An electric field line emerges from a positive point charge +q at angle $\alpha$ to the straight line connecting it to a negative point charge – 2q as shown in figure. At what angle $\beta$ will the field line enter the charge – 2q?

In the immediate proximity of each of point charges, the contribution from the other charge to the total field strength is negligibly small, and hence the electric field lines emerge from (enter) the charge in a spatially homogeneous bundle, their total number being proportional to the magnitude of the charge. Only  fraction of these lines gest into a cone with an angle 2α at the  vertex near the charge +q1​. The ratio o the number of these lines to the total number of the lines emerging from the charge +q1​ is equal to the ratio of the areas of the corresponding spherical segments:

$\frac{2\mathrm{\pi RR}(1-\cos \mathrm{\alpha })}{4{\mathrm{\pi R}}^2}=\frac{1}{2}(1-\cos \mathrm{\alpha })$

Since the electric field lines connect the two charges of equal magnitude, the number of lines merging from the charge +q1​ within the angle 2α is equal to the number of lines entering the charge −q2​ at an angle 2β. Consequently

$\left|{\mathrm{q}}_1\right|(1-\cos \alpha )=\left|{\mathrm{q}}_2\right|(1-\cos \beta )$

$\begin{array}{l}\text{ Solving }\mathrm{\beta }=2{\sin }^{-1}\left[\sqrt{\frac{{\mathrm{q}}_1}{{\mathrm{q}}_2}}\sin \left(\frac{\mathrm{\alpha }}{2}\right)\right]\\ =2{\sin }^{-1}\left[\sqrt{\frac{\mathrm{q}}{2\mathrm{q}}}\sin \left(\frac{\mathrm{\alpha }}{2}\right)\right]\\ =2{\sin }^{-1}\left[\frac{1}{\sqrt{2}}\sin \left(\frac{\mathrm{\alpha }}{2}\right)\right]\end{array}$