**1 Answer**

written 4.8 years ago by |

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)**

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)

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.