A proton accelarted by a potential differnce `V` gets into the unifrom electric field of a paralallel-plate capacitor whose plates extended over a length `l` in the motion direction. The field strenth varies with time as `E = at`, where `a` is a constant. Assuming the proton to be non-relatistic, find the angle between the motion directions of the proton before and after its fight throgh the capacitor, the proton gets in the field at the moment `t = 0`. The edge effects are to be neglected.

Explanation

The electric filed inside the capacitor vaires with time as,
`E = at`
Hence, electric force on the proton,
`F = ea t`
and subsequentely, acceleration of the proton,
`w = (ea t)/(m)`
Now, if is the time elapsed during the motion of the proton between the plates, then `t = (l)/(v_(|\|)`, as no acceleration is effective in this direction, (Here `v_(|\|)` is velocity along the though of the plate.)
From kinematics, `(dv_(_|_))/(dt) = w`
so, `int_(0)^(v_(_|_)) dv_(_|_) = int_(0)^(t) w dt`,
(as initally, the component of velocity in the direction, `_|_` to plates, was zero.)
or, `,v_(_|_) = int_(0)^(t) (ea)/(m) (t^(2))/(2m) = (ea)/(2m) (l^(2))/(v_(|\|)^(2))`
Now, `tan alpha = (v_(_|_))/(v_(|\|)) = (e omega l^(2))/(2m v_(|\|)^(3))`
`= (ea l^(2))/(2m ((2e V)/(m))^(3/2))` as `v_(|\|) = ((2 eV)/(m))^(1/2)`, from energy conservation.
`= (al^(2))/(4) sqrt((m)/(2 eV^(3)))`