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?

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

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    <h3> Statement </h3>
    <p>
        $2.3.38.$ 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?
    </p>

    <h3>Solution</h3>
    <p>
        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.

    <h4>Answer</h4>
    <p>
        $$F=\frac{kqQ}{r^2}$$
    </p>
    <p style="text-align: right; font-style: italic; font-size: 14;">
      Almaskhan Arsen<br>
    </p>


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