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The expression for the electric field of a {\bf point charge} \(q\), located at \(\vec r_0\),
the electric field at \(\vec r\) is
\begin{equation}
\vec E = \frac{q}{4\pi\epsilon_0}
\frac{\vec r-\vec r_)}{|\vec r-\vec r_0|^3}
\end{equation}
we see that the flux expression for an infinitesimal surface element,\eqRef{eq1}, is equal to, see \eqref{SolidAngle}
\begin{equation}
\text{Flux} = \frac{q}{4\pi\epsilon_0} \times \text {Solid angle subtended by the surface at\,} \vec r_0.
\end{equation}
Point Charge at the centre of a sphere
We present a simple case of calculation of a flux of a electric field due to a single point charge \(Q\) through a
sphere of radius \(R\) with the charge at the centre. Now ,
\begin{eqnarray}
\bar{E}=\frac{Q}{4\pi\epsilon_0}\frac{\bar{r}}{r^3} \\ \label{eq3}
\hat{n}=\frac{\bar{r}}{r} \\ \label{eq4}
\bar{E}\cdot\hat{n}=\frac{Q}{4\pi\epsilon_0}\frac{1}{r^2} \label{eq5}
\end{eqnarray}
$\bar{E}\cdot\hat{n}$ has the same value all over the surface. Therefore ;
\begin{eqnarray}
\Sigma_k\bar{E_k}\hat{n_k}&=&\frac{Q}{4\pi\epsilon_0}\frac{1}{r^2}\times
4\pi\epsilon_0 r^2 \label{eq6}\\
&=&Q/\epsilon_0 \label{eq7}
\end{eqnarray}
Other cases when the charge is not at the center, and the case when the surface is not spherical cannot be handled in the direct fashion outlined above.
