A proton of mass m and accelerated by a potential difference Vgets into a uniform electric field of a parallel plate capacitor parallel to plates of length 1 at mid-point of its separation between plates. The field strength in it varies with time as E = at, where a is a positive constant. Find the angle of deviation of the proton as it comes out of the capacitor. (Assume that it does not collide with any of the plates.)

Answer

A, B, D

Explanation

`v_x=vsqrt(((2qV)/m)), t=1/v_x=lsqrt(m/(2qV))`
`a_y=(qE)/m=(qat)/m=(dv_y)/(dt)`
Integrating both sides we get
`v_y=(qat^2)/(2m)`
`or v_y=((qa)/(2m))l^2(m/(2qV))=(al^2)/(4V)`
Now, angle of deviation
`theta=tan^-1(v_y/v_x)=tan^-1((al^2)/(4V)/sqrt((2qV)/m))`
`=tan^-1((al^2)/(4V) sqrt(m/(2eV)))`