
Let us assume at distance y, force is maximum.
Now, force at a distance y is given by, Fnet=2Fcosθ
So, Fnet=(y2+a2)232Kqq0y
For Fnet to be maximum, dydFnet=0
Or Kqq0(y2+a2)3(y2+a2)23−y23×2y(y2+a2)21=0
Simplifying, we get y=2a.
Hence, the value of x=2.