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Luisito edited Jun 14, 2024, 5:01 PM

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

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

		@@ -34,7 +34,7 @@
			<h4>Statement</h4>
			<p>
			$7.1.17^*$
		−	There are two diametrically opposed holes in a non-conductive thin-walled sphere with an uniformly distributed charge $Q$ which has mass $M$. At the initial istant, the spehere is in rest. Certain particle of mass $m$ and charge $q$ (same sign of $Q$) is moved from infinite with speed $v$ over line that join the holes. Determine the time which particle will be inside the sphere.
		+	There are two diametrically opposed holes in a non-conductive thin-walled sphere with an uniformly distributed charge $Q$ which has mass $M$. At the initial istant, the sphere is in rest. Certain particle of mass $m$ and charge $q$ (same sign of $Q$) is moved from infinite with speed $v$ over line that join the holes. Determine the time which particle will be inside the sphere.
			</p>
			<h4>Solution</h4>
			<p class="TxtSolutions">
			Firstly, we consider the situation before $q$-particle penetrates $Q$-sphere. Since both bodies have equal-sign charges, $q$-particle will move with a deacceleration while $Q$-sphere will accelerate.
			</p>
			<p class="TxtSolutions">
			Applying Energy Conservation Law (from the beginning until $q$-particle reaches $Q$-sphere surface)
			</p>
			<center>
			$\frac{mv^2}{2} = \frac{mv_q^2}{2}+\frac{kQq}{R}+\frac{Mv_Q^2}{2} \;(1)$
			</center>
			<p class="TxtSolutions">
			From Momentum Conservation Law
			</p>
			<center>
			$mv = mv_q + Mv_Q \;(2)$
			</center>
			<p class="TxtSolutions">
			Inside of $Q$-sphere, there is no electric field, then both bodies have constant velocities while $q$-particle be inside $Q$-sphere (see Figure). The displacement of $q$-particle is
			</p>
			<center>
			<figure>
			<img src="7.1.17.png"
			loading="lazy" alt="7.1.17" width="250" />
			<figcaption>
			Particle motion inside the sphere
			</figcaption>
			</figure>
			</center>
			<center>
			$\Delta S_q = \Delta S_Q+2R = v_q~T \;(3)$
			</center>
			<p class="TxtSolutions">
			And the sphere's displacement is
			</p>
			<center>
			$\Delta S_Q = v_Q~T \;(4)$
			</center>
			<p class="TxtSolutions">
			From $(4)$ in $(3)$, and separating $T$
			</p>
			<center>
			$T = \frac{2R}{v_q-v_Q} \;(5)$
			</center>
			<p class="TxtSolutions">
			Separating $v_q$ from $(2)$
			</p>
			<center>
			$v_q = v - \frac{M}{m}v_Q \;(6)$
			</center>
			<p class="TxtSolutions">
			Putting $(6)$ into $(1)$ and developing algebraically
			</p>
			<center>
			$0 = \frac{M}{2}\left(1+\frac{M}{m}\right)v_Q^2-Mv~v_Q + \frac{kQq}{R} \;(7)$
			</center>
			<p class="TxtSolutions">
			$(7)$ is a second-degree equation or quadratic equation in $v_Q$ ($AX^2+BX+C=0$). We can solve this by Discriminant method or General-Solution formula for quadratic equations. The Discriminant is
			</p>
			<center>
			$D = B^2-4AC$
			</center>
			<center>
			$D = M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}$
			</center>
			<p class="TxtSolutions">
			The solutions are
			</p>
			<center>
			$X = \frac{-B\pm\sqrt{D}}{2A}$
			</center>
			<center>
			$v_Q = \frac{Mv\pm\sqrt{M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}}}{M\left(1+\frac{M}{m}\right)} \;(8)$
			</center>
			<p class="TxtSolutions">
			Substituting $(6)$ into $(5)$
			</p>
			<center>
			$T = \frac{2R}{v-v_Q\left(1+\frac{M}{m}\right)} \;(9)$
			</center>
			<p class="TxtSolutions">
			Putting $(8)$ into $(9)$
			</p>
			<center>
			$T = \frac{2MR}{\mp\sqrt{M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}}}$
			</center>
			<p class="TxtSolutions">
			Choosing ``plus'' (+) sign, because time is non-negative, taking $k = 1/(4\pi\epsilon_0)$ and developing
			</p>
			<center>
			$\boxed{T = \frac{2R}{v}\left[1-\frac{qQ(M+m)}{2\pi\epsilon_0mMv^2}\right]^{-\frac{1}{2}}}$
			</center>
			<p class="TxtSolutions">
			This is valid for $\frac{\mu v^2}{2}>\frac{qQ}{4\pi\epsilon_0R}$, where $\mu = \frac{mM}{M+m}$ is the reduced mass.
			</p>
			<p class="TxtSolutions" style="text-align: right; font-style: italic; font-size: 14;">
			BSc. Luis Daniel Fernández Quintana<br>
			Physics Department (FCNE)<br>
			Universidad de Oriente, Cuba<br>
			</p>
			</article>
			</main>

