# Distribution of Potential and Kinetic energies of a system consisting of two charges

Say we have two opposite charges with charge $$+q_1$$ and $$-q_2$$, and we'll refer to them by $$+q_1$$ and $$-q_2$$. Originally, they are separated at a distance of $$d$$. Now while keeping only $$-q_2$$ at rest, $$+q_1$$ moves closer to to $$-q_2$$ at a new distance $$d'$$, and the potential energy of the system decrease by $$\Delta U = \frac{(+q_1)(-q_2)}{4\pi\epsilon_0}\left( \frac{1}{d'}-\frac{1}{d} \right) < 0$$. This decrease causes an increase in the kinetic energy of $$+q_1$$ (the sum of the kinetic energies of the system, but for $$-q_2$$ is $$0$$). Now if we look at it mathematically, this simply follows from the work-energy theorem and the fact that the formula $$\Delta U =U_b - U_a= \frac{1}{4\pi\epsilon_0}Q_1Q_2\left(\frac{1}{r_b}-\frac{1}{r_a}\right)$$ is derived when when of the charges $$Q_1$$ or $$Q_2$$ is assumed to be at rest.

However, in my book, it is also stated that when we release both $$+q_1$$ and $$-q_2$$ from a distance $$d$$ to a distance $$d'$$, the sum of the kinetic energy (which is $$\Delta K$$ since originally they are both at rest) both of which are non-zero, is equal to the (negative) change in potential energy. This is intuitively true, but I am looking for a mathematical basis.

## 2 Answers

Model the system of the two charges as an isolated one. There is no external work done on the system, and hence the appropriate reduction of the coservation of energy yields: $$\Delta K + \Delta U_E = \underbrace{W_{\text{ext}}}_\text{= 0} \rightarrow \Delta K + \Delta U_E = 0$$ and hence,$$(K_1- 0) + (K_2-0) = -(U_\text{f}-U_\text{i}) \rightarrow \boxed{K_1 + K_2 = -\Delta U_E}$$

If you intend to only use the work-kinetic energy theorem, then use the fact that the electric force is internal to the same system, and therfore: $$W_\text{int} = -\Delta U_E$$ Then, applying the work-kinetic energy theorem gives: $$W = \Delta K \rightarrow -\Delta U_E = \Delta K$$ and consequently: $$\boxed{K_1 + K_2 = -\Delta U_E}$$

Note that both approaches neglect the gravitational force in between the two charges. This assumption is justified because the electric force is often much stronger than the gravitational force between them.

Hope this helps.

• Ok, I understood everything. But how do we prove that $\Delta U_E = \frac{1}{4\pi\epsilon_0}Q_1Q_2(\frac{1}{d'}-\frac{1}{d})$? As far as I know, the proof of that equation requires that one of the charges is at rest. Jan 4, 2022 at 7:02
• @Adola, Why do you think that non-relativistic motion would affect the potential energy function $U(r) = \frac{k_eq_1q_2}{r_{12}}$? Potential energy is energy stored in a system by virtue of it's configuration, and therby independent of it's motion (which would affect the kinetic energy). Jan 4, 2022 at 7:24
• I do understand that the potential energy function is a function of configuration, but only when one of the charges is at rest as requested by the derivation of the formula. I also (but I am not too sure, I haven't checked manually for a specific case) believe that there is mathematical proof for the question since the fact that the electrical force is conservative is from a mathematical perspective assuming Coulumb's law. Jan 4, 2022 at 8:52
• @Adola, From what I've learnt, the potential energy of a system is a state function, due to it being defined only when conservative forces act, unlike kinetic energy. As for the mathematical proof of it being invariant, I can't really help, because I do not know it. My apologies. Jan 4, 2022 at 17:38

Strictly speaking, the equation $$\Delta U_E = \frac{1}{4\pi\epsilon_0}Q_1Q_2(\frac{1}{d'}-\frac{1}{d})$$ holds for electrostatic potential energy, so this only makes sense when both charges are static. Now, is it usually assumed in these derivations that the charges move so slowly that a quasistatic approximation holds, and therefore you can still use the same equation.

As for a mathematical basis, you can get the equation by computing the line integral for the energy of a system of two charges, check this.