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+ <h2>Solutions of Savchenko Problems in Physics</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../../#2.3">$\leftarrow$Back</a></h3>
+
+ <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
+$$dU=dA\Rightarrow dU=-Fdr\quad(2)$$
+<i>NO: The larger $r$ is, the smaller $F$ is</i>
+</p>
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+ <figure>
+ <img src="draw.png"
+ loading="lazy" width="270" />
+ <figcaption>
+ Forces acting on the system
+ </figcaption>
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+<p>
+
+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.
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$F=\frac{kqQ}{r^2}$$
+ </p>
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