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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.) |
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Answer» `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)))` |
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