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  <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.
    </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>