Updated greek laters @ latex compiling

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en/7.1.17.md +9 −9
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<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)
− $$\frac{mv^2}{2} = \frac{mv_q^2}{2}+\frac{kQq}{R}+\frac{Mv_Q^2}{2} \;(1)$$
+ $$\frac{mv^2}{2} = \frac{mv_q^2}{2}+\frac{kQq}{R}+\frac{Mv_Q^2}{2} \quad(1)$$
From Momentum Conservation Law
− $$mv = mv_q + Mv_Q \;(2)$$
+ $$mv = mv_q + Mv_Q \quad(2)$$
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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− $$\Delta S_q = \Delta S_Q+2R = v_q~T \;(3)$$
+ $$\Delta S_q = \Delta S_Q+2R = v_q~T \quad(3)$$
And the sphere's displacement is
− $$\Delta S_Q = v_Q~T \;(4)$$
+ $$\Delta S_Q = v_Q~T \quad(4)$$
From $(4)$ in $(3)$, and separating $T$
− $$T = \frac{2R}{v_q-v_Q} \;(5)$$
+ $$T = \frac{2R}{v_q-v_Q} \quad(5)$$
Separating $v_q$ from $(2)$
− $$v_q = v - \frac{M}{m}v_Q \;(6)$$
+ $$v_q = v - \frac{M}{m}v_Q \quad(6)$$
Putting $(6)$ into $(1)$ and developing algebraically
− $$0 = \frac{M}{2}\left(1+\frac{M}{m}\right)v_Q^2-Mv~v_Q + \frac{kQq}{R} \;(7)$$
+ $$0 = \frac{M}{2}\left(1+\frac{M}{m}\right)v_Q^2-Mv~v_Q + \frac{kQq}{R} \quad(7)$$
$(7)$ 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
$$X = \frac{-B\pm\sqrt{D}}{2A}$$
− $$v_Q = \frac{Mv\pm\sqrt{M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}}}{M\left(1+\frac{M}{m}\right)} \;(8)$$
+ $$v_Q = \frac{Mv\pm\sqrt{M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}}}{M\left(1+\frac{M}{m}\right)} \quad(8)$$
Substituting $(6)$ into $(5)$
− $$T = \frac{2R}{v-v_Q\left(1+\frac{M}{m}\right)} \;(9)$$
+ $$T = \frac{2R}{v-v_Q\left(1+\frac{M}{m}\right)} \quad(9)$$
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 $k = 1/(4\pi\epsilon_0)$ and developing
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