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The flux of the electric field of a point charge placed at the centre of a sphere is explicitly computed and shown to be
\[\text{Flux} = \frac{q}{\epsilon_0 }\]
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 know that the flux expression for an infinitesimal surface element,\eqref{eq1}, is equal to
\[\text{Flux} = \frac{q}{4\pi\epsilon_0} \times \text {Solid angle subtended by the surface at\,} \vec r_0.\]
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}&=&\label{eq3} \frac{Q}{4\pi\epsilon_0}\frac{\vec{r}}{r^3} \qquad \hat{n}=\frac{\vec{r}}{r} \\
\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}
\sum_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.
