A charged Particle, proton, moves between the p.d of 'V'and enters into the perpendicular uniform electric field of strength E, extended upto distance X. what is the expression for deflection of the charged particle when it is just out of the electric field region.

Asked by  | 3rd Aug, 2012, 08:04: AM

Expert Answer:

Let the particle enters with speed= v
AB will be the trajectory, it will follow. at point A, it will have velocity v in x direction, acceleration in x direction=ax=0  and acceleration in y direction=ay=eE/m=eV/mx, as the particle will get accelerated towards the field.
so the distance covered in x direction:
x=uxt+1/2 ax t2 =vt
so  t=x/v
distance covered in y direction: b=uyt+1/2 ay t2 =(1/2)( eV/x) t2 =eVx/2mv2 , putting value of t.

Answered by  | 19th Nov, 2012, 03:19: PM

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