The procedure for applying Gauss's law to calculate the electric field involves first knowing whether symmetry exists. Once symmetric charge distribution exists, we construct a mathematical closed surface (known as a Gaussian surface). The surface is chosen such that D is normal or tangential to the Gaussian surface. When D is normal to the surface, D • dS = D dS because D is constant on the surface. When D is tangential to the surface, D • dS = 0. Thus we must choose a surface that has some of the symmetry exhibited by the charge distribution. We shall now apply these basic ideas to the following cases.

A. Point Charge

Suppose a point charge Q is located at the origin. To determine D at a point P, it is easy to see that choosing a spherical surface containing P will satisfy symmetry conditions. Thus,a spherical surface centered at the origin is the Gaussian surface in this case and is shown in Figure 1.

Since D is everywhere normal to the Gaussian surface, that is, $D = D_{r}a_{r}$ applying Gauss's law $\left(\Psi=Q_{\text { enclosed }}\right)$ gives

$$ Q=\oint \mathbf{D} \cdot d \mathbf{S}=D_{r} \oint d S=D_{r} 4 \pi r^{2} $$ ...(1)

where $\oint d S=\int_{\phi=0}^{2 \pi} \int_{\theta=0}^{\pi} r^{2} \sin \theta d \theta d \phi=4 \pi r^{2}$ is the surface area of the Gaussian surface. Thus

$\mathbf{D}=\frac{Q}{4 \pi r^{2}} \mathbf{a}_{r}$ ...(2)

B. Infinite Line Charge

Suppose the infinite line of uniform charge $\rho_L$ C/m lies along the z-axis. To determine D at a point P, we choose a cylindrical surface containing P to satisfy symmetry condition as shown in Figure 2. D is constant on and normal to the cylindrical Gaussian surface; that is,$ D = D_{\rho}a_{\rho}$. If we apply Gauss's law to an arbitrary length l of the line

$\rho_{L} \ell=Q=\oint \mathbf{D} \cdot d \mathbf{S}=D_{p} \oint d S=D_{\rho} 2 \pi \rho \ell$ ...(3)

where $\oint d S=2 \pi \rho \ell$ is the surface area of the Gaussian surface. Note that $\int \mathbf{D} \cdot d$ . evalu- ated on the top and bottom surfaces of the cylinder is zero since $\mathbf{D}$ has no z-component; that means that $\mathbf{D}$ is tangential to those surfaces. Thus

$\mathbf{D}=\frac{\rho_{L}}{2 \pi \rho} \mathbf{a}_{\rho}$ ...(4)

C. Infinite Sheet of Charge

Consider the infinite sheet of uniform charge $\rho_s$ C/m2 lying on the z = 0 plane. To determine D at point P, we choose a rectangular box that is cut symmetrically by the sheet of charge and has two of its faces parallel to the sheet as shown in Figure 3. As D is normal to the sheet, $D = D_{z}a_{z}$, and applying Gauss's law gives

$\rho_{S} \int d S=Q=\oint \mathbf{D} \cdot d \mathbf{S}=D_{z}\left[\int_{\mathrm{top}} d S+\int_{\mathrm{bottom}} d S\right]$ ...(5)

Note that D • dS evaluated on the sides of the box is zero because D has no components along $a_x$ and $a_y$. If the top and bottom area of the box each has area A, eq. 5 becomes

$$ \rho_{S} A=D_{z}(A+A) ...(6) $$

and thus $$ \mathbf{D}=\frac{\rho_{S}}{2} \mathbf{a}_{z} $$ or

$\mathbf{E}=\frac{D}{\varepsilon_{\mathbf{o}}}=\frac{\rho_{S}}{2 \varepsilon_{\mathbf{o}}} \mathbf{a}_{z}$ ...(7)

D. Uniformly Charged Sphere

Consider a sphere of radius a with a uniform charge $\rho_v$ C/m3. To determine D everywhere,we construct Gaussian surfaces for cases $r \leq a$ and $r \geq a$ separately. Since the charge has spherical symmetry, it is obvious that a spherical surface is an appropriate Gaussian surface.

For $r \leq a$ , the total charge enclosed by the spherical surface of radius $r,$ as shown in Figure 4. (a), is

$$ \begin{aligned} Q_{\text { enc }} &=\int \rho_{v} d v=\rho_{v} \int d v=\rho_{v} \int_{\phi=0}^{2 \pi} \int_{\theta=0}^{\pi} \int_{r=0}^{r} r^{2} \sin \theta d r d \theta d \phi \\ &=\rho_{v} \frac{4}{3} \pi r^{3} \end{aligned} $$ and $\begin{aligned} \psi &=\oint \mathbf{D} \cdot d \mathbf{S}=D_r \oint d S=D_r \int_{\phi=0}^{2 \pi} \int_{\theta=0}^{\pi} r^{2} \sin \theta d \theta d \phi \\ &=D_{r} 4 \pi r^{2} \\ \text { Hence, } \Psi=Q_{\text { enc }} & \text { gives } \end{aligned}$

$D_{r} 4 \pi r^{2}=\frac{4 \pi r^{3}}{3} \rho_{v}$

or

$\mathbf{D}=\frac{r}{3} \rho_{v} \mathbf{a}_{r} \quad 0\ltr \leqslant \mathbf{a}$

For $r \geq a$ , the Gaussian surface is shown in Figure 4.(b) . The charge enclosed by the surface is the entire charge in this case, that is,

$\begin{aligned} Q_{\text { enc }} &=\int \rho_{v} d v=\rho_{v} \int d v=\rho_{v} \int_{\phi=0}^{2 \pi} \int_{\theta=0}^{\pi} \int_{r=0}^{a} r^{2} \sin \theta d r d \theta d \phi \\ &=\rho_{v} \frac{4}{3} \pi a^{3} \end{aligned}$

while

$$ \Psi=\oint \mathbf{D} \cdot d \mathbf{S}=D_{r} 4 \pi r^{2} $$