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| <h3 id="back-link"><a href="../../#7.1">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#7.1">$\leftarrow$Back</a></h3> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <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 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. | | 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> |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <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. | | 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) | | Applying Energy Conservation Law (from the beginning until $q$-particle reaches $Q$-sphere surface) |
| $$\frac{mv^2}{2} = \frac{mv_q^2}{2}+\frac{kQq}{R}+\frac{Mv_Q^2}{2} \;(1)$$ | | $$\frac{mv^2}{2} = \frac{mv_q^2}{2}+\frac{kQq}{R}+\frac{Mv_Q^2}{2} \;(1)$$ |
| From Momentum Conservation Law | | From Momentum Conservation Law |
| $$mv = mv_q + Mv_Q \;(2)$$ | | $$mv = mv_q + Mv_Q \;(2)$$ |
| 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 | | 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> |
| <p> | | <p> |
| $$\Delta S_q = \Delta S_Q+2R = v_q~T \;(3)$$ | | $$\Delta S_q = \Delta S_Q+2R = v_q~T \;(3)$$ |
| And the sphere's displacement is | | And the sphere's displacement is |
| $$\Delta S_Q = v_Q~T \;(4)$$ | | $$\Delta S_Q = v_Q~T \;(4)$$ |
| From $(4)$ in $(3)$, and separating $T$ | | From $(4)$ in $(3)$, and separating $T$ |
| $$T = \frac{2R}{v_q-v_Q} \;(5)$$ | | $$T = \frac{2R}{v_q-v_Q} \;(5)$$ |
| Separating $v_q$ from $(2)$ | | Separating $v_q$ from $(2)$ |
| $$v_q = v - \frac{M}{m}v_Q \;(6)$$ | | $$v_q = v - \frac{M}{m}v_Q \;(6)$$ |
| Putting $(6)$ into $(1)$ and developing algebraically | | Putting $(6)$ into $(1)$ and developing algebraically |
| $$0 = \frac{M}{2}\left(1+\frac{M}{m}\right)v_Q^2-Mv~v_Q + \frac{kQq}{R} \;(7)$$ | | $$0 = \frac{M}{2}\left(1+\frac{M}{m}\right)v_Q^2-Mv~v_Q + \frac{kQq}{R} \;(7)$$ |
| $(7)$ 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 |
| $$D = B^2-4AC$$ | | $$D = B^2-4AC$$ |
| $$D = M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}$$ | | $$D = M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}$$ |
| The solutions are | | The solutions are |
| $$X = \frac{-B\pm\sqrt{D}}{2A}$$ | | $$X = \frac{-B\pm\sqrt{D}}{2A}$$ |
| $$v_Q = \frac{Mv\pm\sqrt{M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}}}{M\left(1+\frac{M}{m}\right)} \;(8)$$ | | $$v_Q = \frac{Mv\pm\sqrt{M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}}}{M\left(1+\frac{M}{m}\right)} \;(8)$$ |
| Substituting $(6)$ into $(5)$ | | Substituting $(6)$ into $(5)$ |
| $$T = \frac{2R}{v-v_Q\left(1+\frac{M}{m}\right)} \;(9)$$ | | $$T = \frac{2R}{v-v_Q\left(1+\frac{M}{m}\right)} \;(9)$$ |
| Putting $(8)$ into $(9)$ | | Putting $(8)$ into $(9)$ |
| $$T = \frac{2MR}{\mp\sqrt{M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}}}$$ | | $$T = \frac{2MR}{\mp\sqrt{M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}}}$$ |
| 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 |
| $$\boxed{T = \frac{2R}{v}\left[1-\frac{qQ(M+m)}{2\pi\epsilon_0mMv^2}\right]^{-\frac{1}{2}}}$$ | | $$\boxed{T = \frac{2R}{v}\left[1-\frac{qQ(M+m)}{2\pi\epsilon_0mMv^2}\right]^{-\frac{1}{2}}}$$ |
| 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> |
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| <p style="text-align: right; font-style: italic; font-size: 14;"> | | <p 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> |
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