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	<body id="top">		<body id="top">
	<header>		<header>
	<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>

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

	@@ -34,7 +34,7 @@
	<h4>Statement</h4>		<h4>Statement</h4>
	<p>		<p>
	$7.1.17^*$		$7.1.17^*$
	There are two diametrically opposed holes in a non-conductive thin-walled sphere with an uniformly distributed charge $Q$ which has mass $M$. At the initial istant, the spehere is in rest. Certain particle of mass $m$ and charge $q$ (same sign of $Q$) is moved from infinite with speed $v$ over line that join the holes. Determine the time which particle will be inside the sphere.		There are two diametrically opposed holes in a non-conductive thin-walled sphere with an uniformly distributed charge $Q$ which has mass $M$. At the initial istant, the sphere is in rest. Certain particle of mass $m$ and charge $q$ (same sign of $Q$) is moved from infinite with speed $v$ over line that join the holes. Determine the time which particle will be inside the sphere.
	</p>		</p>
	<h4>Solution</h4>		<h4>Solution</h4>
	<p class="TxtSolutions">		<p class="TxtSolutions">
	Firstly, we consider the situation before $q$-particle penetrates $Q$-sphere. Since both bodies have equal-sign charges, $q$-particle will move with a deacceleration while $Q$-sphere will accelerate.		Firstly, we consider the situation before $q$-particle penetrates $Q$-sphere. Since both bodies have equal-sign charges, $q$-particle will move with a deacceleration while $Q$-sphere will accelerate.
	</p>		</p>
	<p class="TxtSolutions">		<p class="TxtSolutions">
	Applying Energy Conservation Law (from the beginning until $q$-particle reaches $Q$-sphere surface)		Applying Energy Conservation Law (from the beginning until $q$-particle reaches $Q$-sphere surface)
	</p>		</p>
	<center>		<center>
	$\frac{mv^2}{2} = \frac{mv_q^2}{2}+\frac{kQq}{R}+\frac{Mv_Q^2}{2} \;(1)$		$\frac{mv^2}{2} = \frac{mv_q^2}{2}+\frac{kQq}{R}+\frac{Mv_Q^2}{2} \;(1)$
	</center>		</center>
	<p class="TxtSolutions">		<p class="TxtSolutions">
	From Momentum Conservation Law		From Momentum Conservation Law
	</p>		</p>
	<center>		<center>
	$mv = mv_q + Mv_Q \;(2)$		$mv = mv_q + Mv_Q \;(2)$
	</center>		</center>
	<p class="TxtSolutions">		<p class="TxtSolutions">
	Inside of $Q$-sphere, there is no electric field, then both bodies have constant velocities while $q$-particle be inside $Q$-sphere (see Figure). The displacement of $q$-particle is		Inside of $Q$-sphere, there is no electric field, then both bodies have constant velocities while $q$-particle be inside $Q$-sphere (see Figure). The displacement of $q$-particle is
	</p>		</p>
	<center>		<center>
	<figure>		<figure>
	<img src="7.1.17.png"		<img src="7.1.17.png"
	loading="lazy" alt="7.1.17" width="250" />		loading="lazy" alt="7.1.17" width="250" />
	<figcaption>		<figcaption>
	Particle motion inside the sphere		Particle motion inside the sphere
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>
	<center>		<center>
	$\Delta S_q = \Delta S_Q+2R = v_q~T \;(3)$		$\Delta S_q = \Delta S_Q+2R = v_q~T \;(3)$
	</center>		</center>
	<p class="TxtSolutions">		<p class="TxtSolutions">
	And the sphere's displacement is		And the sphere's displacement is
	</p>		</p>
	<center>		<center>
	$\Delta S_Q = v_Q~T \;(4)$		$\Delta S_Q = v_Q~T \;(4)$
	</center>		</center>
	<p class="TxtSolutions">		<p class="TxtSolutions">
	From $(4)$ in $(3)$, and separating $T$		From $(4)$ in $(3)$, and separating $T$
	</p>		</p>
	<center>		<center>
	$T = \frac{2R}{v_q-v_Q} \;(5)$		$T = \frac{2R}{v_q-v_Q} \;(5)$
	</center>		</center>
