$7.1.4.$ A particle of mass $m$ with a charge $q > 0$ enters a flat capacitor, the plates of which are metal grids. The field strength in the capacitor is $E$, the distance between the grids is $d$. The initial velocity $v$ of the particle makes an angle $\alpha$ with the plane of the first grid. With what velocity and at what angle to the plane of the second grid will the particle fly out of the capacitor?
Solution
The problem can be solved either using the kinematic approach or the energetic approach. The energetic method of solution is chosen here (it is shorter).
The work done by the electric field goes into changing the kinetic energy of the particle: $$\frac{1}{2}mu^2 - \frac{1}{2}mv^2 = qEd$$
From here, we immediately find the magnitude of the velocity at which the particle exits the capacitor: $$u = v\sqrt{1 + \frac{2qEd}{mv^2}}$$
Let's decompose the initial and final velocities of the particle along the axes: one axis is directed parallel to the plates (perpendicular to the field strength vector), the other axis is parallel to the field. Initial velocity components: $v_{\perp} = v \cos\alpha$ (this component does not change, since no forces act along this axis), $v_{\parallel} = v \sin\alpha$.
Final velocity components: $u_{\perp} = v_{\perp} = v \cos\alpha$, and we find the component $u_{\parallel}$ from the Pythagorean theorem: $$u_{\parallel} = \sqrt{u^2 - u_{\perp}^2} = \sqrt{\left(v^2 + \frac{2qEd}{m}\right) - v^2 \cos^2\alpha} = \sqrt{v^2 \sin^2\alpha + \frac{2qEd}{m}}$$ Or rewritten as: $$u_{\parallel} = v \sin\alpha \sqrt{1 + \frac{2qEd}{mv^2 \sin^2\alpha}}$$
The tangent of the exit angle of the particle from the capacitor ($\beta$) is the ratio of the parallel component to the perpendicular component relative to the plates: $$\tan\beta = \frac{u_{\parallel}}{u_{\perp}} = \frac{v \sin\alpha \sqrt{1 + \frac{2qEd}{mv^2 \sin^2\alpha}}}{v \cos\alpha} = \tan\alpha \sqrt{1 + \frac{2qEd}{mv^2 \sin^2\alpha}}$$