When we calculate electric potential energy of a two particle system, say first I bring $+q_1$ and then I bring $+q_2$ against $q_1$'s electric field. Say I get that $q_2$ charge to a point $r$ distance away from $q_1$. Now potential $V_1$ at that point (where $q_2$ resides) due $q_1$ is $kq_1/r^2$ and hence the potential energy of $q_2$ is $q_2V_1= kq_1q_2/r$.
Similarly, at the point where $q_1$ resides, which is again $r$ distance away from $q_2$, the potential $V_2$ at that point due to $q_2$ is $kq_2/r^2$. Hence the potential energy of $q_1$ becomes $q_1V_2= kq_1q_2/r$.
So, the potential energy of both the charges are same and hence when we talk about the potential energy of the system, it should be $2(kq_1q_2/r)$. But in books it's written that the potential energy of a two particle system is $kq_1q_2/r$. Why is it not twice that?
I agree that the time when only charge $q_1$ arrived (with no other charge present) the potential energy of charge $q_1$ at that time will be zero but after the arrival of charge $q_2$ in the presence of $q_2$, then of course the potential of $q_2$ exists but at the same time potential of the charge $q_1$ will also exist as $q_2$ has its potential created at the point where $q_1$ resides, and hence $q_1$ would also have potential energy. And hence potential energy of the system should be the potential energy of $q_1$ and $q_2$ and it is $2(kq_1q_2/r)$, but why do books claim it is $kq_1q_2/r$?