Electric Angle

$\large \begin{aligned} & \text{In the figure above, } '\text{A}' \text{ is a circular region, which is also the base of cones with apex angles } 2{\color{#3D99F6}\alpha} \text{ and } 2{\color{#D61F06}\beta}. \\ & \text{Let } {\color{#3D99F6}\Omega_\alpha} \text{ and } {\color{#D61F06}\Omega_\beta} \text{ be the solid angles subtended by the region A on the charges } {\color{#3D99F6} +Q_1} \text{ and } {\color{#D61F06} -Q_2} \text{ respectively. Then,} \\ & \text{(Flux entering the region A)}={\color{#3D99F6} \phi_\alpha} = \text{(Flux leaving the region A)}={\color{#D61F06} \phi_\beta} \\ & \implies \dfrac{{\color{#3D99F6}\Omega_\alpha}}{4\pi}\dfrac{\color{#3D99F6} Q_1}{\epsilon_0}=\dfrac{{\color{#D61F06}\Omega_\beta}}{4\pi}\dfrac{\color{#D61F06}Q_2}{\epsilon_0} \\ & \implies \dfrac{2\pi(1-\cos{\color{#3D99F6} \alpha})}{4\pi}\dfrac{\color{#3D99F6} Q_1}{\epsilon_0}=\dfrac{2\pi(1-\cos{\color{#D61F06} \beta})}{4\pi}\dfrac{\color{#D61F06} Q_2}{\epsilon_0} \\ & \implies \boxed{\dfrac{1-\cos{\color{#3D99F6} \alpha}}{1-\cos{\color{#D61F06} \beta}}=\dfrac{\color{#D61F06} Q_2}{\color{#3D99F6} Q_1}} \\ & \text{Putting } Q_1=4, Q_2=1, \alpha=60^{\circ}, \beta=\theta \\ & \implies \dfrac{1-\cos{60^{\circ}}}{1-\cos{\theta}}=\dfrac{1}{4} \\ & \implies \cos{\theta}=-1 \implies \boxed{\theta=180^{\circ}} \end{aligned}$

Notes:

• To see why $\Omega=2\pi(1-\cos{\theta})$ , click here .

• The solid angle of a sphere measured from any point in its interior is $4\pi\text{ sr}$ .

• The flux through a sphere which has charge $Q$ inside it, is $Q/\epsilon_0$ . See Gauss Law .

• $\epsilon_0$ is permittivity of free space .

Standard bookwork tells us that the equation of a field-line between the two charges is given by the equation $4q\cos\theta_1 - q\cos\theta_2 = c$ where $\theta_1$ is the angle formed by the line from a point on the field line to the charge $4q$ and the line joining the charges, and $\theta_2$ is the angle formed by the line from a point on the field line to the charge $-q$ and the line joining the charges.

When the field-line enters the $4q$ charge, we that $\theta_1 = 60^\circ$ and $\theta_2 = 180^\circ$ . Thus $c = 3q$ . When the field-line enters the other charge, we have $\theta_1 = 0^\circ$ , and so $4q - q\cos\theta_2 = 3q$ . Thus $\cos\theta_2 = 1$ , and hence $\theta_2 = 0^\circ$ , and hence $\theta = \boxed{180^\circ}$ .

The flux density emerging from the positive charge is 4 times the flux density entering the negative charge. Integrating the solid angle made by all angles less than 60 degrees out of the positive charge gives an area of pi, or one quarter of a sphere. Thus all flux lines in this region enter the negative charge, while all others do not. The line at 60 degrees from the axis must therefore enter the negative charge at the most distant possible point - 180 degrees

I tried a numerical approach. I.e. plotting the path whose direction is determined by the field strength. This led me to a discovery (obvious when known) then there is a region with almost no net field on the right of -Q and the same distance from -Q as the spatial separation of the charges (and twice that from +4Q). Essentially if $\alpha = 60^o - \epsilon$ the field line approaches from the right with $\beta < 180$ . If $\alpha = 60^o+\epsilon$ then the field line does not travel to -Q. Therefore if $\alpha = 60^o$ then the field line approaches with $\beta = \boxed{180^o}$

It should be noted that $\frac{d\beta}{d\alpha}$ is immense in the region where $\alpha \approx 60^o$

Thanks to Digvijay Singh, whilst I did not solve this with your elegance or rigour. I did gain an insight into what happens to the field lines in regions that are almost nulled out.