GAUSS'S THEOREM OR GAUSS'S LAW :
Question: State and explain Gauss' theorem for electrostatic.
Solution:
Gauss's Theorem: According to Gauss's theorem, the total electric flux ($\phi$) through any closed surface (S) in free space is equal to $1/\epsilon_0$ times the total electric charge ($q$) enclosed by the surface.
$\phi = \oint_S \vec{E} \cdot d\vec{S} = \frac{q}{\epsilon_0}$
$\phi = \frac{q}{\epsilon_0}$
Proof: Consider an isolated point charge $+q$ placed at point $O$. Let surface $S$ be a sphere of radius $r$ around the charge $+q$. This sphere is called Gaussian surface.
What is Gaussian Surface ?
Gaussian surface around a charge (point or continuous distribution) is an imaginary closed surface, such that the intensity of electric field at all points on its surface is same.}
Proof of Gauss's Theorem or Gauss's law :
Electric field intensity due to charge $+q$ at every point on the Gaussian surface is given by :
$\vec{E} = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r} \qquad \dots (i)$
where $\hat{r}$ is a unit vector in the direction of increasing $r$.
Now consider a small element of area $dS$ with area vector $d\vec{S}$ normal to the surface of the area element.
Electric flux through the area element is given by
$d\phi = \vec{E} \cdot d\vec{S}$
$d\phi= \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r} \cdot d\vec{S}$
$d\phi= \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r} \cdot dS \hat{n}$
$d\phi== \frac{1}{4\pi\epsilon_0} \frac{qdS}{r^2} \hat{r} \cdot \hat{n}$
where $\hat{n}$ is unit vector perpendicular to the surface of the area element.
Since $\hat{r}$ and $\hat{n}$ are along the same direction, so $\theta = 0^{\circ}$.
$\hat{r} \cdot \hat{n} = 1 \times 1 \times \cos 0^{\circ} = 1$
Hence,
$d\phi = \vec{E} \cdot d\vec{S}$
$d\phi= \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} dS \qquad \dots (ii)$
Now, electric flux over the entire closed Gaussian surface is given by
$\phi = \oint_S d\phi = \oint_S \vec{E} \cdot d\vec{S}$
$\phi= \oint_S \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} dS$
$\phi = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \oint_S dS \qquad \dots (iii)$
$\oint_S dS$ = surface area of the Gaussian sphere of radius r
$\oint_S dS= 4\pi r^2$
$\therefore \quad \phi = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \times 4\pi r^2$
$\phi = \frac{q}{\epsilon_0} \qquad \dots (iv)$
or
$\phi = \oint_S \vec{E} \cdot d\vec{S} = \frac{q}{\epsilon_0}$
which is the expression of Gauss' law or theorem in electrostatics.
Special Cases:
1. If the closed surface does not enclose any charge i.e., $q = 0$, then eqn. (iv) becomes
$\phi = \oint_S \vec{E} \cdot d\vec{S} = 0$
Thus, electric flux through a closed surface containing no charge is zero.
2. If the closed surface contains two charges of equal magnitude but opposite in sign, then the net charge enclosed by the closed surface $= +q - q = 0$. Hence, from eqn. (iv),
$\phi = \oint_S \vec{E} \cdot d\vec{S} = 0$
In this case, inward electric flux cancels outward electric flux and hence the net electric flux through the closed surface is zero. This is the case of an electric dipole enclosed by a closed surface.
3. Outward electric flux through a closed surface is taken as positive.}
$\phi = \oint_S \vec{E} \cdot d\vec{S} = \frac{+q}{\epsilon_0}$
4. Inward electric flux through a closed surface is taken as negative.
$\phi = \oint_S \vec{E} \cdot d\vec{S} = \frac{-q}{\epsilon_0}$
5. If closed surface has discrete distribution of charges. Let $q_1$, $q_2$, $q_3$, ..., $q_n$ be the discrete charges in a closed surface, then the net or total charge enclosed is given by,
$q = q_1 + q_2 + q_3 + \dots + q_n = \sum_{i=1}^n q_i$
$\phi = \oint_S \vec{E} \cdot d\vec{S} = \frac{1}{\epsilon_0} \sum_{i=1}^n q_i \qquad \dots (v)$
6.Net electric flux through a closed surface is directly proportional to the net amount of charge enclosed within the surface but is independent of the size of the closed surface.}
7. A charge outside a closed surface having no charge inside does not affect the net electric flux of the surface.
Consider a charge $+q$ outside a closed surface. In this case, the number of electric field lines entering the closed surface is equal to the number of electric field lines leaving the closed surface. Hence, the net electric flux of the closed surface is taken as zero.
