If we have two particles (let's just say of equal but opposite charges, $-q$ and $q$) and place them a distance $r$ apart, the potential energy of the system is:

$$ U = k \frac{q^2}{r} $$

Now if we let time continue, the particles start to accelerate toward one another (and the acceleration itself is also increasing, but regardless).

We can find the kinetic energy as

$$ K = \frac{1}{2} mv^2 $$

And the potential energy increase as time goes on, because $r$ (the distance between the particles) decreases over time.

So both $K$ and $U$ are increasing. Where is this energy coming from?

My current thought process is this:

We get the formula for electrical potential energy by integrating the formula for electric force with respect to $r$:

$$ U = kq^2 \int \frac{1}{r^2} dr = -k \frac{q^2}{r} $$

Therefore potential energy is negative and is decreasing over time. But I can't find a negative sign for potential energy in any formulas I've come across.

And anyway, if the charges were repulsive, this would still not conserve energy, as potential energy would increase again.

  • 1
    $\begingroup$ Take a look at the correct formula. When the two charges have opposite sign, their potential energy is negative. $\endgroup$
    – Ghoster
    Oct 5, 2023 at 20:48
  • $\begingroup$ Double check your expression for $U$. $\endgroup$ Oct 5, 2023 at 20:49
  • $\begingroup$ Where did you find your incorrect formula? $\endgroup$
    – Ghoster
    Oct 5, 2023 at 22:13
  • $\begingroup$ @Ghoster I was just under the impression that the sign of each charge doesn't matter, because of how we use the Coulomb's law. But it now makes sense; if the charges are equal, potential energy must act differently if the charges are opposite. $\endgroup$ Oct 6, 2023 at 1:54
  • $\begingroup$ the sign of each charge doesn't matter, because of how we use the Coulomb's law When I use Coulomb’s Law, the sign of the charges matters a lot: Same-sign charges repel, while opposite-sign charges attract. This repulsion or attraction leads to positive or negative electrostatic potential energy. $\endgroup$
    – Ghoster
    Oct 6, 2023 at 5:21

1 Answer 1


The correct definition of the potential energy is $$U=k\frac{q_1 q_2}{r}$$ For equal and opposite charges this becomes $$U=-k\frac{q^2}{r}$$ Thus as KE increases PE decreases.


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