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<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>
−
− <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 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.
+ <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <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 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 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">
+
+ <h3>Solution</h3>
+ <p>
+ 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.
+ Applying Energy Conservation Law (from the beginning until $q$-particle reaches $Q$-sphere surface)
+ $$\frac{mv^2}{2} = \frac{mv_q^2}{2}+\frac{kQq}{R}+\frac{Mv_Q^2}{2} \;(1)$$
From Momentum Conservation Law
+ $$mv = mv_q + Mv_Q \;(2)$$
+ 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>
− $mv = mv_q + Mv_Q \;(2)$
− </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">
+ <p>
+ $$\Delta S_q = \Delta S_Q+2R = v_q~T \;(3)$$
And the sphere's displacement is
− </p>
− <center>
− $\Delta S_Q = v_Q~T \;(4)$
− </center>
− <p class="TxtSolutions">
+ $$\Delta S_Q = v_Q~T \;(4)$$
From $(4)$ in $(3)$, and separating $T$
− </p>
− <center>
− $T = \frac{2R}{v_q-v_Q} \;(5)$
− </center>
− <p class="TxtSolutions">
+ $$T = \frac{2R}{v_q-v_Q} \;(5)$$
Separating $v_q$ from $(2)$
− </p>
− <center>
− $v_q = v - \frac{M}{m}v_Q \;(6)$
− </center>
− <p class="TxtSolutions">
+ $$v_q = v - \frac{M}{m}v_Q \;(6)$$
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">
+ $$0 = \frac{M}{2}\left(1+\frac{M}{m}\right)v_Q^2-Mv~v_Q + \frac{kQq}{R} \;(7)$$
$(7)$ 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">
+ $$D = B^2-4AC$$
+ $$D = M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}$$
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">
+ $$X = \frac{-B\pm\sqrt{D}}{2A}$$
+ $$v_Q = \frac{Mv\pm\sqrt{M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}}}{M\left(1+\frac{M}{m}\right)} \;(8)$$
Substituting $(6)$ into $(5)$
− </p>
− <center>
− $T = \frac{2R}{v-v_Q\left(1+\frac{M}{m}\right)} \;(9)$
− </center>
− <p class="TxtSolutions">
+ $$T = \frac{2R}{v-v_Q\left(1+\frac{M}{m}\right)} \;(9)$$
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">
+ $$T = \frac{2MR}{\mp\sqrt{M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}}}$$
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">
+ $$\boxed{T = \frac{2R}{v}\left[1-\frac{qQ(M+m)}{2\pi\epsilon_0mMv^2}\right]^{-\frac{1}{2}}}$$
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>
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+ BSc. Luis Daniel Fernández Quintana<br>
+ Physics Department (FCNE)<br>
+ Universidad de Oriente, Cuba<br>
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