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.
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<h3> Statement </h3>
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$7.1.17^*$
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.
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<h3>Solution</h3>
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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)
$$\frac{mv^2}{2} = \frac{mv_q^2}{2}+\frac{kQq}{R}+\frac{Mv_Q^2}{2} \quad(1)$$
From Momentum Conservation Law
$$mv = mv_q + Mv_Q \quad(2)$$
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
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<img src="7.1.17.png"
loading="lazy" alt="7.1.17" width="250" />
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Particle motion inside the sphere
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<p>
$$\Delta S_q = \Delta S_Q+2R = v_q\~T \quad(3)$$
And the sphere's displacement is
$$\Delta S_Q = v_Q\~T \quad(4)$$
From $(4)$ in $(3)$, and separating $T$
$$T = \frac{2R}{v_q-v_Q} \quad(5)$$
Separating $v_q$ from $(2)$
$$v_q = v - \frac{M}{m}v_Q \quad(6)$$
Putting $(6)$ into $(1)$ and developing algebraically
$$0 = \frac{M}{2}\left(1+\frac{M}{m}\right)v_Q^2-Mv\~v_Q + \frac{kQq}{R} \quad(7)$$
$(7)$ 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
$$D = B^2-4AC$$
$$D = M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}$$
The solutions are
$$X = \frac{-B\pm\sqrt{D}}{2A}$$
$$v_Q = \frac{Mv\pm\sqrt{M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}}}{M\left(1+\frac{M}{m}\right)} \quad(8)$$
Substituting $(6)$ into $(5)$
$$T = \frac{2R}{v-v_Q\left(1+\frac{M}{m}\right)} \quad(9)$$
Putting $(8)$ into $(9)$
$$T = \frac{2MR}{\mp\sqrt{M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}}}$$
Choosing ``plus'' (+) sign, because time is non-negative, taking $k = 1/(4\pi\varepsilon_0)$ and developing
$$\boxed{T = \frac{2R}{v}\left[1-\frac{qQ(M+m)}{2\pi\varepsilon_0mMv^2}\right]^{-\frac{1}{2}}}$$
This is valid for $\frac{\mu v^2}{2}>\frac{qQ}{4\pi\varepsilon_0R}$, where $\mu = \frac{mM}{M+m}$ is the reduced mass.
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<p 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>
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