In this case it is possible to choose a simple coordinate frame:
The electric field is given by 
and the electric potential by 
.
Because we are in a bidimensional world, we will have equipotential
lines instead of surfaces.
The equation of lines of force contains now a single equality (because
the third coordinate z is not of concern) and can be re-arranged as:
.
This equations turns out to have the solution
For different values of the parameter C the equation describes all the
field lines. The physical meaning of this expression will be given later.
For y=0, 
The two factors that multiply q1 and q2 are precisely
Here a1 and a2
are the angles that describe the polar position of the target point relative
to q1 and q2. Consequently, the two factors should be between -1 and +1,
so that the parameter C must satisfy the inequation:
.
For other C's, the equation has no solution (the parameter does not describe
any line of force).
Inequation (2) is necessary, but not sufficient. The angles a1
and a2 must satisfy 
,
because APB is a triangle (a degenerated one maybe). It can be shown that
extremal values for C are obtained for points of AB; therefore the two
limits of C are the minimum and maximum value in the set { - q1 - q2 ,
q1 - q2, q1 + q2 }
a) If the two charges have like signs, the two extremal values for C are obtained
b) If the two charges have opposite signs, one limit will be on segment AB.
Looking at the images in the first paragraph one can see that
The question answered here is: what are the lines of force that pass through a given charge ? Refer to equation (1) again: since lines of force are controlled by the parameter C, we have to find a range for C.
For a line of force that passes through charge qk , the restriction of function f to that path is continuous in the neighborhood of the point (xk, 0), where qk lies; consequently, on that path the limit.
or, substituting
Take k=1 (for the other charge the results will be similar). Then xk =
-a; the second term poses no problem and can be taken out of the limit,
replacing x by -a and y by 0:
Remembering that the expression under the limit is a cosine, we get the
inequality 
, which
can be broken into 
.
At least one of these conditions will be satisfied automatically (it will
be equivalent to one of the inequalities in the validity inequation (2)).
The conditions for lines of force to pass through q2 are analogous:
The inequalities may or may not be strict: taking the equal sign, some
of them yield to reunions of singular paths.
To study the field lines further, we have to examine each case separately.
a) charges of same sign and different magnitudes: q1 > 0, q2 > 0, |q1| < |q2|.
The conditions become:
The separation between the two families is 
.
For y=0, 
So segment AB is part of the limit curve; the other part is a curve which passes between q1 and q2. Since it only crosses line AB once (within segment AB) and since field lines are infinite, the separation is an infinite curve, bending to the side of q1 (the smallest charge in terms of magnitude).
The two parts of the separation cross where the differential equation
has a singular point, that is at the point where E=0. From the equation
E=0 we can compute the position of this singular point: 
(closer to the smallest charge)
b) charges of opposite signs and different magnitudes: q1 > 0, q2 < 0, |q1| < |q2|
(C must be "large enough")
(all field lines pass through the charge of greatest magnitude)The separation is given by 
;
for y = 0, 
which is the ray starting at B and running away from A. The other part
of the separation is a bounded curve enclosing both charges (because
all field lines passing through q1 are bounded). The two parts cross at
a point where E=0
(outside segment
AB, closer to A, where the smallest charge lies)
c) charges of same sign and same magnitudes, q1 = q2 = q > 0
The configuration is symmetric about the vertical axis.
,
which is segment AB and the perpendicular through its midpoint O. In ,
E becomes zero.d) charges of opposite signs and same magnitudes
q1 = q > 0, q2 = -q < 0
