Fixed 2.3.39

astrosander правка от 30 авг. 2024 г., 13:21

правка #2506 предыдущая #2498 GitHub 39257f5 ← раньше позже →

en/2.3.39.md +2 −2

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			<h2>Solutions of Savchenko Physics Textbook</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>
			$2.3.39.$ 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>
			$$
			u = \frac{kqQ}{r}
			$$

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

			$$
			R = \text{const}
			$$

			$$
			(u_x)' = p
		@@ -66,7 +66,7 @@
			$$

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

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

			$$
			(\text{const})' = 0 \Rightarrow (R)' = 0
		@@ -78,7 +78,7 @@
			$$

			$$
		−	(u')' = (u')'!
		+	(u)' = (u')'!
			$$

			$$
			u = u'!
			$$

			$$
			\text{No, it won't!}
			$$
			</p>

			<h4>Answer</h4>
			<p>
			No.
			</p>

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


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	<h2>Solutions of Savchenko Physics Textbook</h2>		<h2>Solutions of Savchenko Physics Textbook</h2>
	<p class="author">		<p class="author">
	Aliaksandr Melnichenka <br/>		Aliaksandr Melnichenka <br/>
	October 2023		October 2023
	</p>		</p>
	</header>		</header>

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

	<h3> Statement </h3>		<h3> Statement </h3>
	<p>		<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?		$2.3.39.$ 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>

	<h3>Solution</h3>		<h3>Solution</h3>
	<p>		<p>
	$$		$$
	u = \frac{kqQ}{r}		u = \frac{kqQ}{r}
	$$		$$

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

	$$		$$
	R = \text{const}		R = \text{const}
	$$		$$

	$$		$$
	(u_x)' = p		(u_x)' = p
	@@ -66,7 +66,7 @@
	$$		$$

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

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

	$$		$$
	(\text{const})' = 0 \Rightarrow (R)' = 0		(\text{const})' = 0 \Rightarrow (R)' = 0
	@@ -78,7 +78,7 @@
	$$		$$

	$$		$$
	(u')' = (u')'!		(u)' = (u')'!
	$$		$$

	$$		$$
	u = u'!		u = u'!
	$$		$$

	$$		$$
	\text{No, it won't!}		\text{No, it won't!}
	$$		$$
	</p>		</p>

	<h4>Answer</h4>		<h4>Answer</h4>
	<p>		<p>
	No.		No.
	</p>		</p>

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


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