$2.3.39.$ 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>
@@ -49,50 +49,53 @@
<h3>Solution</h3>
<p>
+
Expression for the potential energy
$$
−
u = \frac{kqQ}{r}
+
U = \frac{kqQ}{r}
$$
$$
−
u' = \frac{kqQ}{r} - \frac{kqQ}{R}
+
U' = \frac{kqQ}{r} - \frac{kqQ}{R}
$$
+
By statement, $R$ is a constant distance
$$
R = \text{const}
$$
−
+
Derivatives for potential energy of expressions
$$
−
(u_x)' = p
+
(U_x)' = p
$$
$$
−
(u)' = - \frac{kqQ}{r^2} \, dr
+
(U)' = - \frac{kqQ}{r^2} \, dr
$$
$$
−
(u')' = - \frac{kqQ}{r^2} \, dr
+
(U')' = - \frac{kqQ}{r^2} \, dr
$$
+
The derivative of the constant is zero
$$
(\text{const})' = 0 \Rightarrow (R)' = 0
$$
$$
−
(u)' = (u')'!
+
(U)' = (U')'!
$$
−
+
Thus, the speeds remain unchanged and the motion of the particles is not affected
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
$2.3.39.$ 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?
$2.3.39.$ 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>
</p>
@@ -49,50 +49,53 @@
<h3>Solution</h3>
<h3>Solution</h3>
<p>
<p>
Expression for the potential energy
$$
$$
u = \frac{kqQ}{r}
U = \frac{kqQ}{r}
$$
$$
$$
$$
u' = \frac{kqQ}{r} - \frac{kqQ}{R}
U' = \frac{kqQ}{r} - \frac{kqQ}{R}
$$
$$
By statement, $R$ is a constant distance
$$
$$
R = \text{const}
R = \text{const}
$$
$$
Derivatives for potential energy of expressions
$$
$$
(u_x)' = p
(U_x)' = p
$$
$$
$$
$$
(u)' = - \frac{kqQ}{r^2} \, dr
(U)' = - \frac{kqQ}{r^2} \, dr
$$
$$
$$
$$
(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
(\text{const})' = 0 \Rightarrow (R)' = 0
$$
$$
$$
$$
(u)' = (u')'!
(U)' = (U')'!
$$
$$
Thus, the speeds remain unchanged and the motion of the particles is not affected
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
