Tracing Electric Field Lines

Standard bookwork tells us that the equation of a general field line is $q_1\cos\theta_1 + q_2\cos\theta_2 = c$ for some constant $c$ , where $\theta_1$ is the angle that the line from the particle to the charge $q_1$ makes with the line joining the two charges, and $\theta_2$ is the angle that the line from the particle to the charge $q_2$ makes with the same line.

When the field line leaves $q_1$ we have $\theta_1=45^\circ$ and $\theta_2=180^\circ$ , When the field line enters $q_2$ we have $\theta_1=\theta_2=0^\circ$ . Thus $q_1 + q_2 \; = \; c \qquad \qquad \tfrac{q_1}{\sqrt{2}} - q_2 \; = \; c$ and solving these simultaneous equations gives $\left|\tfrac{q_1}{q_2}\right| \; = \; \tfrac{2\sqrt{2}}{\sqrt{2}-1} \; = \; 2(2+\sqrt{2}) \; = \; 6.828$

Instead of considering just one line, let us consider a numerous amount of Field lines that emerge from $q_1$ and enter $q_2$ at the respective angles $\alpha$ and $\beta$ (thus creating a cone in the vicinity of each charge)

Now,
It is a proven fact, and also an obvious one, that the magnitude of the charges are inversely proportional to the respective Solid angles .

We know,
Solid angle $\displaystyle = 2\pi(1-\cos\theta)$ , where $\theta$ is the half angle of the cone.

Hence,
$\displaystyle |q_1| \propto \frac{1}{2\pi(1-\cos\alpha)}$ and $\displaystyle |q_2| \propto \frac{1}{2\pi(1-\cos\beta)}$

Dividing the two equations,

$\displaystyle \frac{|q_1|}{|q_2|} = \frac{1-\cos\beta}{1-\cos\alpha} = \frac{1-\cos\pi}{1-cos\frac{\pi}{4}} = \frac{2}{1-\frac{1}{\sqrt{2}}} \approx \boxed{6.83}$