$2.3.38.$ Potential energy of electrostatic interaction of point charges $q$ and $Q$ located at a distance $r$ from each other, $U = \frac{kqQ}{r}$. Find the electrostatic force. Which charges have repulsion and which have attraction?
Let us find the derivative of the potential energy of interaction by distance $$\frac{dU}{dr}=\left(\frac{kqQ}{r}\right)'$$ $$dU=-\frac{kqQ}{r^2}dr\quad(1)$$ Also by the law of conservation of energy, the change in that energy is equal to the work to be done $$dU=dA\Rightarrow dU=-Fdr\quad(2)$$ NO: The larger $r$ is, the smaller $F$ is
According to Newton's third law $$\vec{F}_1=-\vec{F}_2=F$$ Substituting $(2)$ into $(1)$ leads to $$-\frac{kqQ}{r^2}dr=-Fdr$$ $$\boxed{F=\frac{kqQ}{r^2}}$$ When $qQ > 0$ — repulsion, when $qQ < 0$ — attraction.
$$F=\frac{kqQ}{r^2}$$
Almaskhan Arsen