A uniform electric field E is created between two parallel charged plates as shown in figure . An electron enters the field symmetrically between the plates with a speed `v_0.` The length of each plate is l. Find the angle of deviation of the path of the electron as it comes out of the field.

Options

A. `theta = tan^(-1) (el/(mv_0^2)).`
B. `theta = tan^(-1) (eEl/v_0^2).`
C. `theta = tan^(-1) (eEl/(mv_0^2)).`
D. `theta = tan^(-1) (eEl/v_0^2).`

Answer

C

Explanation

The acceleration of the electronis a =` (eE)/(m)`in the
upward direction. The horizontal velocity remains v_0 as
there is no acceleration in this direction. Thus, the time
taken in crossing the field is
`t =(l)/(v_0). (i)`
The upward component of the velocity of the electron as
it emerges form the field region is
`v_y = alpha t = (eEI)/(mv_0)`
The horizontal compnent of the velocity remains
`v_x = v_0.`
The angle `theta` made by the resultant velocity with the
original direction is given by
`tan theta =(v_y)/(v_x) = (eEl)/(mv_0).`
Thus, the electron deviates by an angle
`theta = tan^(-1) (eEl)/mv_0^2.`