Show that the path of a …

Force F acting on a charged particle of charge q in electric field E is given by,   F   = (q/m)E

( bold letters F and E are vectors )

If a charged particle moves in the direction of electric field, Then it is accelerated and  will move in same direction of electric field.

But if a charged particle moves in a direction and not in parallel to electric field, it moves in a parabolic path.            

Let v be the velocity and E be the electric field as shown in figure.

Let at t=0,  y-component of velocity be v_y0 

then vertical displacement y after a time duration t is given by,   y = v_y0 t + (1/2) a t²  .................(1)

acceleration a = qE/m    and  t = x/v_x  , where x is  horizontal displacement , v_x component is unaffected by electric field

Hence eqn.(1) is written as,   y = v_y0 (x/v_x)  + (1/2) a (x/v_x)² .......................(2)

vertical position y as given by eqn.(1) is in parabolic curve  of type  y = a x + b x²

Force F experienced by the charge particle of charge q in a magnetic field B is given by,

F = q v × B    ...................(3)

where v is velocity of charged particle and B is magnetic field. Eqn.(3) is cross product of vectors.

Hence magnitude of force |F| = |v| |B| sinθ  , where θ is the angle between magnetic field direction and velocity of charged particle.

If charged particle velocity is parallel to magnetic field, force acting on charged particle will be zero.

Otherwise, force acting on charged particle is perpendicular to veleocity of particle .

Any moving particle subjected to a force in perpendicular direction to its moving direction will move in a circular path