Does the potential energy of a system consisting of two charged bodies depend on the reference point or the nature of force existing between them(attractive or repulsive)?
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Usually the potential energy of a set of particles is defined as the work done to bring them from infinity to their current positions.
If the particles repel, you will have to do some work to overcome the repulsion, so the potential energy is positive.
If they attract, then they actually do work for you as they come together (since they pull you along). In other words, the amount of work you have to do is negative in this case, so the potential energy is negative.
One slight complication is that anyone is free to add a constant number to this definition of potential energy, because experiments always measure an energy difference between two states, and any constant term will cancel out in the difference. (I assume that's what you mean by reference point?)
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$\begingroup$ Can we define the (change in)potential energy set of the particles at particular position as the work required to bring them from any arbitrary point to their current position? $\endgroup$– MrAPCommented Sep 11, 2016 at 18:38
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$\begingroup$ Yes. And if someone else chooses a different arbitrary point, that's equivalent to adding a constant to your definition. $\endgroup$– Paul GCommented Sep 11, 2016 at 21:24
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$\begingroup$ I could not understand the last part of your comment. $\endgroup$– MrAPCommented Sep 12, 2016 at 9:16
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1$\begingroup$ I was just pointing out that the freedom to choose an arbitrary starting point is the same as the freedom to add a constant to the potential. (Because, if somebody chooses a starting point that is different from yours, his work calculations will all differ from yours by a constant.) $\endgroup$– Paul GCommented Sep 12, 2016 at 15:13
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$\begingroup$ One more query. The potential energy of a set of particles is defined as the work required to bring them from infinity to their current positions. If the force between the particles is attractive, they do work on their own and no work is done by us. Then how are we doing negative work? $\endgroup$– MrAPCommented Sep 13, 2016 at 18:24