r from Q is given by

$\mathbf{D}=\hat{\mathbf{r}} \frac{Q}{4 \pi r^{2}}$ ...(1) enter image description here

Suppose, we place a frame enclosing area A at location P . Then the electric displacement passing through the frame indeed will depend upon the area of cross-section A of the frame but it will also depend upon the orientation of the frame with respect to D. If the direction of D is normal to the area A, the electric displacement through the frame will be $D A .$ On the other hand, if the direction of D is tangential to the area A, the electric displacement through the frame will be zero. In general we can see that the electric displacement through the frame is proportional to the projection of the area A perpendicular to $\mathbf{D} .$ So if the angle between D and the normal to the area $\hat{\mathbf{n}}$ is $\theta,$ the projected area will be $A \cos \theta .$ The electric displacement through the frame therefore will be $D A \cos \theta .$ In vector algebra this is the dot product of $\mathbf{D}$ and the area $\mathbf{A},$ where the direction of an area is defined by its normal. Therefore, Electric displacement

$=\mathbf{D} \cdot \mathbf{A}$

Let us now consider a closed surface S surrounding the charges $Q_{1}, Q_{2}, Q_{3}$ as shown in Fig. 2. Then, through an incremental area da on the surface, the outward electric displacement will be $\mathbf{D}$-da. The total electric displacement can be obtained by integrating over the surface S. Then according to the Gauss's law

$\oint_{S} \mathbf{D} \cdot \mathbf{d a}=Q_{1}+Q_{2}+\ldots+Q_{n}$ ...(2)

enter image description here

Instead of discrete charges $Q_{1}, Q_{2}, \ldots Q_{n}$ if there is a continuous distribution of charge inside the closed surface, we have to find total charge by integrating over the enclosed volume. The distributed charges can be correctly represented by a charge density $\rho$ which in general is a function of space. The $\rho$ is the charge per unit volume having units Coulomb/m. The charge in a small volume $d v$ will be $\rho d v$ . The total charge enclosed by the volume can be obtained by integrating over the volume V. For distributed charges $\operatorname{Eqn}(3.32)$ then becomes

$$\oint_{S} \mathbf{D} \cdot \mathbf{d a}=\iiint_{\mathbf{V}} \rho d v$$...(3)

This is the Gauss law in the integral form.

The equivalent differential form can be obtained by applying the Divergence theorem to Eqn (3) .The LHS of Eqn $(3)$ can be written as

$\oint_{S} \mathbf{D} \cdot \mathbf{d a}=\iiint_{V}(\nabla \cdot \mathbf{D}) d v$ ...(4)

Substituting in Eqn $(3)$ and bringing all the terms on the left hand side of the equality sign we get

$\iiint_{V}\{(\nabla \cdot \mathbf{D})-\rho\} d v=0$ ...(5)

Equation ( 3.35 ) has to be valid for any arbitrary volume. This can happen provided the integrated is identically zero making

$$ \nabla \cdot \mathbf{D}=\rho ...(6) $$

This is the differential form of the Gauss's law.

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