A particle of mass m and charge (-q) enters the region between the two charged plates initially moving along x-axis with speed `v_x` (like particle 1 in figure. The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is `qEL^(2)//(2m v_(x)^(2))`.
Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10 of Class XI Textbook of Physics.

Explanation

Consider vertically upward electric field `vecE` between two horizontal plates as shown in the figure.
A particle of mass m and charge - q, enters this electric field along X-axis with constant horizontal velocity vx at time t = 0. Because of negative charge, it gets deflected towards positive plate. Suppose it passes through points `P_(1)` and `P_2` at time `t_(1)` and `t_2` respectively. At time `t_2`, horizontal displacement of given charge is equal to L. Here, no force is exerted on this charge along horizontal direction and so its horizontal velocity `v_(x)` reminas constant.
`therefore v_(x) = L/t_(2) rArr t_(2) = L/v_(x)`..........(1)
Here electric force exerted on given charge is,
`F_(e) = (-q)E = ma_(y) rArr a_(y) = (-qE)/(m)`.......(2)
At time `t_2` if vertical displacement of given charge is,
`y = v_(0y)t_(2) + 1/2a_(y)t_(2)^(2)`
`therefore y =-1/2((qE)/(m))L^(2)/v_(x)^(2) rArr |y| = 1/2 ((qE)/(m))L^(2)/v_(x)^(2)`......(3)
Above equation gives required result. Here, displacement of given charge is towads - Y-axis and so y is negative. Comparison with projectile motion : Here, at the end of time `t_(1)` horizontal displacement of given charge is `x_1` and so,
`v_(x) = x_(1)/t_(1) rArr t_(1) = x_(1)/v_(1)`...........(4)
At the end of time `t_(1)` vertical displacement of given charge is,
`y_(1) = (v_(0))t_(1) + 1/2 a_(y)t_(1)^(2)`
`therefore y_(1) = -(qE)/(2mv_(x)^(2))x_(1)^(2) y_(1) = x_(1)^(2)` ...............(5)
Above equation is similar to equation of projectile motion.