Angle of electric field lines leaving a postive charge and entering a negative charge in dipole 
A positive and negative charge are some distance d away from each other (Electric dipole configuration) Consider the electric field lines in between them.
That is, there is a straight line electric field vector that points from (+) to (-) along the d direction.
There is also an arbitrary electric field line exiting the (+) charge by some angle alpha with respect to the d direction and it enters the (-) charge by some angle beta with respect to the d direction.
Assume alpha does not equal beta. What are the angles?

This was a question posed by my professor to think about. At first I thought the angles would be equal so I didn't really understand his question. I think the case that they would be equal would be when the positive charge magnitude is equal to the negative charge magnitude (I may be wrong). He did not tell us what the magnitude of the (+) and (-) charges so I think this may be the main point. Would there be a way to calculate the angle? Maybe a ratio?
This is mainly for my own curiosity any insight would be appreciated!

 A: Your intuition is correct $-$ an asymmetry between  $\alpha$ and $\beta$ is only possible if the two charges are different.
If you want a quantitative relationship between the two angles, the correct (read: the only viable) approach is via Gauss's law for the electric flux. From the geometry of the field, which is symmetric about the inter-charge axis, it is relatively easy to see that if you take the surface of revolution generated by the field line about the axis, then you get a cylinder-with-two-conical-ends which goes from one charge to the other:

By definition, the electric field is tangential to this surface at every point, which means that no field lines leave it, and the electric flux is confined inside it. That means, therefore, that the electric flux that leaves charge 1 into the surface must equal the electric flux that arrives at charge 2.
Moreover, we know how to relate these electric fluxes to the angles: close to charge 1, we can ignore the effect of charge 2, and then we just have the flat integral, i.e., the electric flux is the product of the charge times the solid angle spanned by the cone at its apex,
$$
\Phi_1 = q_1 \: \Omega_1,
$$
where the solid angle can be calculated explicitly as
\begin{align}
\Omega_1 
& = \int_0^\alpha \int_0^2\pi \sin(\theta)\mathrm d\phi \:\mathrm d\theta
\\ & = 2\pi \big[-\cos(\theta)\big]_0^\alpha
\\ & = 2\pi(1-\cos(\alpha)).
\end{align}
Assuming that $q_1>0>q_2$, we can set $\Phi_1+\Phi_2=0$ and therefore
\begin{align}
 2\pi(1-\cos(\alpha))|q_1| & =  2\pi(1-\cos(\beta))|q_2|
\\ \implies
\frac{1-\cos(\alpha)}{1-\cos(\beta)} & = \frac{|q_2|}{|q_1|}
.
\end{align}
This relationship then allows you to find any of the relevant quantities in terms of the other three, which is the most that you can do here.
A: 

I hope you understand my handwriting. This question is one of the most interesting one I've seen in a while. Thanks.
A: Here is my best attempt, I couldnt solve for the angle as a function of the other angle so I am not sure if it is completely correct or if there is a simpler way to solve.
