Updated greek laters @ latex compiling

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en/7.1.6.md +9 −9
@@ -66,25 +66,25 @@
<h3>Solution</h3>
<p>
Initial kinetic energy is defined as
− $$K_0 = \frac{mv_0^2}{2} \;(1)$$
+ $$K_0 = \frac{mv_0^2}{2} \quad(1)$$
If we apply Newton's Second Law, acceleration can be defined,
− $$a_y = \frac{qE}{m} \;(2)$$
+ $$a_y = \frac{qE}{m} \quad(2)$$
As there are no forces in horizontal direction, the component of velocity in that direction is unaffected
− $$v_0\cos{\alpha}=v_1\cos{\beta} \;(3)$$
+ $$v_0\cos{\alpha}=v_1\cos{\beta} \quad(3)$$
For the horizontal path until exit's point
− $$l = v_0\cos{\alpha}~t_v \;(4)$$
+ $$l = v_0\cos{\alpha}~t_v \quad(4)$$
The time expended by particles during the path through the capacitor can be found from acceleration definition
$$\vec{a_y} = \frac{\Delta \vec{v_y}}{\Delta t} = \frac{\vec{v_{1y}}-\vec{v_{0y}}}{t_v}$$
$$a_y = \frac{v_1\sin{\beta}+v_0\sin{\alpha}}{t_v}$$
− $$t_v = \frac{v_1\sin{\beta}+v_0\sin{\alpha}}{a_y} \;(5)$$
+ $$t_v = \frac{v_1\sin{\beta}+v_0\sin{\alpha}}{a_y} \quad(5)$$
Substituting $v_1$ from $(3)$ into $(5)$ and developing algebraically
− $$t_v = v_0\cos{\alpha}\frac{\tan{\alpha}+\tan{\beta}}{a_y} \;(6)$$
+ $$t_v = v_0\cos{\alpha}\frac{\tan{\alpha}+\tan{\beta}}{a_y} \quad(6)$$
Putting $(6)$ into $(4)$ and separating $v_0$
− $$v_0^2 = \frac{a_yl}{\cos^2{\alpha}(\tan{\alpha}+\tan{\beta})}$$
+ $$v_0^2 = \frac{a_yl}{\cos^2 {\alpha}(\tan{\alpha}+\tan{\beta})}$$
and according to $(2)$,
− $$v_0^2 = \frac{qEl}{m\cos^2{\alpha}(\tan{\alpha}+\tan{\beta})} \;(7)$$
+ $$v_0^2 = \frac{qEl}{m\cos^2 {\alpha}(\tan{\alpha}+\tan{\beta})} \quad(7)$$
Finally, substituting $(7)$ into $(1)$
− $$\boxed{K_0 = \frac{qEl}{2\cos^2{\alpha}(\tan{\alpha}+\tan{\beta})}}$$
+ $$\boxed{K_0 = \frac{qEl}{2\cos^2 {\alpha}(\tan{\alpha}+\tan{\beta})}}$$
</p>
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