In all the problems, which involve distribution of charge, we choose an element of charge dq to find the element of the field dE produced at the given location. Then we sum all such dEs to find the total field £ at that location.
We must note the symmetry of the situation. For each element dq located at positive x-axis. There is a similar dq located at the same negative value of x. The dE_x produced by one dq is cancelled by the dE_x in the opposite direction due to the other dq. Hence, all the dE_xcomponents add to zero. So, we need to sum only the dE_y components, a scalar sum since they all point in the same direction. The element charge is dq = λ dx

The integral on the right hand side can be evaluated by substituting x =  and dx = r sec² α d α

Let the charge Q be kept along the line joining two charges such that it is at a distance x from q and at a distance (d-x) from 2q.
For the net force on q and 2q to be 0 the system of charges should be in equilibrium.

Force acting on Q due to q= $\frac{1}{4{\mathrm{\pi \epsilon }}_{\circ }}\frac{qQ}{x^2}$
Force acting on Q du to 2q= $\frac{1}{4{\mathrm{\pi \epsilon }}_{\circ }}\frac{2qQ}{(d-x{)}^2}$
For, the system to be in equilibrium 

$$\begin{gathered} \frac{1}{4{\mathrm{\pi \epsilon }}_{\circ }}\frac{qQ}{x^2} = \frac{1}{4\pi {\epsilon }_{\circ }}\frac{2qQ}{(d-x{)}^2} \\ \Rightarrow \frac{1}{x^2}=\frac{2}{(d-x{)}^2} \\ \Rightarrow 2x^2 =(d-x{)}^2 \\ \Rightarrow \sqrt{2}x=(d-x) \\ \Rightarrow x=\frac{d}{(\sqrt{2}+1)} \end{gathered}$$

 from the two charges P and Q respectively, as shown in the figure below:

For equilibrium of q, the forces on it exerted by P and Q must be co-linear, equal and opposite.
Force on q by P
       
Force on q by Q
       towards Q
               
                     
Hence charge q should be equidistant from P and Q.
For the system to be in equilibrium, the charges P and Q must also be in equilibrium. Now,

Since F_pq and F_PQ are oppositely directed along the same line, we have, for equilibrium,
             
or                   
Similarly for the equilibrium of Q, we would get
 Thus 
in magnitude of either charge P or Q.
Stability: A slight displacement of q towards P increases the magnitude of F_qP and decreases the magnitude of F_qQ. Consequently, the displacement of q is increased. Thus the three charges are no longer in equilibrium. Hence the original equilibrium is unstable for displacement along the axis on which the charges are located. For a displacement of q along a direction normal to the line PQ, the resultant of the two forces of attraction F_qp and F_qQ will bring the charge q back to its original position. Thus the equilibrium is stable for displacement in the vertical direction.