[NOTES/EM-02004]-Applications of Gauss Law

In words, the Gauss law states that
the electric flux through any closed surface is proportional to the enclosed electric charge.

Defining flux as $\Phi$, mathematically the Gauss law is;
\begin{eqnarray}
\Phi=\int\bar{E}\cdot dS=\frac{q_\text{enclosed}}{\epsilon_0} \label{eq0}
\end{eqnarray}

We shall now compute $\bar{E}$ in various cases using the Gauss law. This invariably requires the use of symmetry arguments in each case.

1. Plane sheet of charge

Let the charge density be $\sigma$. To find electric field at a point P ,
consider a gaussian surface which is cylindrical as shown in figure.

Clearly $\bar{E}$ is normal to S and S' Therefor ;
\begin{eqnarray}
\nonumber \text{Flux through S} &=& EA
\nonumber\text{ Flux through S'} &=& EA
\nonumber\text{ Flux through the curved surface} &=& 0
\nonumber \text{Total outward flux} &=& 2EA
\nonumber \text{Charge enclosed} &=& \sigma A
\end{eqnarray}

Therefore using the Gauss law
\begin{eqnarray}
\nonumber 2EA &=& \sigma/\epsilon_0
\bar{E} &=& \sigma/2\epsilon_0 \label{eq1}
\end{eqnarray}

2. Line charge with uniform charge density

Let the charge density be $\lambda$. The field due to line charge at any point P
is perpendicular to the line, the parallel components cancel. This can be
observed by dividing the line charge into 2 small pieces of length dl. Here the
Gaussian surface is a cylinder of radius r and height h passing through P and
the line charges as the axis. $\bar{E}$ is normal to the curved surface and
constant over it.
Therefore the flux through the curved surface $\phi_{cs}$ is
\begin{eqnarray}
\nonumber\phi_{cs}&=& area \mid\bar{E}\mid
&=&\bar{E}\cdot2\pi \bar{r} \times h \label{eq2}
\end{eqnarray}
And the flux through the plane top and bottom surfaces $\phi_{top}$ and
$\phi_{bottom}$ is
\begin{eqnarray}
\nonumber\phi_{top}&=&\phi_{bottom}=0
\end{eqnarray}
This is because the electric field through the two plane surfaces is
perpendicular to the normal $\hat{n}$ at the surfaces and
$\bar{E}\cdot\hat{n}=0$
The total flux $\phi$ is
\begin{eqnarray}
\nonumber\phi &=& \phi_{cs}+\phi_{top}+\phi_{bottom}
&=&\bar{E}\cdot2\pi \bar{r}h \label{eq3}
\end{eqnarray}
Charge enclosed q is
\begin{eqnarray}
q=\lambda h \label{eq4}
\end{eqnarray}
Hence using Gauss law ;
\begin{eqnarray}
\nonumber\bar{E}(2\pi \bar{r})h&=&(\lambda h)/\epsilon_0
\bar{E}&=&\lambda/2\pi\epsilon_0\bar{r} \label{eq5}
\end{eqnarray}


3. Spherical shell with uniform charge density

The electric field $\bar{E}$ is radially outward. The guassian surface is a
sphere of radius.
$\mid\bar{E}\mid$ is constant over the guassian surface and $\bar{E}$ is
parallel to the outward normal $\hat{n}$.

Therefore the flux $\phi$ is ;
\begin{eqnarray}
\nonumber\phi&=&\int\bar{E}\cdot\hat{n}dS
\nonumber&=&\int\mid\bar{E}\mid dS
\nonumber&=&\mid\bar{E}\mid\int dS
&=&\bar{E}4\pi r^2 \label{eq6}
\end{eqnarray}

We know by the Guass law that the charge enclosed is q/$\epsilon_0$

Therefore the electric field of the spherical shell at a distance of r is
\begin{eqnarray}
\bar{E}=q/4\pi\epsilon_0 r^2 \label{eq7}
\end{eqnarray}

For a point inside the shell the guassian surface does not enclose any charge.
Therefore
\begin{eqnarray}
\phi&=&0
\nonumber\bar{E} \times 4\pi r^2& = &0
\bar{E}&=&0 \label{eq8}
\end{eqnarray}

4. Uniformly charged solid sphere

The total charge is q. Guassian surface is a sphere. Charge density of the solid sphere $\rho$ is charge per unit volume.
\begin{eqnarray}
\nonumber\rho&=&q/V
&=&3q/4\pi a^3 \label{eq9}
\end{eqnarray}
Where a is the radius of the gaussian spherical surface from the centre of the solid sphere and r is the radius of the solid sphere.

$a>r$

The point is outside the solid sphere.So the charge enclosed is q. Flux through the guassian surface ;
\begin{eqnarray}
\nonumber\phi&=&\bar{E}\cdot 4\pi r^2 = q/\epsilon_0
\bar{E}=q/4\pi\epsilon_0 r^2 \label{eq10}
\end{eqnarray}

$a< r$

The point is inside the solid sphere. The flux is
\begin{eqnarray}
\phi&=&\bar{E}\cdot 4\pi r^2 \label{eq11}
\end{eqnarray}
Charge enclosed q' is;
\begin{eqnarray}
q'=\rho \frac{4}{3} \pi r^3 \label{eq12}
\end{eqnarray}
Using the Gauss law ;
\begin{eqnarray}
\nonumber\bar{E}\cdot 4\pi r^2&=& \rho \frac{4}{3} \pi r^3/\epsilon_0
\nonumber\bar{E}&=&\rho r/3\epsilon_0
&=&qr/4\pi\epsilon_0 a^3 \label{eq13}
\end{eqnarray}