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if there are 2 charges generally potential energy of the system should be the sum of individual potential energy. but why the potential energy is half of the sum of every individual charge's potential energy.

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The potential energy possessed by the system of two charges is $$U=\dfrac{Q_1Q_2}{4\pi\varepsilon_0r}$$ You can rearrange the equation into$$U=\frac{1}{2}\dfrac{Q_2}{4\pi\varepsilon_0r}Q_1+\frac{1}{2}\dfrac{Q_1}{4\pi\varepsilon_0r}Q_2$$ which then becomes$$U=\frac{1}{2}Q_1V_1+\frac{1}{2}Q_2V_2$$ where $V_1$ and $V_2$ are respectively the potentials created by $Q_2$ at $Q_1$ and $Q_1$ at $Q_2$.Finally, this relation is valid for any number of charges( $U=\sum\frac{1}{2}Q_iV_i $ , with $V_i$ being the potential created at $Q_i$ by all other charges ) and can be proved in a similar way.

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    $\begingroup$ Stated another way: The potential energy comes from the interactions between particles, not from the particles themselves. What you count is the interactions. Two particles have one interaction, four have six, and $N$ have $\frac{1}{2}N(N-1)$. $\endgroup$
    – Endulum
    Apr 4 '16 at 17:27
  • $\begingroup$ ok is this correct that potential energy of every particle is Q*(potential created by all other charges) but when we find the total we should divide them by 2? $\endgroup$ Apr 4 '16 at 18:23
  • $\begingroup$ @SelvaratnamLavinan Yes this is completely true. Think of it this way: when you sum all the $Q_iV_i$'s ,then you are counting each interaction twice(i.e the interaction between some $Q_m$ and $Q_n$ is included in both terms $Q_mV_m$ and $Q_nV_n$), so you need in the end to divide the sum by $2$ to obtain the total energy. $\endgroup$
    – Tofi
    Apr 5 '16 at 3:55
  • $\begingroup$ Then a statement that total potential energy of a system is equal to the sum of every particle's potential energy would be a false one. am i right? $\endgroup$ Apr 5 '16 at 4:01
  • $\begingroup$ @SelvaratnamLavinan You are right. $\endgroup$
    – Tofi
    Apr 5 '16 at 8:06

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