Two charge particles are arranged as shown in figure. The height of electric field line at mid point is given by  $h=\frac{\sqrt{n}}{3}a$ . Find the value of n 
 

Solid angle at charge +q due to plane angle of  ${60}^o$  is equal to  $2\pi (1-\cos {60}^o)=\pi$ .
Thus, all field lines amanating from charge +q in the region of solid angle  $\pi$  as shown in the figure will also pass through the disc of radius h. This bending of field lines is due to presence of other charge –q
 
Thus,
 $$\begin{gathered} \left(\begin{array}{l}No.offieldlines\\ emanatingfrom+q\\ insolidangle\pi \end{array}\right)=\left(\begin{array}{l}Totalno.of\\ fieldlinecros\sin g\\ thedisc\end{array}\right) \\ \Rightarrow \frac{q}{4\pi {\epsilon }_0}\times 2\pi (1-\cos {60}^o)=\frac{2\times q}{4\pi {\epsilon }_0}\times 2\pi [1-\frac{a}{\sqrt{a^2+h^2}}] \\ \Rightarrow h=\frac{\sqrt{7}a}{3} \end{gathered}$$