homework and exercises - Trajectory of an electric field line in a charge arrangement emerging at a given angle from a source charge

Consider a system of 2 particles with charge +q and -2q as source of electric field. The charges are fixed at A and B respectively. A field line from 1st charge emerges at an angle 150 degree with AB vector. Will this field line reach the 2nd charge or would travel to infinity? If it reaches the 2nd charge, what will be its angle with BA vector at location of the 2nd charge?

Answer

This problem is made more difficult because it is three dimensional and that makes imagining the situation and drawing diagrams harder so first look at the problem in two dimensions.

The electric field lines very close to a point charges (compared to the separation of the charges) are radially outwards as shown in the diagram below and not affected by the other charge.

The electric field lines are shown as a visual aid and in this case to show that the magnitude of charge $q_1$ id greater than the magnitude of charge $-q_2$ by having more electric field lines leaving charge $q_1$ than arriving at charge $-q_2$.

Given that electric field lines start at a positive charge and finish at a negative charge and also never cross then all the five red field lines which leave charge $q_1$ must finish as the five red field lines arriving at charge $q_2$.

Stating this in a more refined way means that the flux of electric field through arc $AB$ must be the same as the flux of electric field through arc $A'B'$.

If the circles which are drawn have a radius $r$ and the circle is taken to be a Gaussian "surface" then the proportion of the electric flux which passes through arc $AB$ compared with the whole circumference is $\dfrac {AB}{2 \pi r} = \dfrac {2r\alpha}{2\pi r} = \dfrac{ \alpha }{\pi}$.

Using Gauss's law the total flux through the circumference is $\dfrac{q_1}{\epsilon_{\rm o}}$ so the flux through arc $AB$ is $\dfrac{ \alpha }{\pi}\,\dfrac{q_1}{\epsilon_{\rm o}}$.
A similar analysis cab be done to show that the flux through area $A'B'$ is $\dfrac{ \beta}{\pi}\,\dfrac{q_2}{\epsilon_{\rm o}}$.

Equating the two fluxes will give a relationship between the two angles.

In three dimensions the analysis is a little trickier in part because you need to use the area of a cap of a sphere.

The area of a spherical cap is $A = 2 \pi r^2(1-\cos \,\alpha)$.

Knowing the area of the spherical cap the analysis is similar to the two-dimensional case except that now the Gaussian surface has an area of $4\pi r^2$.
You can thus find a relationship between the two angles.