One research institute decided to use an expression for the potential energy of point charges in the form U' = rac{kqQ}{r} - rac{kqQ}{R}, where R is a constant distance set once and for all. Will using U' instead of U = rac{kqQ}{r} affect the results of particle motion calculations?

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/#2.3">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $2.3.39.$ One research institute decided to use an expression for the potential energy of point charges in the form $U' = \frac{kqQ}{r} - \frac{kqQ}{R}$, where $R$ is a constant distance set once and for all. Will using $U'$ instead of $U = \frac{kqQ}{r}$ affect the results of particle motion calculations?
    </p>

    <h3>Solution</h3>
    <p>
        Expression for the potential energy
        $$

U = \frac{kqQ}{r}
$$

$$U' = \frac{kqQ}{r} - \frac{kqQ}{R}$$

By statement, $R$ is a constant distance
$$R = \text{const}$$
Derivatives for potential energy of expressions
$$(U_x)' = p$$

$$(U)' = - \frac{kqQ}{r^2} \, dr$$

$$(U')' = - \frac{kqQ}{r^2} \, dr$$

The derivative of the constant is zero
$$(\text{const})' = 0 \Rightarrow (R)' = 0$$

$$(U)' = (U')'!$$
Thus, the speeds remain unchanged and the motion of the particles is not affected
$$V = V'!$$

    </p>

    <h4>Answer</h4>
    <p>
        $

\text{No, it won't!}
$

    <p style="text-align: right; font-style: italic; font-size: 14;">   
      Almaskhan Arsen<br>
    </p>


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