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Given there are $2$ non-moving like charges in space. the total energy is the sum of potential energies plus kinetic (zero).

The potential energy of charge $A$ due to charge $B$ is $kq_1q_2/r$ and the potential energy of charge $B$ due to charge $A$ is the same, then is the total energy twice the value we have calculated or it is just $kq_1q_2/r$?

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It's just $$U=\frac{1}{4\pi \epsilon_0}\frac{q_1q_2}{r}$$ because the potential energy is associated with system not to individual charge.

In another word, we define the potential energy of a system of charges as the work necessary to bring them in from infinity. Thus We just have to do work in bringing $q_1$ near the $q_2$ or vice versa.

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You are basically asking the definition of Electrostatic potential energy. No you shouldn't add it twice the potential energy you have calculated already contains contributions from both the charges! Whenever you have conceptual issues like this, it is best to rethink how/what the definition is.

The electrostatic potential energy, $U_E$, of one point charge $q$ at position $r$ in the presence of an electric field $E$ is defined as the negative of the work $W$ done by the electrostatic force to bring it from the reference position to that position $r$. $$U_E(r)=-W_{r_{ref}{\to}r}=-\int_r^{r_{ref}}{qE(r').dr'}$$

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The potential energy is $k q_1 q_2 \over r$, or ${1 \over 4 \pi \epsilon_0} {q_1q_2 \over r}$ if you prefer: it does not 'belong' to A or B individually but to the joint AB system. You can't ascribe potential energy to individual objects in the same way that you can for kinetic energy.

(Even the gravitational potential energy of a ball is, strictly speaking, a joint property of the ball and the earth, though when the objects are of such different magnitudes you can get away with regarding the earth as fixed.)

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