	<p class="TxtSolutions">		<p class="TxtSolutions">
	Separating $v_q$ from $(2)$		Separating $v_q$ from $(2)$
	</p>		</p>
	<center>		<center>
	$v_q = v - \frac{M}{m}v_Q \;(6)$		$v_q = v - \frac{M}{m}v_Q \;(6)$
	</center>		</center>
	<p class="TxtSolutions">		<p class="TxtSolutions">
	Putting $(6)$ into $(1)$ and developing algebraically		Putting $(6)$ into $(1)$ and developing algebraically
	</p>		</p>
	<center>		<center>
	$0 = \frac{M}{2}\left(1+\frac{M}{m}\right)v_Q^2-Mv~v_Q + \frac{kQq}{R} \;(7)$		$0 = \frac{M}{2}\left(1+\frac{M}{m}\right)v_Q^2-Mv~v_Q + \frac{kQq}{R} \;(7)$
	</center>		</center>
	<p class="TxtSolutions">		<p class="TxtSolutions">
	$(7)$ is a second-degree equation or quadratic equation in $v_Q$ ($AX^2+BX+C=0$). We can solve this by Discriminant method or General-Solution formula for quadratic equations. The Discriminant is		$(7)$ is a second-degree equation or quadratic equation in $v_Q$ ($AX^2+BX+C=0$). We can solve this by Discriminant method or General-Solution formula for quadratic equations. The Discriminant is
	</p>		</p>
	<center>		<center>
	$D = B^2-4AC$		$D = B^2-4AC$
	</center>		</center>
	<center>		<center>
	$D = M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}$		$D = M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}$
	</center>		</center>
	<p class="TxtSolutions">		<p class="TxtSolutions">
	The solutions are		The solutions are
	</p>		</p>
	<center>		<center>
	$X = \frac{-B\pm\sqrt{D}}{2A}$		$X = \frac{-B\pm\sqrt{D}}{2A}$
	</center>		</center>
	<center>		<center>
	$v_Q = \frac{Mv\pm\sqrt{M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}}}{M\left(1+\frac{M}{m}\right)} \;(8)$		$v_Q = \frac{Mv\pm\sqrt{M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}}}{M\left(1+\frac{M}{m}\right)} \;(8)$
	</center>		</center>
	<p class="TxtSolutions">		<p class="TxtSolutions">
	Substituting $(6)$ into $(5)$		Substituting $(6)$ into $(5)$
	</p>		</p>
	<center>		<center>
	$T = \frac{2R}{v-v_Q\left(1+\frac{M}{m}\right)} \;(9)$		$T = \frac{2R}{v-v_Q\left(1+\frac{M}{m}\right)} \;(9)$
	</center>		</center>
	<p class="TxtSolutions">		<p class="TxtSolutions">
	Putting $(8)$ into $(9)$		Putting $(8)$ into $(9)$
	</p>		</p>
	<center>		<center>
	$T = \frac{2MR}{\mp\sqrt{M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}}}$		$T = \frac{2MR}{\mp\sqrt{M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}}}$
	</center>		</center>
	<p class="TxtSolutions">		<p class="TxtSolutions">
	Choosing ``plus'' (+) sign, because time is non-negative, taking $k = 1/(4\pi\epsilon_0)$ and developing		Choosing ``plus'' (+) sign, because time is non-negative, taking $k = 1/(4\pi\epsilon_0)$ and developing
	</p>		</p>
	<center>		<center>
	$\boxed{T = \frac{2R}{v}\left[1-\frac{qQ(M+m)}{2\pi\epsilon_0mMv^2}\right]^{-\frac{1}{2}}}$		$\boxed{T = \frac{2R}{v}\left[1-\frac{qQ(M+m)}{2\pi\epsilon_0mMv^2}\right]^{-\frac{1}{2}}}$
	</center>		</center>
	<p class="TxtSolutions">		<p class="TxtSolutions">
	This is valid for $\frac{\mu v^2}{2}>\frac{qQ}{4\pi\epsilon_0R}$, where $\mu = \frac{mM}{M+m}$ is the reduced mass.		This is valid for $\frac{\mu v^2}{2}>\frac{qQ}{4\pi\epsilon_0R}$, where $\mu = \frac{mM}{M+m}$ is the reduced mass.
	</p>		</p>
	<p class="TxtSolutions" style="text-align: right; font-style: italic; font-size: 14;">		<p class="TxtSolutions" style="text-align: right; font-style: italic; font-size: 14;">
	BSc. Luis Daniel Fernández Quintana<br>		BSc. Luis Daniel Fernández Quintana<br>
	Physics Department (FCNE)<br>		Physics Department (FCNE)<br>
	Universidad de Oriente, Cuba<br>		Universidad de Oriente, Cuba<br>
	</p>		</p>
	</article>		</article>
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