Question

A particle of mass m and charge (–q) enters the region between thetwo charged plates initially moving along x-axis with speed vx (likeparticle 1 in Fig. 1.33). The length of plate is L and an uniformelectric field E is maintained between the plates. Show that thevertical deflection of the particle at the far edge of the plate isqEL2/(2m vx2).Compare this motion with motion of a projectile in gravitational fielddiscussed in Section 4.10 of Class XI Textbook of Physics.

Answer 1

The mass of the particle is m and charge is −q.

Speed of the particle is v x .

Length of the plate is L and magnitude of the electric field between the plates is E.

Formula for the mechanical force on the particle is,

F=Mass( m )×Acceleration( a ) a= F m a= qE m [ ∵F=qE ] (1)

Time taken by the particle to cross the field of length is calculated as,

Time= Distance Velocity = L v x (2)

The equation for the deflection of the particle is,

s=ut+ 1 2 a t 2

Substitute the values from equation (1) and (2) in the above equation.

s=( 0 )t+ 1 2 ( qE m ) ( L v x ) 2                    [ Initial velocity is 0 ] = qE L 2 2m v x 2

Hence, the vertical deflection of the particle is qE L 2 2m v x 2 .

This equation is similar to the motion of horizontal projectiles under gravity.