Say we have charge $Q_1$ and charge $Q_2$, both exerting a potential $V_1$ and $V_2$ respectively, at a separation $r$.

The interaction energy is $\frac{Q_1 Q_2}{4\pi \epsilon_0 r}$.

The energy is als $qV$ in general, shouldn't it be $Q_1 V_2 + Q_2 V_1$ here, ending up with a factor of $2$ to the expression above?


The energy of a system isn't the sum of the individual energies, ie. energy is not extensive in general. The energy of the system is the energy required to create this system out of nothing. To do it, you can bring the first charge from nowhere to its position: it cost you nothing, since there is no potential at that time in space, except the potential it creates. No, you can bring the second charge, from infinity to its position: it costs you $Q_2 V_1(r) = \frac{Q_1Q_2}{4\pi\epsilon_0 r}$.

In general, you can show that for any given $Q_1,\dots,Q_n$ charges, then the energy of the system is

$$ E = \frac{1}{2}\sum\limits_{i=1}^{n} Q_iV_i$$

where $V_i$ is the potential created at the position of $Q_i$ by all the other charges. To do this, consider the result true with $n$ charges, and show that it is still true with $n+1$ charges.

  • $\begingroup$ Great yes sorry forgot how this went $\endgroup$ – SuperCiocia May 14 '17 at 22:36

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