**1 Answer**

written 4.8 years ago by |

Let us consider any point P , shown in Fig.1 located by a cartesian coordinate system. The value of D at the point P may be expressed in cartesian components, $\mathbf{D}_{0}=D_{x 0} \mathbf{a}_{x}+D_{y 0} \mathbf{a}_{y}+D_{z 0} \mathbf{a}_{z}$ . We choose as our closed surface the small rectangular box, centered at $P,$ having sides of lengths $\Delta x, \Delta y,$ and $\Delta z$ and apply Gauss's law,

$\oint_{S} \mathbf{D} \cdot d \mathbf{S}=Q$

In order to evaluate the integral over the closed surface, the integral must be broken up into six integrals, one over each face,

$$ \oint_{S} \mathbf{D} \cdot d \mathbf{S}=\int_{\text { front }}+\int_{\text { back }}+\int_{\text { left }}+\int_{\text { right }}+\int_{\text { bottom }}+\int_{\text { bottom }} $$

Consider the first of these in detail. Since the surface element is very small, $\mathbf{D}$ is essentially constant (over this portion of the entire closed surface) and $\begin{aligned} \int_{\text { front }} & \doteq \mathbf{D}_{\text { front }} \cdot \Delta \mathrm{S}_{\text { front }} \\ & \doteq \mathbf{D}_{\text { front }} \cdot \Delta y \Delta z \mathbf{a}_{x} \\ & \doteq D_{\mathrm{x}, \text { front }} \Delta y \Delta z \end{aligned}$

where we have only to approximate the value of $D_{x}$ at this front face. The front face is at a distance of $\Delta x / 2$ from $P,$ and hence

$\begin{aligned} D_{x, \text { front }} & \doteq D_{x 0}+\frac{\Delta x}{2} \times \text { rate of change of } D_{x} \text { with } x \\ & \doteq D_{x 0}+\frac{\Delta x}{2} \frac{\partial D_{x}}{\partial x} \end{aligned}$

where $D_{x 0}$ is the value of $D_{x}$ at $P$ , and where a partial derivative must be used to express the rate of change of $D_{x}$ with x, since $D_{x}$ in general also varies with y and z. This expression could have been obtained more formally by using the constant term and the term involving the first derivative in the Taylor's-series expansion for $D_{x}$ in the neighborhood of P. We have now $\int_{\text { front }} \doteq\left(D_{x 0}+\frac{\Delta x}{2} \frac{\partial D_{x}}{\partial x}\right) \Delta y \Delta z$ Consider now the integral over the back surface, $\begin{aligned} \int_{\text { back }} & \doteq \mathbf{D}_{\text { back }} \cdot \Delta \mathbf{S}_{\text { back }} \ & \doteq \mathbf{D}_{\text { back }} \cdot\left(-\Delta y \Delta z \mathbf{a}_{x}\right) \ &=-D_{x, \text { back }} \Delta y \Delta z \ D_{x, \text { back }} & \doteq D_{x 0}-\frac{\Delta x}{2} \frac{\partial D_{x}}{\partial x} \ & \int_{\text { back }} \doteq\left(-D_{x 0}+\frac{\Delta x}{2} \frac{\partial D_{x}}{\partial x}\right) \Delta y \Delta z \end{aligned}$ If we combine these two integrals, we have $\int_{\mathrm{front}}+\int_{\mathrm{back}} \doteq \frac{\partial D_{x}}{\partial x} \Delta x \Delta y \Delta z$ By exactly the same process we find that $\int_{\mathrm{right}}+\int_{\mathrm{left}} \doteq \frac{\partial D_{y}}{\partial y} \Delta x \Delta y \Delta z$ and $\int_{\mathrm{top}}+\int_{\mathrm{bottom}} \doteq \frac{\partial D_{z}}{\partial z} \Delta x \Delta y \Delta z$ and $\int_{\mathrm{top}}+\int_{\mathrm{bottom}} \doteq \frac{\partial D_{z}}{\partial z} \Delta x \Delta y \Delta z$ and these results may be collected to yield $$ \oint_{S} \mathbf{D} \cdot d \mathbf{S} \doteq\left(\frac{\partial D_{x}}{\partial x}+\frac{\partial D_{y}}{\partial y}+\frac{\partial D_{z}}{\partial z}\right) \Delta x \Delta y \Delta z $$

or

$\oint_{S} \mathbf{D} \cdot d \mathbf{S}=Q \doteq\left(\frac{\partial D_{x}}{\partial x}+\frac{\partial D_{y}}{\partial y}+\frac{\partial D_{z}}{\partial z}\right) \Delta v$

The expression is an approximation which becomes better as $\Delta v$ becomes smaller, and in the following section we she volume $\Delta v$ approach zero. For the moment, we have applied Gauss's law to the closed surface surrounding the volume element $\Delta v$ and have as a result the approximation stating that

Charge enclosed in volume $\Delta v \doteq\left(\frac{\partial D_{x}}{\partial x}+\frac{\partial D_{y}}{\partial y}+\frac{\partial D_{z}}{\partial z}\right) \times$ volume $\Delta v$