A particle having charge $q=8.85\mu C$ is placed on the axis of a circular ring of radius $R=30 cm$. Distance of the particle from centre of the ring is $a=40 cm$. Calculate electrical flux passing through the ring.

Electric field strength at a point in plane of ring depends upon its distance from centre of the ring. Magnitude of electric field is same at all those points which are equidistant from the centre and co-planar with the ring. Therefore, consider a coplanar and concentric ring of radius $x$ and radial thickness $dx$ as shown in figure below.

Its area is $dS=2\pi xdx$

Distance of every point of this ring from point

Change is $r=\sqrt{a^2+x^2}$

$\therefore$Electric field strength at circumference of this ring is $E=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}$

Inclination $\theta$ of $\overrightarrow{E}$ with the normal to surface of the ring considered is given by $\cos \theta =\frac{a}{r}$.

Flux passing through this ring is

 $$\begin{gathered} d\varphi =\overrightarrow{E}\overrightarrow{dS} \\ d\varphi =EdS\cos \theta \\ =\left[\frac{1}{4\pi {\epsilon }_0}\frac{q}{\left(a^2+x^2\right)}\right]\left(2\pi xdx\right)\left(\frac{a}{\sqrt{a^2+x^2}}\right) \end{gathered}$$

Hence, total flux passing through the given ring is

$$\begin{gathered} \varphi ={\int }_{x=0}^{x=R}\frac{aq}{2{\epsilon }_0}\frac{xdx}{{\left(a^2+x^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}=\frac{qa}{2{\epsilon }_0}\left[\frac{1}{a}-\frac{1}{\sqrt{a^2+R^2}}\right] \\ \varphi ={10}^5 N{C}^{-1}{m}^2 \end{gathered}$$