Particles with mass m and charge q enter in a capacitor of length l under an angle lpha respect to plate's plane and exit forming an angle eta. Find the initial kinetic energy of particles, if the field intensity inside the capacitor is E.
<h3 id="back-link"><a href="/#7.1">$\leftarrow$Back</a></h3>
<h3> Statement </h3>
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$7.1.6$
Particles with mass $m$ and charge $q$ enter in a capacitor of length $l$ under an angle $\alpha$ respect to plate's plane and exit forming an angle $\beta$. Find the initial kinetic energy of particles, if the field intensity inside the capacitor is $E$.
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For problem 7.1.6
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<h3>Solution</h3>
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Initial kinetic energy is defined as
$$K_0 = \frac{mv_0^2}{2} \\;(1)$$
If we apply Newton's Second Law, acceleration can be defined,
$$a_y = \frac{qE}{m} \\;(2)$$
As there are no forces in horizontal direction, the component of velocity in that direction is unaffected
$$v_0\cos{\alpha}=v_1\cos{\beta} \\;(3)$$
For the horizontal path until exit's point
$$l = v_0\cos{\alpha}\~t_v \\;(4)$$
The time expended by particles during the path through the capacitor can be found from acceleration definition
$$\vec{a_y} = \frac{\Delta \vec{v_y}}{\Delta t} = \frac{\vec{v_{1y}}-\vec{v_{0y}}}{t_v}$$
$$a_y = \frac{v_1\sin{\beta}+v_0\sin{\alpha}}{t_v}$$
$$t_v = \frac{v_1\sin{\beta}+v_0\sin{\alpha}}{a_y} \\;(5)$$
Substituting $v_1$ from $(3)$ into $(5)$ and developing algebraically
$$t_v = v_0\cos{\alpha}\frac{\tan{\alpha}+\tan{\beta}}{a_y} \\;(6)$$
Putting $(6)$ into $(4)$ and separating $v_0$
$$v_0^2 = \frac{a_yl}{\cos^2{\alpha}(\tan{\alpha}+\tan{\beta})}$$
and according to $(2)$,
$$v_0^2 = \frac{qEl}{m\cos^2{\alpha}(\tan{\alpha}+\tan{\beta})} \\;(7)$$
Finally, substituting $(7)$ into $(1)$
$$\boxed{K_0 = \frac{qEl}{2\cos^2{\alpha}(\tan{\alpha}+\tan{\beta})}}$$
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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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