Trajectory of electric field lines I came across an interesting problem of electric field which is as follows:

An electric field line from a charge $q$ is as shown. It enters a negative charge $–q$ as shown. Find the maximum height of the field line from the $x$-axis.


I know the tangent to electric field lines at any point gives the direction of the electric field.  By symmetry, the highest point in the trajectory should be the midpoint of the line joining the charges. But I have no idea how to calculate the maximum height. I could not figure out the use of the 60° angle given as at those points magnitude of electric field due to the nearer charge would be tending to infinity.
Any help would be greatly appreciated!
 A: Contrary to the comments, this problem can be solved analytically without the use of differential equations.  Since this is a "homework-like" problem, but one requiring an unusual technique, I will describe the method in a bit more detail than I would otherwise.  I will still leave the details of the calculations to you.
Let $x = 0$ denote the midpoint of the charges.  Consider a closed surface consisting of three parts:

*

*A small spherical cap centered at $+q$ with radius $\epsilon$;

*A circle in the plane $x = 0$, of radius $h$;  and

*The surface of revolution obtained by rotating the given electric field line (for $x < 0$) about the $x$-axis.


There is no charge enclosed by this surface, and by definition there is no flux through portion #3 of the surface since it follows the field lines. So by Gauss's Law we must have
$$
\iint_1 \mathbf{E} \cdot d \mathbf{a} + \iint_2 \mathbf{E} \cdot d \mathbf{a} = 0.
$$
In the limit $\epsilon \to 0$, the first integral will be dominated by the inward flux from the $+q$ charge, with the flux due to the $-q$ charge becoming negligible;  so this flux can be calculated easily.  The outward flux through surface #2 can also be calculated in terms of the unknown height $h$.  Here we do have to take the effects of both charges into account;  however, it is not hard to see that the fluxes through surface #2 due to the $+q$ charge and the $-q$ charge are equal, so that simplifies matters somewhat.
Take it from there.
A: Since the field line indicates the direction  of the field, The field at each point must be tangent to the line.  In an xy system, dy/dx = $E_y/E_x$.  Putting the (+q) at the origin of an xy system you can use Coulomb's law the find an expression for the resultant $E_y$ and $E_x$ as a function of (x) and (y). That gives you (dy) as a function of (dx).  I was not able to separate variables in order to do the integration. However, with the given numbers, this situation lends itself to a numeric simulation.  Start with a very small x and choose a y that puts the resulting point at 60 degrees from the line joining the charges. To simplify your expressions, at each step, calculate the denominators, then the field components, and then the slope. After the first two steps,  I adjusted the slope by adding half the increment from the previous two.  After 80 steps that brings the curve into the (-q) at 60 degrees plus a very small fraction. My maximum (y) rounds down to 2.
